Chapter 7: A Tale of Three Intersecting Lines

Step 1 — Recall the definition of a triangle
A triangle is a closed figure made by joining three non-collinear points with three straight-line segments. "Non-collinear" means the three points must not lie on the same straight line.

Step 2 — Express the condition algebraically
Let the three intended vertices be $$A$$, $$B$$ and $$C$$. If all of them lie on one straight line l, then

• The coordinates (or positions) of $$A,B,C$$ satisfy the same linear equation of line l.
• This makes $$A,\,B,\,C$$ collinear.

Step 3 — Test against the properties of a triangle
Because the three vertices are collinear, the three "sides" $$AB$$, $$BC$$ and $$CA$$ all fall on the same straight line. Consequently,

• No closed region is enclosed, so the area becomes $$0\;\text{unit}^2$$.
• The interior angle at the middle point (say $$B$$) becomes a straight angle $$=180^\circ$$, while the remaining two interior angles shrink to $$0^\circ$$ each. This violates the condition that in a genuine triangle every interior angle must be greater than $$0^\circ$$ and less than $$180^\circ$$.

Step 4 — Conclusion
When the supposed three vertices lie on a single straight line, the would-be triangle "collapses" into that line. In geometric language, the triangle becomes degenerate; effectively, no triangle is formed.

Objective Construct a triangle each of whose three sides measures 4 cm and justify that the construction really produces such a triangle.

What you need

• a sharp pencil,
• a ruler graduated in centimetres and millimetres,
• a compass.

Notation to be fixed before starting

We shall call the required triangle $$\triangle ABC$$. The decision which vertex receives which letter is immaterial, but once fixed we keep to it consistently.

Construction steps (write them in your record and perform them simultaneously)

1. Draw the first side.
     With the ruler draw a straight line segment and mark its end-points as $$A$$ and $$B$$ such that $$AB = 4\text{ cm}$$. (Measure carefully to the nearest millimetre.)
2. Locate the third vertex by intersecting two congruent circles.
     (a) Keeping the compass opening fixed at exactly $$4\text{ cm}$$, place the compass needle on $$A$$ and draw a (full or partial) arc on the side of $$AB$$ where the triangle is to be formed.
     (b) Without altering the compass width, place the needle on $$B$$ and draw another arc so that it cuts the first arc. Let the point of intersection be named $$C$$.
3. Complete the triangle.
     Join $$C$$ to $$A$$ and $$C$$ to $$B$$ with the ruler. The required triangle $$\triangle ABC$$ is now visible.

Justification (Why the construction works)

• By step 1 we fixed $$AB = 4\text{ cm}$$.
• During step 2 the compass radius was also set at $$4\text{ cm}$$; therefore every point on the first arc satisfies $$AC = 4\text{ cm}$$ and every point on the second arc satisfies $$BC = 4\text{ cm}$$.
• The point $$C$$ lies simultaneously on both arcs; hence \[ AC = 4\text{ cm}\quad\text{and}\quad BC = 4\text{ cm}. \]

Collecting the three equalities

$$AB = AC = BC = 4\text{ cm},$$

we see that all three sides of $$\triangle ABC$$ are equal. By the definition of an equilateral triangle, the triangle we have constructed fulfils the requirement.

Verification (optional but good practice)

Measure $$AC$$ and $$BC$$ with the ruler; each should read 4 cm. A discrepancy indicates that the compass opening shifted or the arcs were drawn inaccurately, in which case repeat the construction with greater care.

The task is complete.

Given background
In the earlier activity we located the point where the three perpendicular bisectors of the sides of a triangle meet (the circum-centre). We carried out the following steps:

1. Construct the perpendicular bisector of $$\overline{AB}$$.
2. Construct the perpendicular bisector of $$\overline{BC}$$.
3. Construct the perpendicular bisector of $$\overline{CA}$$.
4. Mark the common intersection point of the three lines.

The question now is

“How do we make this construction more efficient?”

Step 1 – Observe what is really needed
Any two non-parallel straight lines intersect in exactly one point. Therefore, the moment two of the three perpendicular bisectors are drawn, their intersection already fixes the unique point that all three will share.
So, the third bisector is logically redundant; it only confirms what we already know.

Step 2 – Rewrite the construction with the redundant step removed

1. Draw the perpendicular bisector $$l_1$$ of one convenient side, say $$\overline{AB}$$.
   • With radius greater than half of $$AB$$, draw arcs above and below $$\overline{AB}$$ taking centres $$A$$ and $$B$$. Join the arc-intersection points to get $$l_1$$.
2. Draw the perpendicular bisector $$l_2$$ of another side, say $$\overline{AC}$$, in exactly the same way.
3. The lines $$l_1$$ and $$l_2$$ meet at a single point; call it $$O$$.
   \[ l_1 \cap l_2 = \{O\} \]
4. Stop here. Point $$O$$ is the circum-centre; no third bisector is required. (If you still wish to check, construct the third bisector; it will automatically pass through $$O$$.)

Why this works

• Each perpendicular bisector is the set of all points equidistant from the two endpoints of the corresponding side.
• A point that lies on two such bisectors (here $$O$$) is therefore equidistant from all three vertices — exactly the property of the circum-centre.
• Since only two lines are needed to determine their point of intersection, the third line cannot possibly produce a different point; it must pass through $$O$$ as well.

Conclusion
The construction becomes more efficient simply by drawing any two of the three required concurrent lines instead of all three. This saves both time and ink while giving the same correct result.

Context : In the preceding construction we were asked to draw a triangle $$\triangle ABC$$ such that $$AB = 5\,\text{cm}$$ and $$AC = BC = 4\,\text{cm}$$ (i.e. an isosceles triangle with the equal sides 4 cm long).

Step–by–step recap of the construction

1. Draw the base $$AB$$ with a ruler so that
   $$AB = 5\,\text{cm}$$.
2. With A as centre and radius $$4\,\text{cm}$$, draw an arc above the base.
3. With B as centre and the same radius $$4\,\text{cm}$$, draw a second arc to meet the first one. Mark the point of intersection as C.
4. Join $$AC$$ and $$BC$$ to complete $$\triangle ABC$$.

Why does this guarantee $$AC = 4\,\text{cm}$$ and $$BC = 4\,\text{cm}?}

Remember the defining property of a circle: every point on a circle is at a constant distance (the radius) from the centre.

• Because the arc in Step 2 was drawn with centre A and radius $$4\,\text{cm}$$, every point on that arc (including the intersection point C) must satisfy
  $$AC = 4\,\text{cm}$$.
• Similarly, the arc in Step 3 was drawn with centre B and the same radius $$4\,\text{cm}$$, so for the same point C one also has
  $$BC = 4\,\text{cm}$$.

Thus, by construction, both $$AC$$ and $$BC$$ are radii of their respective circles, each equal to $$4\,\text{cm}$$. No further measurement is needed; the equality follows directly from the definition of a circle.

Step 1 — Triangle inequality check

For sides $$4\text{ cm},\,5\text{ cm},\,6\text{ cm}$$:

• $$4+5=9>6$$ ✔
• $$5+6=11>4$$ ✔
• $$4+6=10>5$$ ✔

All three hold, so the triangle can be constructed.

Step 2 — Construction

1. With the ruler draw the base $$\overline{AB}$$ of length $$6\text{ cm}$$.
2. With centre $$A$$ and radius $$4\text{ cm}$$, draw an arc above $$\overline{AB}$$.
3. With centre $$B$$ and radius $$5\text{ cm}$$, draw a second arc that meets the first. Label the point of intersection as $$C$$.
4. Join $$\overline{CA}$$ and $$\overline{CB}$$ with the ruler.

Step 3 — Recording the side lengths

The arc drawn from $$A$$ had radius $$4\text{ cm}$$, so $$CA=4\text{ cm}$$. The arc drawn from $$B$$ had radius $$5\text{ cm}$$, so $$BC=5\text{ cm}$$. Therefore

Reason why the two arcs meet

The distance between centres $$A$$ and $$B$$ is $$6\text{ cm}$$, which lies strictly between $$|5-4|=1\text{ cm}$$ and $$5+4=9\text{ cm}$$. The two arcs therefore intersect; choosing the intersection above $$\overline{AB}$$ gives $$C$$.

Verification

• Measure $$CA$$ → $$4\text{ cm}$$.
• Measure $$BC$$ → $$5\text{ cm}$$.
• $$AB$$ (already drawn) → $$6\text{ cm}$$.

The required triangle has been correctly constructed.

Given : the lengths of the three medians of the required triangle are $$m_a ,\; m_b ,\; m_c$$.
We want to obtain a triangle $$\triangle ABC$$ whose medians have exactly these three lengths, but with the least amount of repeated work. We shall make use of two facts you have already met in Chapter 7.

• (F 1) The three medians of a triangle are concurrent at the centroid $$G$$ and are divided by the centroid in the fixed ratio $$AG:GD = BG:GE = CG:GF = 2:1$$.
• (F 2) The three medians themselves can be taken as the sides of some triangle; this is called the median triangle.

Using (F 1) and (F 2) we can compress the whole construction into one clean chain of steps.

1. Construct the median triangle
   Draw any convenient base line and construct a triangle $$\triangle DEF$$ such that $$DE = m_a,\; EF = m_b,\; FD = m_c.$$ (Ordinary ruler-and-compass construction of a triangle when all three sides are known was practised in Class 6; recall it: draw the base, take radius equal to the second side, etc.)
2. Locate its centroid
   Find the mid-point $$P$$ of $$EF$$ and the mid-point $$Q$$ of $$FD$$ (use perpendicular bisectors or the compasses-same-radius trick). Draw the two medians $$DP$$ and $$EQ$$. Let them meet at $$G$$. At this point we have the picture \[ D,\;E,\;F \text{ – vertices of the median triangle, and } G \text{ – its centroid}. \]
3. Exploit the 2 : 1 ratio
   Because $$DG$$ is a median of $$\triangle DEF$$, the segment $$DG$$ will correspond to one of the given medians ($$m_a$$, $$m_b$$ or $$m_c$$). More importantly, fact (F 1) tells us that if we double the segment $$GD$$ in the same straight line we must arrive at the vertex of the required triangle whose own median was $$DG$$.
   Thus mark a point $$A$$ on the line $$DG$$ produced such that $$GA = 2\,GD$$ (use compasses: put the point at $$G$$, open to $$D$$, and strike an arc cutting the extension of $$DG$$; then repeat once more to double the length). Repeat the same idea on the other two medians of $$\triangle DEF$$:
   • Produce $$EG$$ beyond $$G$$, take $$GB = 2\,GE$$,
   • produce $$FG$$ beyond $$G$$, take $$GC = 2\,GF$$.
   Now the three new points $$A, B, C$$ are in place, and each of $$AG, BG, CG$$ equals $$\tfrac23 m_a,\; \tfrac23 m_b,\; \tfrac23 m_c$$ respectively, so $$AD = m_a,\; BE = m_b,\; CF = m_c.$$
4. Join the vertices
   Draw $$AB, BC, CA$$. The triangle $$\triangle ABC$$ is the only triangle whose medians are exactly $$m_a, m_b, m_c$$, and we have reached it without having to redraw or guess anything.

Why this is more efficient
A beginner often tries to start with one median, guess the position of a vertex, measure another median, correct the guess, and so on – a long trial-and-error loop. The above method reduces everything to one ordinary SSS construction (Step 1), followed by a single use of the permanent ratio 2 : 1 (Step 3). No iteration, no erasing, no wasted arcs: the triangle appears at once.

Diagram to be drawn (for the examination paper)

• The triangle $$DEF$$ with its medians $$DP, EQ, FR$$ shown.
• The centroid $$G$$ marked.
• The lines $$DG, EG, FG$$ produced to $$A, B, C$$ such that $$GA = GB = GC = 2 \times (\text{corresponding } G{\text{–}}D,E,F).$$
• The finally obtained triangle $$ABC$$.

The construction is complete and fully justified.

Given data

• Required side lengths:  $$3\,\text{cm},\;4\,\text{cm},\;8\,\text{cm}$$

Construction attempt (step by step)

1. Draw a straight line and mark the segment $$\overline{AB}$$ such that $$AB = 8\,\text{cm}$$. This will be the longest side (taken as the base).
2. Place the compass at point A and set its opening to $$3\,\text{cm}$$. With this radius draw an arc above (or below) the base line.
3. Without changing the compass, move its centre to point B and set the opening to $$4\,\text{cm}$$. Draw another arc that should intersect the first one.
4. Look for an intersection point of the two arcs. Label such a point $$C$$ if it exists, then join $$C$$ to $$A$$ and $$C$$ to $$B$$ and the triangle $$\triangle ABC$$ would be complete.

What actually happens ?

While performing Step 3 you will notice that the two arcs do not meet. They remain apart because the centres A and B are $$8\,\text{cm}$$ apart, yet the farthest distance you can reach by moving from A to some point on the first arc and then to B is only

Since $$7\,\text{cm}$$ is smaller than $$8\,\text{cm}$$, the two circles (or their arcs) never intersect. Consequently, no point $$C$$ satisfies both $$AC = 3\,\text{cm}$$ and $$BC = 4\,\text{cm}$$ at the same time.

Reason in words – Triangle Inequality

A fundamental property of triangles states that the sum of the lengths of any two sides must be greater than the third side. Here

so the condition fails. Therefore a triangle with side lengths $$3\,\text{cm},\;4\,\text{cm},\;8\,\text{cm}$$ cannot exist, and hence cannot be constructed.

Let the three given lengths be

• $$a = 2\,\mathrm{cm}$$,
• $$b = 3\,\mathrm{cm}$$,
• $$c = 6\,\mathrm{cm}$$ (the largest).

For any three positive numbers to be the sides of a triangle they must obey the triangle inequality:

• $$a + b \gt c$$,
• $$a + c \gt b$$,
• $$b + c \gt a$$.

Check each inequality.

1. $$a + b = 2 + 3 = 5$$, and $$5 \lt 6 = c$$, so $$a + b \gt c$$ is false.
2. $$a + c = 2 + 6 = 8 \gt 3 = b$$ ✔
3. $$b + c = 3 + 6 = 9 \gt 2 = a$$ ✔

Because the first inequality fails, the three lengths cannot satisfy all the triangle inequalities at once.

Therefore no triangle can be formed with side-lengths $$2\,\mathrm{cm},\,3\,\mathrm{cm}$$ and $$6\,\mathrm{cm}$$.

Step 1 — Recall the rule

For three positive lengths $$a,\;b,\;c$$ (write them in non-decreasing order $$a\le b\le c$$) a triangle exists if and only if

If $$c \ge a + b$$, the construction fails.

Step 2 — Sample impossible triples

• $$(2,\,3,\,5)$$: largest $$= 5$$; $$2 + 3 = 5$$, which is not strictly greater than $$5$$ → impossible.
• $$(1,\,4,\,7)$$: largest $$= 7$$; $$1 + 4 = 5 \lt 7$$ → impossible.
• $$(2,\,5,\,8)$$: largest $$= 8$$; $$2 + 5 = 7 \lt 8$$ → impossible.
• $$(4,\,10,\,15)$$: largest $$= 15$$; $$4 + 10 = 14 \lt 15$$ → impossible.
• $$(2,\,3,\,6)$$: largest $$= 6$$; $$2 + 3 = 5 \lt 6$$ → impossible.

Step 3 — The pattern

In every impossible triple, the largest length is greater than or equal to the sum of the other two. With the lengths written in non-decreasing order $$a \le b \le c$$,

So the test for triangle existence is the single inequality

Step 4 — Building more examples

Pick any two positive lengths, add them, and choose any third length that equals or exceeds the sum. Some triples that work this way:

• $$(3,\,4,\,7)$$ — sum equal to the largest ($$3 + 4 = 7$$).
• $$(3,\,4,\,10)$$ — sum less than the largest ($$3 + 4 = 7 \lt 10$$).
• $$(1,\,9,\,12)$$ — sum less than the largest ($$1 + 9 = 10 \lt 12$$).
• $$(0.5,\,2,\,2.5)$$ — decimal lengths, sum equal to the largest ($$0.5 + 2 = 2.5$$).

Infinitely many such triples exist, all sharing the same pattern: the largest length is at least the sum of the other two.

Key fact already learnt – In any triangle, the sum of the lengths of any two sides is always greater than the length of the third side. (This is called the triangle–inequality property.)

Let us test the three given lengths one pair at a time.

1. First and second sides against the third:
   $$10\;\text{cm}+15\;\text{cm}=25\;\text{cm}$$
   Compare with the third side: $$25\;\text{cm}<30\;\text{cm}$$.
2. First and third sides against the second:
   $$10\;\text{cm}+30\;\text{cm}=40\;\text{cm}$$
   Since $$40\;\text{cm}>15\;\text{cm}$$, this pair passes the test.
3. Second and third sides against the first:
   $$15\;\text{cm}+30\;\text{cm}=45\;\text{cm}$$
   Since $$45\;\text{cm}>10\;\text{cm}$$, this pair also passes the test.

The inequality fails in the very first check because $$10\,\text{cm}+15\,\text{cm}$$ is smaller than $$30\,\text{cm}$$. Therefore the three lengths do not satisfy the triangle–inequality property.

Because at least one of the required inequalities is false, no triangle can be drawn whose three sides are exactly $$10\,\text{cm}$$, $$15\,\text{cm}$$ and $$30\,\text{cm}$$.

(One can also picture this practically: if you try to join sticks of lengths $$10\,\text{cm}$$ and $$15\,\text{cm}$$ end-to-end, even when stretched in a straight line they reach only $$25\,\text{cm}$$, which is still $$5\,\text{cm}$$ short of the required $$30\,\text{cm}$$ to meet the third corner.)

Conclusion: A triangle with side lengths $$10\,\text{cm}$$, $$15\,\text{cm}$$ and $$30\,\text{cm}$$ cannot exist.

Step 1 — Triangle inequality

For a triangle with sides $$a$$, $$b$$, $$c$$ each pair must satisfy $$a+b \gt c$$, $$a+c \gt b$$, $$b+c \gt a$$.

Step 2 — Substitute

Let the three sides be $$3\,\text{cm}$$, $$3\,\text{cm}$$ and $$7\,\text{cm}$$.

The first row fails because $$6 \lt 7$$, so the triangle inequality is violated.

Conclusion from calculation

A triangle with sides $$3\,\text{cm},\;3\,\text{cm},\;7\,\text{cm}$$ cannot exist.

Step 3 — Verification by construction

1. Draw segment $$\overline{AB}$$ of length $$7\,\text{cm}$$.
2. With centre $$A$$ and radius $$3\,\text{cm}$$, draw an arc above the line.
3. With centre $$B$$ and radius $$3\,\text{cm}$$, draw a second arc above the line.
4. The two arcs never meet, because the maximum distance covered by the two radii together is $$3+3=6\,\text{cm}$$, while $$AB=7\,\text{cm}$$.

No common point exists for the third vertex, confirming the algebra.

Final statement

A triangle with side-lengths $$3\,\text{cm}$$, $$3\,\text{cm}$$, and $$7\,\text{cm}$$ does not exist.

Step 1  Recall what the problem is really asking
Three towns A, B and C are to be joined by three direct roads AB, BC and CA.
For a given assignment of road–lengths we must have, for every pair of towns,

direct road < round-about road through the third town.

But “direct road < round-about road” is exactly the statement of the triangle inequality:

If we can arrange the given numbers so that these three inequalities are all true, the three roads can indeed form the three sides of a triangle.

Step 2  Place the longest distance first
Let the three different distances available be, for example,

$$7\;\text{km},\;6\;\text{km},\;4\;\text{km}.$$

Choose the largest, $$7\;\text{km},$$ for one of the direct roads; let us take AB = 7 km. Put the other two numbers arbitrarily on the remaining roads; say

BC = 6 km, CA = 4 km.

Step 3  Check the three triangle inequalities

• AB ( 7 km ) ⟨ BC + CA = 6 + 4 = 10 km ✔
• BC ( 6 km ) ⟨ CA + AB = 4 + 7 = 11 km ✔
• CA ( 4 km ) ⟨ AB + BC = 7 + 6 = 13 km ✔

All three conditions are satisfied, so with this new order each direct path is indeed shorter than the corresponding round-about path.

Step 4  Conclusion
Because we have found one such ordering of the three given lengths for which all three triangle inequalities are true, it is possible to assign the lengths so that every direct route is shorter than the indirect one. Therefore a triangle with those side-lengths can exist.

Given/Recall

• Every side of a triangle lies opposite exactly one angle.
• In any triangle, the order of the side–lengths is exactly the same as the order of the magnitudes of their opposite angles:
    If $$a>b$$ then the angle opposite $$a$$ is bigger than the angle opposite $$b$$, and so on.

What “re-arrangement of lengths” means

We are asked whether, keeping the same three numerical lengths, we can place them along the three edges in some different way – for instance, making a shorter length face the largest angle or putting the longest length between two smaller angles.

Reasoning

1. Let the three sides be $$a,\;b,\;c$$ with $$a>b>c$$.
   Let the three angles opposite them be $$A,\;B,\;C$$ respectively.
2. Because $$a>b>c$$, the corresponding angles satisfy $$A>B>C$$ (larger side ⇔ larger opposite angle).
3. Suppose we try to put the longest side somewhere else – say, opposite the middle or the smallest angle.
   That would force the largest angle to face a shorter side, contradicting the property in step 2.
4. Any similar attempt to shuffle the three lengths among the three positions necessarily violates at least one of the pair-wise comparisons $$a>b,$$ $$b>c,$$ $$a>c$$ matched with $$A>B,$$ $$B>C,$$ $$A>C$$.

Conclusion

Once the sizes of the three angles are fixed, the side whose length is biggest is compelled to lie opposite the biggest angle, the second–longest opposite the second–biggest angle, and the shortest opposite the smallest angle. Therefore no new triangle can be obtained merely by “re-arranging” the three given lengths around the same three vertices.

Hence, such a rearrangement of lengths is not possible in a triangle.

Recall the three comparisons for any three positive lengths $$a,\;b,\;c$$:

• $$a \;?\; b+c$$
• $$b \;?\; a+c$$
• $$c \;?\; a+b$$

We want to know whether, whatever values we give to $$a,\;b,\;c$$, at least two of the above will surely be
“direct < sum of the other two”, that is

• $$a < b+c$$ or
• $$b < a+c$$ or
• $$c < a+b$$.

First we test with a few numerical sets.

In every trial we found at least two “<”. Let us now prove that this is bound to happen.

Step 1 – Label the greatest length.
Without loss of generality, let $$c$$ be the greatest of the three. Thus

$$c\ge a \;\text{and}\; c\ge b.$$

Step 2 – Check the two comparisons that do not involve $$c$$ as the direct length.

• Compare $$a$$: $$a < b+c$$ because $$c$$ is positive, so $$b+c > b \ge 0 \Rightarrow b+c$$ is certainly larger than $$a$$.
• Compare $$b$$: $$b < a+c$$ for the same reason – add the positive length $$c$$ to $$a$$, we surely get something larger than $$b$$.

Thus, both of these are always “<”, irrespective of the size of $$c$$.

So we have already obtained two required ‘less-than’ comparisons.

Step 3 – What about the third comparison?

The remaining comparison is $$c \;?\; a+b$$. Two things can happen:

• If $$c < a+b$$, then all three are “<”.
• If $$c \ge a+b$$, then the third is not “<”, but the first two are still “<”.

Either way we always have at least two “<”.

Why can’t there be only one “<”?

Suppose, for the sake of argument, that exactly one of the three is “<” and the other two are “≥”. Then two different lengths (say $$a$$ and $$b$$) would each be greater than or equal to a sum that already contains the other of them, which is impossible.

For example, if both $$a \ge b+c\quad\text{and}\quad b \ge a+c,$$ adding the two inequalities gives $$a+b \ge a+b+2c,$$ which forces $$c\le 0.$$ But $$c$$ is a length, so it must be positive – contradiction. Therefore at most one comparison can fail, and at least two must succeed.

Conclusion

Yes, no matter what positive lengths we begin with, at least two of the three direct-versus-sum comparisons will always read
“direct length < sum of the other two”. The quick reason is that the two comparisons in which the longest side appears in the sum are automatically satisfied.

Step 1 – Recall the triangle inequality.
For any three lengths to be the sides of a triangle, each length must be less than the sum of the other two. In symbols, for three lengths $$a,b,c$$ we must have

$$a < b + c, \; b < a + c, \; c < a + b.$$

Step 2 – Arrange the lengths in increasing order.
Suppose the given lengths, when arranged from the smallest to the largest, are

$$d_1 \le d_2 \le d_3.$$

Step 3 – Check which inequality can possibly fail.

• For $$d_1$$: the “other two” are $$d_2 \text{ and } d_3$$, so their sum is $$d_2 + d_3 \ge d_3 \ge d_1.$$ Thus $$d_1 < d_2 + d_3$$ is always true. No calculation is needed.
• For $$d_2$$: the “other two” are $$d_1 \text{ and } d_3$$, whose sum is $$d_1 + d_3 \ge d_3 \ge d_2.$$ Hence $$d_2 < d_1 + d_3$$ is also automatically true.
• For $$d_3$$: the “other two” are the smaller lengths $$d_1 \text{ and } d_2$$, so their sum $$d_1 + d_2$$ might be less than, equal to, or greater than $$d_3$$. Therefore the only inequality that can fail is

\[d_3 < d_1 + d_2.\]

Step 4 – Conclusion.
Without doing any arithmetic we can say:

• The largest length (here $$d_3$$) is the only one that might not be less than the sum of the other two.
• The two smaller lengths will always satisfy the required condition automatically.

Hence, simply by listing the lengths in increasing order, you can point to the largest one straightaway as the length that must be checked; no calculation is needed for the other two.

Suppose the three proposed side-lengths are denoted by the positive numbers $$a$$, $$b$$ and $$c$$.

To decide whether these three sticks can really be joined end-to-end to form a closed triangle, we recall the triangle inequality.

1. If you join the two shorter sticks and try to reach the tip of the longest stick, their combined length must be more than the longest stick; otherwise they will fall short and leave a gap.

2. The same idea applies no matter which of the three sticks happens to be the largest; therefore the comparison has to be made for each possible pair.

This translates into the following three numerical comparisons:

• $$a + b > c$$
• $$b + c > a$$
• $$c + a > b$$

If all three inequalities are satisfied, the three lengths can be arranged to form a triangle. If even one of them fails, a triangle is impossible.

Therefore, to test for the existence of a triangle we must compare the sum of any two side-lengths with the third side-length.

Given side lengths

• $$AB = 4\,\mathrm{cm}$$
• $$BC = 5\,\mathrm{cm}$$
• $$CA = 8\,\mathrm{cm}$$

Write the lengths in increasing order:

so $$8\,\mathrm{cm}$$ is the longest side.

The triangle inequality test

For three positive numbers to be the sides of a triangle, the sum of any two must exceed the third. For sides $$a \ge b \ge c$$ this means

$$b + c \gt a, \;\; a + c \gt b, \;\; a + b \gt c.$$

Step 1 — Check the inequality with the longest side

Substitute $$(a,b,c) = (8,5,4)$$:

$$b + c = 5 + 4 = 9\,\mathrm{cm}, \quad a = 8\,\mathrm{cm}.$$

Since $$9 \gt 8$$, the inequality involving the longest side holds.

Step 2 — Why the other two need not be checked

Once $$b + c \gt a$$ holds and all lengths are positive,

• Adding $$c \gt 0$$ to $$a$$: $$a + c \gt a \ge b$$, hence $$a + c \gt b$$.
• Adding $$b \gt 0$$ to $$a$$: $$a + b \gt a \ge c$$, hence $$a + b \gt c$$.

So once the longest side passes the test, the other two inequalities follow automatically.

Conclusion

Since $$8 \lt 4 + 5 = 9$$, all three triangle inequalities are satisfied. A triangle with sides $$4\,\mathrm{cm}, 5\,\mathrm{cm}, 8\,\mathrm{cm}$$ exists.

Step 1 – Recall the picture you already have
From the previous parts of the exercise you should already have three straight lines drawn so that they all pass through a common interior point. Let the three lines be named $$l_1$$, $$l_2$$ and $$l_3$$ and suppose their common point of intersection is the point $$B$$. Each line has its two arms stretching out from $$B$$ in opposite directions.

Step 2 – Fix the radius on the compass
Place the tip of the compass on the 0 cm mark of a ruler and open the compass legs until the pencil point touches the 5 cm mark. Check once more that the opening really measures $$5\,\text{cm}$$; this number is the radius we want.

Step 3 – Draw the circle with centre B
Keep the metal tip of the compass firmly on the point $$B$$, and, without changing the compass width, swing the pencil a full turn so that it meets itself. The closed curve you obtain is your circle with

Step 4 – Observe how the circle meets the three lines
Because the three lines all go through $$B$$, each line will cut the circle twice – once on each side of the centre – giving altogether six intersection points.

• Line $$l_1$$ meets the circle at points we could call $$P_1$$ and $$P_1'$$.
• Line $$l_2$$ meets the circle at points $$P_2$$ and $$P_2'$$.
• Line $$l_3$$ meets the circle at points $$P_3$$ and $$P_3'$$.

Step 5 – How to sketch the rough diagram in your notebook

1. Lightly draw three straight lines crossing in the middle of your page; label the common point $$B$$.
2. With a compass set to a radius of 5 cm draw a circle centred at $$B$$ so that the circle clearly cuts each of the three lines twice.
3. Mark the six intersection points and, if you like, label them $$P_1$$, $$P_1'$$, $$P_2$$, $$P_2'$$, $$P_3$$, $$P_3'$$.

Given three positive lengths $$a,\,b,\,c$$ such that each pair sums to more than the third:

$$a + b \gt c,\;\; b + c \gt a,\;\; c + a \gt b.$$

This set of three inequalities is called the triangle inequality.

The question: if these inequalities are true, can we actually draw a triangle whose three sides measure $$a,\,b,\,c$$?

Yes — and we show this by giving an explicit ruler-and-compass construction that always works.

Step-by-step construction

1. Draw a straight line and mark on it a segment $$\overline{PQ}$$ of length $$a$$. This will be the base.
2. With the compass width set to $$b$$, place its point on $$P$$ and draw an arc above the line.
3. Set the compass width to $$c$$, place its point on $$Q$$ and draw a second arc on the same side of $$\overline{PQ}$$.
4. Let the two arcs intersect at $$R$$. Join $$R$$ to $$P$$ and to $$Q$$. Triangle $$\triangle PQR$$ has sides $$a$$, $$b$$, $$c$$.

Why do the arcs meet?

Every point on the first arc is at distance $$b$$ from $$P$$; every point on the second arc is at distance $$c$$ from $$Q$$. For the two arcs to share a point, the distance between the centres $$P$$ and $$Q$$ (the base, $$a$$) must be less than the sum of the two radii:

This is exactly the triangle inequality $$b + c \gt a$$. Because it is true, the two circles overlap, so their intersection point $$R$$ exists.

The other two inequalities $$a + b \gt c$$ and $$c + a \gt b$$ guarantee that none of the sides is too long for the other two; together they ensure that the figure is a true triangle and not a degenerate straight line.

Conclusion

Whenever three positive numbers satisfy the triangle inequality, the construction succeeds, so a triangle with exactly those side lengths exists.

Triangles exist for, and only for, sets of lengths that satisfy the triangle inequality.

Let the three given lengths be arranged in ascending order

\(r_1\), \(r_2\) (the two smaller lengths) and \(AB\) (the greatest length).

To try to make a triangle we keep the longest side fixed as the base \(AB\).
At the end A we draw a circle of radius \(r_2\); at the end B we draw a circle of radius \(r_1\).
Any point where the two circles meet will be the third vertex of the required triangle.

Denote the fixed distance between the two centres by

$$d = AB.$$

From geometry of two circles the number of their intersection points depends on how \(d\) compares with the sum of the radii:

1. Two intersection points (a non-degenerate triangle)
   The circles cut each other iff \[r_1 + r_2 > AB.\] In this situation there are exactly two possible positions of the third vertex (above and below \(AB\)).
2. Exactly one common point (a degenerate triangle)
   If the two circles just touch externally, \[r_1 + r_2 = AB.\] The “triangle” collapses into a straight line – the only point of contact lies on \(AB\).
3. No common point (no triangle possible)
   When the distance between the centres exceeds the total reach of the two radii, \[r_1 + r_2 < AB.\] Because the circles fail to meet, we cannot get a third vertex and hence cannot construct any triangle.

Note that the inequality involving the difference of the radii, \(|r_1-r_2|\), is automatically true here because each radius is itself smaller than \(AB\). Therefore the deciding factor is only the comparison between \(r_1 + r_2\) and \(AB\).

Conclusion

The relation between the two smaller lengths (regarded as radii) and the largest length \(AB\) is summarised as:

• \(r_1 + r_2 > AB\) → two possible non-degenerate triangles,
• \(r_1 + r_2 = AB\) → one degenerate (collinear) triangle,
• \(r_1 + r_2 < AB\) → no triangle can be formed.

Given : Two circles are drawn with centres A and B respectively. Let the radius chosen for both circles be $$r$$ and let the distance between the two centres be the length of the given line segment $$AB$$.

We want Case 2 to occur – the two circles should not intersect internally (in fact they will not have any common point at all).

Step 1 ▪︎ Express the condition for intersection

For a point P to lie on both circles we must have $$AP = BP = r$$.

Applying the Triangle Inequality to triangle $$\triangle ABP$$, a necessary condition for such a point P to exist is \[ AB \le AP + BP = r + r = 2r \]

Thus, if the two circles are to meet, the distance between their centres must satisfy $$AB \le 2r$$.

Step 2 ▪︎ Reverse the inequality for no intersection

To ensure that no common point P can be found, we must violate the above condition; that is, we must have $$AB > 2r$$.

Step 3 ▪︎ Write the required relation

Re-arranging, the requirement becomes \[ r < \dfrac{1}{2}\,AB \]

Conclusion

Therefore, for Case 2 (the two circles with centres A and B do not intersect internally) the radius chosen for each circle must be less than half the length of AB.

Given. Three positive numbers $$a$$, $$b$$ and $$c$$ such that

$$a+b>c,\;b+c>a,\;c+a>b$$.

We already know (proved earlier in the chapter) that every triangle must obey these three inequalities. Now we prove the converse: if the inequalities hold, one can really draw a triangle having the three given lengths as its sides.

1. Choose the largest of the three numbers
   Without loss of generality let $$c$$ be the greatest. Draw a straight line and mark two points $$A$$ and $$B$$ on it so that $$AB=c$$.
2. Draw two circles (arcs)
   With centre $$A$$ and radius $$b$$ draw an arc above the line.
   With centre $$B$$ and radius $$a$$ draw another arc above the same side of $$AB$$.
3. Why do the arcs meet?
   Because $$a+b>c$$, the sum of the two radii is greater than the distance between the centres $$A$$ and $$B$$. Therefore the two circles intersect in exactly one or two points. Mark one such intersection as $$P$$.
4. Forming the triangle
   Join $$PA$$ and $$PB$$. The three segments are
   • $$PA=b$$ (radius of the circle centred at $$A$$),
   • $$PB=a$$ (radius of the circle centred at $$B$$),
   • $$AB=c$$ (constructed in Step 1).
   Thus $$\triangle PAB$$ has the required side lengths $$a$$, $$b$$ and $$c$$.
5. The other two inequalities are automatically satisfied
   The construction has already produced the side $$c$$ as the largest. The remaining inequalities $$b+c>a$$ and $$c+a>b$$ guarantee that neither $$a$$ nor $$b$$ is as long as (or longer than) the sum of the other two sides, so the figure really is a triangle and not a straight line.

Hence, whenever three lengths obey the triangle inequality, the above compass-and-ruler construction succeeds, proving that a triangle with those side lengths exists.

Conclusion. Yes — checking the triangle inequalities is both a necessary and a sufficient test for the existence of a triangle made with the three given lengths.

Objective. We are given three line-segments whose lengths are $$a\text{ cm}$$, $$b\text{ cm}$$, $$c\text{ cm}$$. We must decide, before drawing, whether they can form the three sides of a triangle.

The required test is the triangle inequality. Here is a complete procedure.

1. Write down the three lengths. Let them be $$a$$, $$b$$, $$c$$ (using the same unit).
2. Check positivity. If any one of $$a$$, $$b$$, $$c$$ equals $$0$$ or is negative, stop — no triangle exists.
3. Arrange in non-increasing order. Relabel so that $$a \ge b \ge c$$; here $$a$$ is the largest length.
4. Apply the triangle inequality. The triangle exists iff the largest side is less than the sum of the other two:

With our ordering this single condition already implies $$b \lt a + c$$ and $$c \lt a + b$$ (both follow because $$a \ge b \ge c$$ and lengths are positive).

5. State the conclusion.
   • If $$a \lt b + c$$, a triangle with sides $$a, b, c$$ can be constructed.
   • If $$a \ge b + c$$, no triangle exists.

Quick examples.

The procedure above gives a complete check for any three given lengths.

Required data

• Side $$AB = 5\,\mathrm{cm}$$
• Side $$AC = 4\,\mathrm{cm}$$
• Included angle $$\angle A = 45^\circ$$ (between $$AB$$ and $$AC$$)

Instruments : ruler, sharp pencil, compass (and an optional pro-tractor only to check).

Construction steps

1. Draw the known side $$AB$$
   With the ruler mark two points A and B so that the distance between them is exactly $$5\,\mathrm{cm}$$. Join them with a straight line segment. $$AB = 5\,\mathrm{cm}$$ is now fixed.
2. Construct a right angle at A
   Set the compass on A. With any convenient radius draw an arc that cuts $$AB$$ at a point P.
   Without changing the radius, put the compass point on P and draw an arc above AB. Call its intersection with the first arc Q.
   Join A to Q. Thus $$AQ$$ is perpendicular to $$AB$$ and $$\angle PAQ = 90^\circ$$.
3. Bisect the right angle to obtain $$45^\circ$$
   With centre P and any radius, draw an arc inside the 90° angle; with the same radius, centre Q, draw another arc that cuts the previous one at R.
   Join A to R. The ray $$AR$$ now makes an angle of $$45^\circ$$ with $$AB$$, i.e. $$\angle BAR = 45^\circ$$.
4. Mark off the second side $$AC = 4\,\mathrm{cm}$$
   Open the compass to exactly $$4\,\mathrm{cm}$$ using the ruler. Keeping the centre at A and the pencil on the 45° ray $$AR$$, cut the ray at a point C so that $$AC = 4\,\mathrm{cm}$$.
5. Complete the triangle
   Join C to B with a straight line segment. Triangle ABC is now formed.

Verification (optional)

• Measure $$AC$$ with the ruler; it should read $$4\,\mathrm{cm}$$.
• Place a pro-tractor at A; the angle between $$AB$$ and $$AC$$ should read $$45^\circ$$.

Hence, the required triangle $$\triangle ABC$$ satisfying $$AB = 5\,\mathrm{cm}$$, $$AC = 4\,\mathrm{cm}$$ and $$\angle A = 45^\circ$$ has been successfully constructed.

Given. Two lengths  $$AB = a\;( >0)$$,  $$AC = b\;( >0)$$ and their included angle $$\angle BAC = \theta$$.

Construction steps (SAS method).

1. Draw a ray AX. On it mark $$AB = a$$. Point B is now fixed.
2. At A construct the required angle $$\theta = \angle BAX$$. This produces a second ray AY.
3. On ray AY mark $$AC = b$$. Point C is now fixed.
4. Join $$BC$$. The figure ABC obtained is the required triangle (see the note below for when this fails).

What do we observe?

• Points B and C are distinct because both given sides are positive in length.
• The three points A, B, C will be non-collinear provided the angle you opened at step 2 satisfies $$0^{\circ} < \theta < 180^{\circ}$$. In that situation the segment $$BC$$ is automatically determined and a genuine triangle is formed.
• If $$\theta = 0^{\circ}$$ the two rays AX and AY coincide; then $$C$$ falls on the same line as $$A$$ and $$B$$, so the figure collapses into a straight segment and no triangle exists.
• If $$\theta = 180^{\circ}$$ the two rays become opposite straight lines; $$A,\,B,\,C$$ once again lie on a single line and a triangle cannot be obtained.

Algebraic check (Law of Cosines). The third side produced by the construction has length

Because $$\cos\theta$$ is defined only for $$0^{\circ}\le \theta\le 180^{\circ}$$ we see:

• For $$0^{\circ}<\theta<180^{\circ}$$, the right–hand side is positive, so $$BC>0$$ and the triangle is valid.
• For $$\theta=0^{\circ}$$ or $$\theta=180^{\circ}$$, the right–hand side becomes $$(a\!-\!b)^{2}$$ or $$(a\!+\!b)^{2}$$; geometrically this forces the three points to be collinear, giving a degenerate (invalid) triangle.

Conclusion. A triangle can always be drawn from two given side lengths and their included angle except when that included angle is $$0^{\circ}$$ or $$180^{\circ}$$. Hence, apart from these two extreme cases, every set of “two sides and the included angle” produces one and only one triangle.

Given data

• Side $$AB = 5\,\mathrm{cm}$$
• Angle $$\angle A = 45^\circ$$
• Angle $$\angle B = 80^\circ$$

Because of the angle–sum property of a triangle, the third angle is

\[\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 45^\circ - 80^\circ = 55^\circ.\]

This shows that only one triangle satisfies the three conditions, so if we construct it correctly the answer is unique.

Instruments required

• Ruler (centre–zero scale)
• Compass (for transferring lengths)
• Protractor (for measuring the two given angles; a compass-only construction is possible but longer)
• Sharp pencil

Construction steps

1. Draw the base.
   Place the ruler, mark points $$A$$ and $$B$$ so that $$AB = 5\,\mathrm{cm}$$. Join them with a straight line. Keep the labelling exactly as required (left-to-right, or right-to-left; any one is fine).
2. Construct $$\angle A = 45^\circ$$.
   Place the centre of the protractor at $$A$$, match the base line with $$AB$$, read the 45-degree mark and put a small dot there. Remove the protractor and, using the ruler, draw a ray $$AX$$ through the dot. Ray $$AX$$ is now such that $$\angle BAX = 45^\circ$$. (If you prefer a pure–compass method, first draw a right angle and then bisect it.)
3. Construct $$\angle B = 80^\circ$$.
   Similarly, put the protractor at $$B$$ with the baseline along $$BA$$, mark the 80-degree point on the correct scale and draw ray $$BY$$ through it. Thus $$\angle ABY = 80^\circ$$.
4. Locate point $$C$$.
   Extend rays $$AX$$ and $$BY$$ with the ruler until they meet. Mark the point of intersection as $$C$$. (If they do not meet, extend the rays further; they must intersect because their interior angle sum is less than 180°.)
5. Complete the triangle.
   Join $$C$$ to $$A$$ and $$C$$ to $$B$$ with straight lines. Triangle $$ABC$$ is obtained.

Verification

• The side $$AB$$ was drawn exactly $$5\,\mathrm{cm}$$ long — matches the given data.
• $$\angle A$$ was laid off as $$45^\circ$$ and $$\angle B$$ as $$80^\circ$$ — again matches.
• By the angle–sum property, the third angle must be $$55^\circ$$; measuring $$\angle C$$ with a protractor confirms this, so no construction error has crept in.

Hence the required triangle $$ABC$$ has been constructed.

Problem restatement

We are given two angles and the side included between them. We want to find when a triangle can be formed.

Notation

• The required triangle is $$\triangle ABC$$.
• The given side is the side included between the two given angles, so we take it as $$BC = a$$. This is the base.
• The two given angles are $$\angle ABC = B$$ and $$\angle BCA = C$$ (the angles at the ends of the base).

Step 1 — Draw the given side

Draw a straight segment $$BC$$ of the given length $$a$$. Mark its end-points $$B$$ and $$C$$.

Step 2 — Construct the two given angles

1. At $$B$$ construct $$\angle CBX = B$$, with $$BC$$ as one arm. The new ray $$BX$$ is the second arm.
2. At $$C$$ construct $$\angle BCY = C$$, with $$CB$$ as one arm. The new ray $$CY$$ is the second arm.

($$X$$ and $$Y$$ are points only to illustrate the direction of the rays; they are not new vertices.)

Step 3 — Locate the third vertex

The third vertex $$A$$ must lie on both rays $$BX$$ and $$CY$$. So $$A$$ is the intersection of these two rays.

Step 4 — When do the rays meet?

• Case 1: $$B + C \lt 180^{\circ}$$. The rays drawn on the same side of $$BC$$ tilt towards each other and meet at a unique point $$A$$. A unique triangle is formed.
• Case 2: $$B + C = 180^{\circ}$$. The rays become parallel (or form a straight line on $$BC$$ itself); they never meet, so no triangle exists.
• Case 3: $$B + C \gt 180^{\circ}$$. The rays spread away from each other and never meet, so no triangle exists.

Step 5 — The condition

For a triangle the three angles must sum to $$180^{\circ}$$. The third angle is $$A = 180^{\circ} - (B + C)$$. For $$A$$ to be a positive angle we need

$$B + C \lt 180^{\circ}.$$

So the necessary and sufficient conditions are

Step 6 — Uniqueness (ASA)

When the conditions above hold, the triangle is unique up to congruence by the ASA criterion (two angles and the included side determine a single triangle).

Conclusion

No, triangles do not exist for every combination of two angles and their included side. They exist only when the sum of the two given angles is less than $$180^{\circ}$$ (and the side has positive length). In that case exactly one triangle can be constructed.

Recall of the condition

When we are given two angles and the side included between them (symbolically: $$A-S-A$$), we first draw the given side, then construct the two given angles at its end–points. The two rays will meet and give the third vertex only when the sum of the two given angles is less than $$180^{\circ}$$. Algebraically, if the given angles are $$\angle A = x^{\circ}$$ and $$\angle B = y^{\circ}$$, then a triangle can exist ⇔

$$x^{\circ} + y^{\circ} < 180^{\circ}.$$

Therefore, to obtain examples where a triangle cannot be drawn we must deliberately choose the two angles so that their sum is $$\ge 180^{\circ}$$. Two convenient choices are shown below.

Example 1

1. Let the included side be $$AB = 5\,\text{cm}$$.
2. Give $$\angle A = 110^{\circ}$$ and $$\angle B = 80^{\circ}$$.
3. Test the condition: $$110^{\circ}+80^{\circ}=190^{\circ}>180^{\circ}$$.
4. The third (would–be) angle would have to be $$180^{\circ}-190^{\circ}=-10^{\circ},$$ which is impossible.

Hence no triangle can be constructed with these data.

Example 2

1. Let the included side be $$AB = 4\,\text{cm}$$.
2. Give $$\angle A = 60^{\circ}$$ and $$\angle B = 120^{\circ}$$.
3. Now $$60^{\circ}+120^{\circ}=180^{\circ}$$ exactly, so the two rays lie in the same straight line and never form a closed figure.

Again, a triangle is impossible.

Thus any pair of angles whose sum is $$\ge 180^{\circ}$$ (with any positive length for the included side) furnishes the required counter-example.

Step 1 : Recall the angle–sum property of a triangle
For every triangle, the sum of its three interior angles is
\[\angle A + \angle B + \angle C = 180^\circ\]

Step 2 : Let the two given angles be $$x^\circ$$ and $$y^\circ$$
Assume these are to be taken as any two interior angles of some triangle.
Denote the third (unknown) angle by $$z^\circ$$.

Step 3 : Express the third angle using the angle–sum property
Substituting $$\angle A = x^\circ, \; \angle B = y^\circ, \; \angle C = z^\circ$$ in the relation from Step 1 gives
$$x^\circ + y^\circ + z^\circ = 180^\circ$$

Transposing, the third angle is
$$z^\circ = 180^\circ - (x^\circ + y^\circ)$$

Step 4 : Impose the basic condition on every angle of a triangle
Every interior angle of a triangle must be strictly greater than $$0^\circ$$. Hence
$$0^\circ < z^\circ$$

Substituting for $$z^\circ$$ from Step 3, we require
$$0^\circ < 180^\circ - (x^\circ + y^\circ)$$

Step 5 : Rearrange the inequality
Adding $$(x^\circ + y^\circ)$$ to both sides gives
$$(x^\circ + y^\circ) < 180^\circ$$

Step 6 : State the complete rule

(Of course, each of the given angles must itself be greater than $$0^\circ$$.)

Thus the rule analogous to the triangle–inequality for sides is:
“A triangle is possible only when the sum of the two given angles is less than $$180^\circ$$.”

What is being asked?
Earlier you learnt a rule for recognising parallel lines that used the equality of certain angles (for example, “if a pair of corresponding angles are equal, the two lines are parallel”).  The question now is: can we frame a similar rule by looking at the sum of a pair of angles instead of their equality?

Step 1 : Set-up of the figure
Let two distinct lines $$l_1$$ and $$l_2$$ be cut by a transversal $$t$$ at points $$A$$ and $$B$$ respectively.
The two angles that lie between the lines and on the same side of the transversal are usually labelled $$ \angle 4$$ (at $$A$$) and $$\angle 5$$ (at $$B$$).

Given condition to test
Assume $$\angle 4 + \angle 5 = 180^{\circ}\;(1)$$ In other words, the two interior angles are supplementary.

Step 2 : Draw a definite parallel through one point
Through point $$A$$ draw a line $$m$$ such that $$m \parallel l_2$$.  (By the Parallel Postulate this is always possible and unique.)

Step 3 : Relate the new angle to the given angle
Because $$m \parallel l_2$$, the angle that $$m$$ makes with the transversal at $$A$$—call it $$\angle(m,t)$$—is a corresponding angle of $$\angle 5$$. Hence $$\angle(m,t) = \angle 5 \;(2)$$

Step 4 : Use the linear-pair fact at point A
On the straight line through $$A$$, the angles $$\angle 4$$ and $$\angle(m,t)$$ form a linear pair, so $$\angle 4 + \angle(m,t) = 180^{\circ}\;(3)$$

Substitute (2) into (3): $$\angle 4 + \angle 5 = 180^{\circ}$$ which is exactly the given condition (1).  Thus (3) is satisfied automatically.

Step 5 : Conclude that the two lines through A coincide
At point $$A$$ the transversal $$t$$ now makes the same angle with $$l_1$$ (namely $$\angle 4$$) and with $$m$$ (because $$\angle(m,t)=180^{\circ}-\angle 4$$, the supplement of $$\angle 4$$, and a straight line determines a unique supplement). Therefore the only possibility is that $$m$$ and $$l_1$$ are the same line.

Step 6 : Final deduction
Since $$m \parallel l_2$$ and $$m$$ coincides with $$l_1$$, we obtain the key result: \[ l_1 \;\parallel\; l_2 \] Hence the condition (1) is sufficient to guarantee that the two lines are parallel.

Framing the rule
Yes, we can use the sum of the two angles to formulate the rule:

• If the sum of a pair of interior angles on the same side of a transversal is $$180^{\circ}$$, then the two lines are parallel.
• Conversely, when two lines are parallel, every such pair of interior angles is supplementary.

Given data

• Two angles: $$60^\circ$$ and $$70^\circ$$ (their sum is less than $$180^\circ$$).
• The side included between these two angles is first taken as $$5\,\text{cm}$$.

I. Construction when the included side is $$5\,\text{cm}$$

1. Draw a straight line and mark its end–points as A and B so that $$AB = 5\,\text{cm}$$.
2. With the help of a protractor make $$\angle XAB = 60^\circ$$. Draw the ray $$AX$$.
3. Again with a protractor make $$\angle YBA = 70^\circ$$. Draw the ray $$BY$$.
4. Let rays $$AX$$ and $$BY$$ meet at a point C. Triangle $$\triangle ABC$$ is obtained.

II. Measuring the third angle

Use a protractor to measure $$\angle ACB$$. You will find

$$\angle ACB = 50^\circ.$$

(Reason: by the angle–sum property of a triangle,

$$\angle A + \angle B + \angle C = 180^\circ$$

so

$$60^\circ + 70^\circ + \angle C = 180^\circ \;\;\Rightarrow\;\; \angle C = 50^\circ.)$$

III. Changing the base length to $$7\,\text{cm}$$

1. Draw another line segment $$DE = 7\,\text{cm}$$.
2. At D construct $$\angle ZDE = 60^\circ$$; draw ray $$DZ$$.
3. At E construct $$\angle WED = 70^\circ$$; draw ray $$EW$$.
4. Let the two rays meet at F. Triangle $$\triangle DEF$$ is formed.
5. Measure $$\angle DFE$$ with a protractor. You will again obtain $$50^\circ$$.

IV. Explanation — Why the third angle does not change

For any triangle, the three interior angles always satisfy

$$\angle 1 + \angle 2 + \angle 3 = 180^\circ.$$

Here two of the angles are fixed at $$60^\circ$$ and $$70^\circ$$, so the third must be

$$180^\circ - (60^\circ + 70^\circ) = 50^\circ$$

no matter how long the side between those two angles is taken. Changing the base merely rescales the triangle; it cannot alter its angles.

Result

• The third angle is always $$50^\circ$$.
• This measure remains the same even if the base is changed from $$5\,\text{cm}$$ to $$7\,\text{cm}$$ (or to any other length).

Step 1 : Recall the angle–sum property of a triangle
For every triangle, the three interior angles always add up to $$180^{\circ}$$.
Therefore, if two of the angles are already known, the third angle can be found from

Step 2 : Choose two convenient angles and verify

• Take $$\angle A = 40^{\circ}$$ and $$\angle B = 60^{\circ}$$.
• Using (1), $$\angle C = 180^{\circ}-(40^{\circ}+60^{\circ}) = 80^{\circ}$$.

Now construct two different triangles having the same angles $$40^{\circ},60^{\circ},80^{\circ}$$ but different side AB (the side included between the chosen angles):

1. In the first triangle, keep AB = 4 cm.
2. In the second triangle, keep AB = 7 cm.

In both constructions you will still measure $$\angle C = 80^{\circ}$$, showing that the third angle does not change when the length of the included side changes.

Step 3 : Try a second pair of angles

• Let $$\angle P = 55^{\circ}$$ and $$\angle Q = 85^{\circ}$$.
• Equation (1) gives $$\angle R = 180^{\circ}-(55^{\circ}+85^{\circ}) = 40^{\circ}$$.

Again draw two triangles with PQ = 5 cm and PQ = 9 cm, keeping the two angles fixed at the ends of PQ. In both cases the remaining angle will still be exactly $$40^{\circ}$$.

Step 4 : General argument
Whatever be the chosen pair of angles, let them be $$x^{\circ}$$ and $$y^{\circ}$$. Substituting in (1)

Equation (2) involves only the two angles. No side length appears in it. Hence the length of the side between the two given angles has no influence on the value of the third angle.

Conclusion
Once two angles of a triangle are fixed, the third angle is uniquely decided by the angle–sum property and remains the same, no matter what length the included side is chosen to be.

Goal of the experiment
We want to discover a numerical relation that links the three interior angles of any triangle. To do this we shall draw several different triangles, measure their three angles and then compare the sums of two angles with the third angle.

Step 1 – Material required

• A ruler and a sharp pencil (to draw triangles)
• A pro-tractor (to measure each angle to the nearest degree)
• An observation sheet (loose ruled paper or a maths notebook)

Step 2 – What to measure / keep track of

1. Label the vertices of every triangle as A, B, C.
2. Measure the size of each interior angle:   $$\angle A\,,\;\angle B\,,\;\angle C$$
3. For every triangle immediately calculate the three possible sums of two angles:   $$\angle A+\angle B,\;\angle B+\angle C,\;\angle C+\angle A$$
4. Optionally find the total of all three angles:   $$\angle A + \angle B + \angle C$$

Step 3 – How to organise the data
Make a neat table. A convenient layout is:

Step 4 – What to look for while filling the table

• Check if, for every row, each of the three sums in columns 5–7 is greater than the remaining third angle.
• Check whether the entries in the last column are always the same number (they should add to $$180^{\circ}$$).

Expected observation
The completed table will show two facts for every triangle you draw:

• The sum of the three interior angles is always $$180^{\circ}$$.
• The sum of any two interior angles is always greater than the third interior angle.

That is the relation you were asked to explore, and the table above is the systematic way to collect and organise the required data.

We know the Angle Sum Property of a triangle:

In any triangle, the sum of the three interior angles is $$180^{\circ}$$.

For $$\triangle ABC$$ the three angles are $$\angle A,\, \angle B,\, \angle C$$. Therefore

$$\angle A + \angle B + \angle C = 180^{\circ}$$

Substitute the given measures $$\angle B = 50^{\circ}$$ and $$\angle C = 70^{\circ}$$:

$$\angle A + 50^{\circ} + 70^{\circ} = 180^{\circ}$$

Add the known angles on the left–hand side:

$$\angle A + 120^{\circ} = 180^{\circ}$$

To isolate $$\angle A$$, subtract $$120^{\circ}$$ from both sides:

$$\angle A = 180^{\circ} - 120^{\circ}$$

Compute the subtraction:

$$\angle A = 60^{\circ}$$

Thus, the measure of $$\angle A$$ in triangle $$ABC$$ is

Step 1 – Recall what the three given lines do
Through the common point A we already had the two lines $$AB$$ and $$AC$$. A third line (call it $$AXY$$) has now been drawn through the same point. Thus three distinct straight lines pass through A.

This divides the space round A into six angles. We are concentrating on the three small angles that lie on the same side of the straight line $$XY$$ :

• $$\angle BAX$$ (already given in the story – let us keep its value as $$p^{\circ}$$),
• $$\angle XAY$$ (already given – keep its value as $$q^{\circ}$$),
• $$\angle YAC$$ (whose value we have to find).

Step 2 – Use the straight-angle property
Along the straight line $$XY$$ (remember, a straight angle measures $$180^{\circ}$$) \[ \angle BAX+\angle XAY+\angle YAC = 180^{\circ}. \]
In symbols, substituting the given values $$p^{\circ}$$ and $$q^{\circ}$$, $$p+q+\angle YAC = 180.$$ Therefore
$$\angle YAC = 180^{\circ}-\bigl(p+q\bigr).$$

Step 3 – Find the third angle between the three lines
Exactly the same reasoning, applied along the straight line $$AB$$, gives the second required angle. The three angles that lie on one side of $$AB$$ are

• $$\angle CAB$$ (already supplied – call its measure $$r^{\circ}$$),
• $$\angle BAX$$ (measure $$p^{\circ}$$),
• $$\angle XAB$$ (wanted).

Step 4 – Put in the numerical data
From the earlier part of the tale we already know (measurements supplied in the textbook)
$$p = 40^{\circ}, \qquad q = 25^{\circ}, \qquad r = 15^{\circ}.$$

Substituting those values:

Step 5 – State the final values
\[ \boxed{\angle XAB = 125^{\circ}\quad\text{and}\quad\angle YAC = 115^{\circ}.} \]

Step 1 : Draw a convenient figure

Draw $$\triangle ABC$$. Through the vertex $$A$$ draw a straight line $$AE$$ so that $$AE$$ is parallel to the base $$BC$$. (Because of the ruler–parallel line postulate, we can always draw one and only one line through a point parallel to a given line.)

Thus

• $$AE \parallel BC$$
• $$AB$$ and $$AC$$ act as transversals cutting these two parallel lines.

Step 2 : Relate the angles made by each transversal

1. When the transversal $$AB$$ cuts the parallel lines $$AE$$ and $$BC$$, the angle

$$\angle B$$ (inside the triangle) is alternate interior to the angle $$\angle BAE$$ (outside on the straight line $$AE$$).

Hence  $$\angle B = \angle BAE$$.

2. When the transversal $$AC$$ cuts the same pair of parallel lines $$AE$$ and $$BC$$, the angle

$$\angle C$$ is alternate interior to $$\angle CAE$$.

Hence  $$\angle C = \angle CAE$$.

Step 3 : Use the straight–angle (linear–pair) property on line $$AE$$

At the point $$A$$ the two rays $$AB$$ and $$AC$$ lie on one side of the straight line $$AE$$, so together the three angles

• $$\angle BAE$$,
• $$\angle BAC$$ (that is, $$\angle A$$ of the triangle),
• $$\angle CAE$$

form a straight angle. By the Linear Pair Axiom, the measure of a straight angle is $$180^{\circ}$$, therefore

Step 4 : Replace each outside angle with the corresponding interior angle

From Step 2 we already have

• $$\angle BAE = \angle B$$,
• $$\angle CAE = \angle C$$.

Substituting these equalities into equation (1):

Step 5 : Conclude

Therefore, in every triangle the sum of the three interior angles is a straight angle, i. e. $$180^{\circ}$$.

Hence proved.

First draw \(\triangle ABC\). Extend side \(BC\) beyond C to a point D. This creates the exterior angle \(\angle ACD\).

1. State the given data
   • $$\angle A = 50^\circ$$
   • $$\angle B = 60^\circ$$

2. Recall the Exterior-Angle Theorem
   For any triangle, the exterior angle is equal to the sum of the two interior opposite angles:
   $$\angle ACD = \angle A + \angle B$$

3. Substitute the given measures
   $$\angle ACD = 50^\circ + 60^\circ$$

4. Add the angles
   \[\angle ACD = 110^\circ\]

Therefore, the measure of the exterior angle at C is $$110^\circ$$.

Given a triangle $$\triangle ABC$$ with side $$BC$$ produced to $$D$$.
The exterior angle at vertex $$C$$ is $$\angle ACD$$.

1  Proof with properties

1. Angle–sum property of a triangle:
   $$\angle A + \angle B + \angle ACB = 180^{\circ} \quad(1)$$
2. Linear pair at $$C$$:
   $$\angle ACB + \angle ACD = 180^{\circ} \quad(2)$$

Subtract (1) from (2):

$$ (\angle ACB + \angle ACD) - (\angle A + \angle B + \angle ACB) = 0 $$
$$ \Rightarrow \; \angle ACD = \angle A + \angle B $$

Thus the exterior angle equals the sum of the two interior opposite (remote) angles.

2  Numerical check

In every example $$\angle ACD = \angle A + \angle B$$.

3  Conclusion

Exterior Angle Property:
\[\boxed{\text{Exterior angle at a vertex of a triangle} = \text{sum of the two interior opposite angles}}\]

Diagram to draw: Sketch $$\triangle ABC$$, extend side $$BC$$ to a point $$D$$, label $$\angle ACD$$ outside and the three interior angles $$\angle A,\,\angle B,\,\angle ACB$$ inside.

Step 1  Name the triangle

Take the paper triangle and call its three vertices $$A,\;B,\;C$$ in such a way that the side $$BC$$ is the base and the vertex $$A$$ lies above it.

Step 2  How the fold is made

Keep the side $$BC$$ pressed flat on the table so that it cannot move. Now lift the portion of the triangle that lies above $$BC$$ and bend it till you obtain a sharp crease that passes through the top vertex $$A$$. (While bending, the edge $$BC$$ continues to touch the table and therefore stays exactly where it was.) Let the crease meet the base at the point $$D$$. Thus the fold-line is the segment $$AD$$.

Step 3  What happens to the base while folding?

Because $$BC$$ is held tight, every point of $$BC$$ is still in the same place after the paper is folded. In other words, the base $$BC$$ coincides with its own mirror image produced by the reflection (fold) in the line $$AD$$.

Step 4  A fact about reflections of a straight line

For any straight line to look exactly the same after reflection in another line, only two possibilities exist:
(i) the reflecting line is the line itself, or
(ii) the reflecting line cuts the given line at a right angle (is perpendicular to it).

The first possibility is ruled out here because the crease $$AD$$ contains the vertex $$A$$, whereas the base $$BC$$ does not. Hence only the second possibility can hold:

\[AD \;\perp\; BC\]

Step 5  Why the crease is an altitude

The crease passes through the vertex $$A$$ and we have just proved that it is perpendicular to the base $$BC$$. By definition, the segment drawn from a vertex of a triangle perpendicular to the opposite side is an altitude. Therefore, $$AD$$ – the crease obtained by folding – is indeed an altitude of $$\triangle ABC$$.

Conclusion

When the triangle is folded as described, the resulting crease must be perpendicular to the fixed base. Hence the paper-folding procedure correctly produces an altitude from the top vertex to the base.

Objective : To draw the altitude from vertex A of an arbitrary $$\triangle ABC$$ to the side (base) $$BC$$.

What you need : ruler, compass, sharp pencil (optional: set–square or pro-tractor for checking).

1. Draw the triangle
   1. With the ruler draw any convenient straight line segment and mark its end-points B and C. This is the base $$BC$$.
   2. Choose any point not lying on the line $$BC$$ and label it A.
   3. Join $$AB$$ and $$AC$$ with the ruler. $$\triangle ABC$$ is now constructed.
2. Construct the altitude from A to BC
   (We are going to drop a perpendicular from the external point A to the line BC.)
   1. Place the compass needle at A, open it so that the pencil end reaches the line $$BC$$ and draw an arc cutting $$BC$$ at two distinct points. Label the intersection points $$P$$ (nearer to B) and $$Q$$ (nearer to C).  (Same radius throughout the next two steps.)
   2. Without altering the compass width, place the compass at $$P$$ and draw an arc below the line $$BC$$ (on the side opposite A).
   3. Keeping exactly the same radius, put the compass at $$Q$$ and draw another arc so that it intersects the previous one. Mark the point of intersection of the two arcs as $$R$$.
   4. With the ruler join $$A$$ to $$R$$ and extend the line if necessary to meet $$BC$$ at a point $$D$$.
3. Result obtained
   Because points $$R$$ are equidistant from $$P$$ and $$Q$$, the segment $$PR = QR$$. Hence $$R$$ lies on the perpendicular bisector of $$PQ$$, and therefore $$AR$$ is perpendicular to $$PQ$$ and to the straight line $$BC$$ (since $$P, Q$$ lie on $$BC$$). Thus \[ AD \perp BC \] The segment $$AD$$ is the required altitude (height) of the triangle from vertex A to the base BC.
4. (Optional check) Place a set–square/pro-tractor at point $$D$$; you will read an angle of $$90^{\circ}$$ between $$AD$$ and $$BC$$, confirming the construction.

Construction finished: $$AD$$ is the altitude from vertex A to side BC in $$\triangle ABC$$.

Objective. From a given vertex of a triangle, draw the altitude (a line perpendicular to the opposite side) accurately with the help of an ordinary ruler and a set square.

Given. A triangle – let its three vertices be $$A$$, $$B$$ and $$C$$. Suppose the altitude has to be drawn from the vertex $$A$$ to the side $$BC$$.

Idea behind the construction. A set square has one corner that is exactly a right angle ($$90^{\circ}$$). If we make one of the short edges of the set square rest exactly on the base $$BC$$, the other short edge automatically becomes perpendicular to $$BC$$ (because the two short edges themselves form a right angle). Sliding the set square in this position until the second edge passes through $$A$$ therefore gives the required altitude.

Step–by–step construction.

1. Draw (or take, if already drawn) the triangle $$\triangle ABC$$ on your sheet.
2. Pick up the set square. Identify its right–angled corner; the two short edges that meet at this corner are mutually perpendicular.
3. Place one of these short edges exactly along the side $$BC$$. Make sure the entire edge sits flush on the segment $$BC$$ — this locks the set square so that its other short edge is now perpendicular to $$BC$$.
4. Without turning the set square, slide it along $$BC$$. During the sliding keep the first edge in continuous contact with the line $$BC$$. Keep sliding until the other short edge of the set square passes through the vertex $$A$$.
5. Hold the set square firmly. Take the ruler and align it with the second short edge (the one passing through $$A$$). With a sharp pencil draw the line through $$A$$ along this edge. Continue the line until it meets $$BC$$ at, say, the point $$D$$.
6. Mark the intersection point $$D$$. The segment $$AD$$ is the required altitude.

Why it works. Because the two short edges of the set square are exactly at $$90^{\circ}$$ to each other, and one of them was kept coincident with $$BC$$, the other is forced to be perpendicular to $$BC$$. Hence $$\angle ADB = 90^{\circ}$$ and $$AD$$ is the altitude from $$A$$.

Check (optional). You may verify the right angle with a protractor: measure $$\angle ADB$$; it should read $$90^{\circ}$$, confirming the construction.

Given idea  An altitude of a triangle is a line segment drawn from one vertex perpendicular to the opposite side (or its extension). We have to decide whether a whole side of the triangle itself can satisfy this condition.

Step 1 – Look for a right angle
If two sides of a triangle are already perpendicular, the triangle is right-angled. Label the triangle $$\triangle ABC$$ with the right angle at $$A$$ so that $$\angle A = 90^\circ$$. Then

• side $$AB$$ is perpendicular to side $$AC$$, because they meet at the right angle.

Step 2 – Check the definition of altitude
Take the vertex $$B$$. The altitude from $$B$$ must be a line through $$B$$ drawn perpendicular to the opposite side $$AC$$.

• Because $$AB \perp AC$$ and the point $$B$$ already lies on $$AB$$, the very segment $$AB$$ itself meets both conditions: it starts from $$B$$ and is perpendicular to $$AC$$.

Hence $$\displaystyle AB$$ acts simultaneously as

• a side of $$\triangle ABC$$, and
• the altitude drawn from vertex $$B$$ to side $$AC$$.

The same reasoning shows that the other leg $$AC$$ is the altitude from vertex $$C$$ to side $$AB$$.

Step 3 – Conclusion
Therefore a triangle in which a side is also an altitude does exist: any right-angled triangle provides such an example. In triangles that are not right-angled, no two sides are perpendicular, so no side can coincide with an altitude.

Diagram to draw (for students): Draw a right-angled triangle ABC with the right angle at A. Mark AB ⟂ AC. Then shade AB and label it both “side” and “altitude from B”.

The four names already used in the chapter classify triangles in two different ways.

• By the lengths of the three sides
    • Equilateral   (all three sides equal)
    • Isosceles   (exactly two sides equal)
    • Scalene   (no two sides equal)
• By the sizes of the three angles
    • Right-angled   (one angle is $$90^\circ$$)

But a triangle can also be named from its angles in two other ways:

1. Acute-angled triangle
     Every angle is less than $$90^\circ$$; for example, an equilateral triangle with $$60^\circ$$, $$60^\circ$$, $$60^\circ$$ is an acute-angled triangle.
2. Obtuse-angled triangle
     Exactly one angle is greater than $$90^\circ$$; the other two angles are then each less than $$90^\circ$$ because the three angles must add up to $$180^\circ$$.

Thus, besides the four kinds already mentioned, we can also “spot”

• an acute-angled triangle, and
• an obtuse-angled triangle.

In total, triangles are classified as

• Equilateral, Isosceles, Scalene (by sides)  and
• Acute-angled, Right-angled, Obtuse-angled (by angles).

Triangles can be classified in two independent ways:

• By the lengths of their sides  →  scalene, isosceles, equilateral (already discussed).
• By the measures of their interior angles (asked here).

Recall that the three interior angles of any triangle always add up to

$$180^{\circ}$$

On the basis of angle measures, every triangle falls into exactly one of the three kinds described below.

1. Right-angled triangle
     • Exactly one interior angle is a right angle, i.e. $$90^{\circ}$$.
     • The other two angles together add up to $$90^{\circ}$$, because $$90^{\circ}+90^{\circ}=180^{\circ}$$.
2. Acute-angled triangle
     • Each of the three interior angles is strictly less than $$90^{\circ}$$.
     • Since all three are acute, their sum is still $$180^{\circ}$$.
3. Obtuse-angled triangle
     • Exactly one interior angle is obtuse, i.e. greater than $$90^{\circ}$$ but less than $$180^{\circ}$$.
     • The other two angles must then be acute, because \[\text{(obtuse angle)} + \text{(sum of the other two)} = 180^{\circ}\]

No triangle can have two right angles or two obtuse angles, because that would force the angle-sum to exceed $$180^{\circ}$$, which is impossible for a triangle.

Step 1 — Recall the basic facts

• Every triangle has exactly three interior angles.
• Their sum is $$180^\circ$$.
• An acute angle measures less than $$90^\circ$$.
• A right angle measures exactly $$90^\circ$$.
• An obtuse angle measures more than $$90^\circ$$ and less than $$180^\circ$$.

Step 2 — How many acute angles each kind of triangle has

Step 3 — Why "one acute angle" is not enough

From the table, every right-angled triangle and every obtuse-angled triangle already contains acute angles. If we defined an acute-angled triangle as "a triangle with one acute angle", the definition would wrongly include those other two kinds as well.

Step 4 — Correct definition

A triangle is called acute-angled if and only if all three of its interior angles are acute. In symbols, with the three angles $$\alpha, \beta, \gamma$$:

Conclusion

• An acute-angled triangle is a triangle whose every interior angle measures less than $$90^\circ$$.
• It cannot be defined as "a triangle with one acute angle", because right-angled and obtuse-angled triangles also have at least one acute angle.

Given figure (to be drawn)
Draw a circle with centre $$O$$. Through $$O$$ draw three straight lines so that each pair makes an angle of $$60^{\circ}$$ with the next. These three lines meet the circle in six points. Name them successively $$A,\,B,\,C,\,D,\,E,\,F$$ while going round the circle. (Thus $$A,B,C,D,E,F$$ form a regular hexagon on the circle.)

We have to use the seven special points $$A,B,C,D,E,F$$ (on the circle) and $$O$$ (its centre) to obtain all possible isosceles triangles and justify why every triangle listed is indeed isosceles.

1. Isosceles triangles having the centre as a vertex

The circle’s radii are all equal, so

$$OA = OB = OC = OD = OE = OF \; (= r\text{ say}).$$

Therefore, whenever the two sides that start from $$O$$ are used together in a triangle, those two sides are equal ⇒ the triangle is isosceles.

Count so far = 6

2. Isosceles triangles made only with points on the circle

Because $$A,B,C,D,E,F$$ form a regular hexagon, the six arcs $$\widehat{AB},\widehat{BC},\dots ,\widehat{FA}$$ are equal.
For equal arcs the corresponding chords are equal; hence

$$AB = BC = CD = DE = EF = FA \; (= s\text{ say}).$$

(a) Triangles whose vertices are three consecutive points

New triangles found = 6
Cumulative count = 12

(b) Triangles whose vertices are alternate points

Take every second point on the hexagon:

• $$A\,,C\,,E$$ form $$\triangle ACE$$
• $$B\,,D\,,F$$ form $$\triangle BDF$$

Their arcs are each double (i.e. $$120^{\circ}$$), so the three corresponding chords are equal. Hence both triangles are equilateral, certainly isosceles.

New triangles found = 2
Final count = 6 + 6 + 2 = 14

Complete list of the 14 isosceles triangles

1. $$\triangle OAB$$ 8. $$\triangle BCD$$
2. $$\triangle OBC$$ 9. $$\triangle CDE$$
3. $$\triangle OCD$$ 10. $$\triangle DEF$$
4. $$\triangle ODE$$      11. $$\triangle EFA$$
5. $$\triangle OEF$$      12. $$\triangle FAB$$
6. $$\triangle OFA$$      13. $$\triangle ACE$$
7. $$\triangle ABC$$      14. $$\triangle BDF$$

All fourteen triangles satisfy the definition of an isosceles triangle because each has at least two equal sides, as proved above.

Notation fixed for both figures

• The two circles in Fig. (i) have common centres $$A$$ and $$B$$ and intersect in the points $$P$$ (upper) and $$Q$$ (lower).
• The three circles in Fig. (ii) have centres $$A,\,B,\,C$$. Besides the centres, every pair of circles meets in a second point: let these be
  AB–circle : $$P$$ (other than $$C$$),
  BC–circle : $$Q$$ (other than $$A$$),
  CA–circle : $$R$$ (other than $$B$$).
• All the circles are congruent, so we write their common radius as $$r$$:
  $$AP = AQ = BP = BQ = AB = AC = BC = BR = CR = AR = r.$$

The only fact used in the whole exercise is:
“Every point that lies on a circle is at distance $$r$$ from its centre.”

===================================================================================

I.  Isosceles and equilateral triangles in Fig. (i)

1. $$\triangle ABP$$
   AP = BP (radii) ⇒ is isosceles;
   in addition AB = r (because A is on the circle with centre B), so all the three sides are equal ⇒  equilateral.
2. $$\triangle ABQ$$ — exactly the same reason as above; therefore equilateral.
3. $$\triangle APQ$$
   AP = AQ ⇒ isosceles (generally not equilateral because PQ ≠ r).
4. $$\triangle BPQ$$
   BP = BQ ⇒ isosceles.

Hence Fig. (i) supplies two equilateral (ABP, ABQ) and four isosceles (ABP, ABQ, APQ, BPQ) triangles.

===================================================================================

II.  Isosceles and equilateral triangles in Fig. (ii)

Step 1 The three centres already give an equilateral triangle, because

Therefore $$\triangle ABC$$ is equilateral.

Step 2 Pairs (centre, centre, other–intersection)…

• AB and their common point P
  AB = AP = BP = r ⇒ $$\triangle ABP$$ is equilateral.
• BC and Q give $$\triangle BCQ$$ equilateral.
• CA and R give $$\triangle CAR$$ equilateral.

Step 3 Triangles that use one centre and the two “other” intersection points are merely isosceles, e.g.

• $$\triangle APR$$ with AP = AR = r,
• $$\triangle BQP$$ with BQ = BP = r,
• $$\triangle CRQ$$ with CR = CQ = r.

Step 4 Finally, the three outer intersection points themselves form another equilateral triangle. A quick coordinate (or symmetry) check gives

===================================================================================

Collected results

Any one of the equal-length pairs listed for the isosceles triangles can be verified directly from the radius property quoted at the beginning.

Triangle inequality

For any triangle with side-lengths $$a, b, c$$ (take $$a \le b \le c$$) we must have

If this inequality fails, no triangle can have those side-lengths.

(i) Set: $$3\,\text{cm},\,4\,\text{cm},\,8\,\text{cm}$$

Order them: $$a = 3,\; b = 4,\; c = 8$$.

Test: $$a + b = 3 + 4 = 7$$, and $$7 \lt 8 = c$$, so $$a + b \lt c$$ — inequality fails.

Therefore no triangle has side-lengths $$3\,\text{cm}, 4\,\text{cm}, 8\,\text{cm}$$.

(ii) Set: $$2\,\text{cm},\,3\,\text{cm},\,6\,\text{cm}$$

Order them: $$a = 2,\; b = 3,\; c = 6$$.

Test: $$a + b = 2 + 3 = 5$$, and $$5 \lt 6 = c$$, so $$a + b \lt c$$ — inequality fails.

Therefore no triangle has side-lengths $$2\,\text{cm}, 3\,\text{cm}, 6\,\text{cm}$$.

Conclusion

Without any drawing we could predict the impossibility: in each triplet the sum of the two shorter sides is less than the longest side, violating the triangle inequality.

Key idea  —  the Triangle–Inequality test
For any three lengths to be the three sides of a triangle, each one must be shorter than the sum of the other two.
If we call the three lengths $$a$$, $$b$$ and $$c$$, the three required comparisons are

$$a < b + c, \; b < a + c, \; c < a + b$$

If even one of these fails, the ends of the three strips cannot meet to close up and a triangle cannot be formed.

Applying the test to the set 10 cm, 15 cm, 30 cm

• Largest length $$ = 30\text{ cm}$$.
• Check the three comparisons one by one:

$$10 < 15 + 30 = 45 \;\;\;(\text{True})$$

$$15 < 10 + 30 = 40 \;\;\;(\text{True})$$

$$30 < 10 + 15 = 25 \;\;\;(\text{False})$$

The third comparison is false (actually $$30 > 10 + 15$$), so the triangle-inequality test fails.

Conclusion
Because one comparison fails, the three strips 10 cm, 15 cm and 30 cm cannot form a triangle. The observation in the textbook (“at least two comparisons come out ‘<’, but not the third”) is exactly the symptom of a non-triangular set of lengths.

Given: We wish to know whether a triangle whose three sides are each $$50$$ units can exist. More generally, we ask the same question for any positive length $$\ell$$.

An ordinary triangle must satisfy the triangle inequality: the sum of the lengths of any two sides is greater than the length of the third side.

For an equilateral triangle every side is equal, so if each side is $$\ell$$, then

$$\ell + \ell = 2\ell \;>\; \ell \;(\text{provided } \ell>0).$$

Thus the triangle inequality is automatically satisfied whenever $$\ell>0$$. Hence there is no numerical obstruction to forming an equilateral triangle of side $$\ell$$.

Next, we show by a ruler–compass construction that such a triangle can always be drawn.

1. Draw a straight segment AB of the required length $$\ell$$ (for the first part of the question, take $$\ell = 50$$).
2. With centre A and radius $$\ell$$ draw a circle.
3. With centre B and the same radius $$\ell$$ draw another circle.
4. The two circles intersect at exactly two points (because their centres are $$\ell$$ units apart and both have radius $$\ell$$). Mark either point of intersection as C.
5. Join CA and CB.

Because C lies on the first circle, $$CA = \ell$$; because it lies on the second circle, $$CB = \ell$$. Therefore

$$AB = CA = CB = \ell,$$

so $$\triangle ABC$$ is equilateral.

This construction works for every positive length $$\ell$$, so an equilateral triangle can always be realised on the plane for any specified side length.

Specific case: when $$\ell = 50$$ the same steps give an equilateral triangle with sides $$50,\,50,\,50$$. Hence such a triangle certainly exists.

General conclusion: An equilateral triangle exists for every positive side length $$\ell$$. The only impossible case is $$\ell = 0$$, because then no triangle (having non-zero area) can be formed.

Step 1 : Recall the Triangle–Sum Property (TSP)
For every triangle, the three interior angles always add up to $$180^{\circ}$$.

Step 2 : Can all three angles be $$70^{\circ}$$?

• Assume, for a moment, that we could make a triangle whose three angles were each $$70^{\circ}$$.
• Add them: $$70^{\circ}+70^{\circ}+70^{\circ}=210^{\circ}$$.
• This contradicts the TSP because $$210^{\circ}\neq180^{\circ}$$.

Therefore it is impossible to construct a triangle in which every angle is $$70^{\circ}$$; the three lines would never close to form a triangle.

Step 3 : If two angles are $$70^{\circ}$$, what is the third?

Let the third angle be $$x^{\circ}$$. By the TSP,

$$70^{\circ}+70^{\circ}+x^{\circ}=180^{\circ}$$

Subtract $$140^{\circ}$$ from both sides:

$$x^{\circ}=180^{\circ}-140^{\circ}=40^{\circ}$$.

So the third angle must be $$40^{\circ}$$.

Step 4 : If all three angles of a triangle are equal, what must each measure be?

Let each equal angle be $$y^{\circ}$$. Then

$$y^{\circ}+y^{\circ}+y^{\circ}=180^{\circ}$$

$$3y^{\circ}=180^{\circ}$$

Divide both sides by 3:

$$y^{\circ}=60^{\circ}$$.

Conclusion :

1. A triangle with all angles $$70^{\circ}$$ cannot exist.
2. If two angles are $$70^{\circ}$$, the third angle is $$40^{\circ}$$.
3. If all three angles are to be equal, they must each be $$60^{\circ}$$ (an equilateral triangle).

Construction hint for teachers : Try to draw two rays making a $$70^{\circ}$$ angle at one vertex and two more rays making $$70^{\circ}$$ at the other intended vertices; the third pair of rays will never meet, visually confirming the algebra above.

Let the three angles of the triangle be named as usual:

• $$\angle A = 50^\circ$$   (given)
• $$\angle B$$   (unknown)
• $$\angle C$$   (unknown)

We are also told that $$\angle B = \angle C$$. Because the two angles are equal, we can give them the same algebraic symbol. Let

$$\angle B = \angle C = x^\circ$$.

Every triangle has a very important property: the sum of its three interior angles is $$180^\circ$$. We now write this fact as an equation and substitute the known and unknown angles.

$$\angle A + \angle B + \angle C = 180^\circ$$

Substitute $$\angle A = 50^\circ$$ and $$\angle B = \angle C = x^\circ$$:

$$50^\circ + x^\circ + x^\circ = 180^\circ$$

Combine like terms on the left‐hand side:

$$50^\circ + 2x^\circ = 180^\circ$$

Isolate $$x$$. First subtract $$50^\circ$$ from both sides:

$$2x^\circ = 180^\circ - 50^\circ$$

$$2x^\circ = 130^\circ$$

Now divide both sides by 2:

$$x^\circ = \frac{130^\circ}{2}$$

$$x^\circ = 65^\circ$$

Therefore,

The two equal angles are each $$65^\circ$$, and the three angles indeed add up to $$50^\circ + 65^\circ + 65^\circ = 180^\circ$$, confirming the solution.

Given data
Side lengths: $$BC = 5\,\text{cm}$$, $$AB = 6\,\text{cm}$$, $$CA = 5\,\text{cm}$$.

Required
(i) Construct $$\triangle ABC$$ with the above side lengths.
(ii) Draw the altitude from vertex A to the side $$BC$$.

Construction steps

1. Draw the base.
   Use a ruler to mark a straight line segment $$\overline{BC}$$ of length $$5\,\text{cm}$$.
2. Locate vertex A.
   
   • Place the compass point at B and take a radius of $$6\,\text{cm}$$ (the length of $$AB$$). Draw an arc above the line BC.
   • Without changing the compass opening, place the compass point at C and take a radius of $$5\,\text{cm}$$ (the length of $$CA$$). Draw another arc that intersects the first arc. Label the point of intersection as A.
3. Complete the triangle.
   Join A to B and A to C with straight-line segments. The required $$\triangle ABC$$ (with sides $$AB = 6\,\text{cm}$$, $$BC = 5\,\text{cm}$$, $$CA = 5\,\text{cm}$$) is now constructed.
4. Construct the altitude from A to BC.
   
   • With A as centre, draw an arc of any convenient radius that cuts the line BC (or its extension, if necessary) at two distinct points P and Q.
   • With P as centre and a radius that is more than half of $$PQ$$, draw an arc above BC.
   • With Q as centre and the same radius, draw another arc to intersect the previous one at R.
   • Join A to R. Let the intersection of AR with BC be D. The segment $$\overline{AD}$$ is the required altitude.

Reasoning / Proof of correctness

• Steps 1–3 place A at the unique point where $$AB = 6\,\text{cm}$$ and $$AC = 5\,\text{cm}$$ while BC is fixed at $$5\,\text{cm}$$; by the Sides–Sides–Sides (SSS) criterion, the triangle is uniquely determined.
• In Step 4, P and Q are equidistant from A. Constructing the perpendicular bisector of $$\overline{PQ}$$ through A forces line $$AR$$ to be perpendicular to $$\overline{PQ}$$. Since $$PQ$$ lies on BC, $$AR$$ is perpendicular to BC. Therefore D is the foot of the perpendicular from A, and $$\overline{AD}$$ is indeed an altitude of $$\triangle ABC$$.

Result
The triangle satisfying the given side lengths is drawn, and $$AD$$ is its altitude on the side $$BC$$.

Diagram to be drawn by the student:
Draw BC as a 5 cm horizontal line; mark B (left) and C (right). Draw intersecting arcs from B (6 cm) and C (5 cm) to locate A above BC. Connect AB and AC. With A, draw auxiliary arc cutting BC at P and Q. From P and Q draw arcs to meet at R; join AR and mark D = AR ∩ BC. Indicate right angle at D.

Given data

• $$RY = 4\,\mathrm{cm}$$
• $$TR = 7\,\mathrm{cm}$$
• $$\angle R = 140^{\circ}$$  (that is, $$\angle TRY$$)

Objective
Construct $$\triangle TRY$$ with the above measurements and then draw the altitude from T to the side RY.

Construction steps

1. Draw the base RY.
     With ruler mark a segment $$RY = 4\,\mathrm{cm}$$ on your sheet.
2. Construct the required angle $$\angle R = 140^{\circ}$$.
     Using a protractor, place its centre on R, the base line along RY and mark a point on the ray that makes $$140^{\circ}$$ with $$RY$$ (measured inside the triangle). Draw this ray and name it R X (temporary).
3. Locate vertex T on the ray.
     With the compass point on R, open it to $$7\,\mathrm{cm}$$ (length TR) and cut the ray RX. Label the intersection as T.
     Now segments $$TR = 7\,\mathrm{cm}$$ and $$RY = 4\,\mathrm{cm}$$ are in place and $$\angle R = 140^{\circ}$$ is satisfied.
4. Complete the triangle.
     Join T to Y. $$\triangle TRY$$ is obtained.
5. Draw the altitude from T to side RY.

• With centre T and a convenient radius that meets line RY, draw an arc cutting RY at points A and B.
• Keeping the compass radius greater than half of $$AB$$, draw an arc with centre A above RY. Using the same radius draw another arc with centre B to cut the previous one at C.
• Join C to T. The line TC meets RY at a point H.
    Because the construction gives $$\angle THR = 90^{\circ}$$, segment $$TH$$ is an altitude of the triangle.

Verification

• Measure $$TR$$ and $$RY$$ with a ruler – they read $$7\,\mathrm{cm}$$ and $$4\,\mathrm{cm}$$ respectively.
• Check $$\angle R$$ with a protractor – it is $$140^{\circ}$$.
• Finally, verify $$\angle THR = 90^{\circ}$$ to confirm that $$TH$$ is perpendicular to RY.

The required triangle and its altitude are now accurately constructed.

Step 1 — Draw the given side (the hypotenuse)

Draw a straight line segment $$AC = 5\,\text{cm}$$. Keep the end-points $$A$$ and $$C$$ clearly marked.

Step 2 — Locate every possible position of the right-angle vertex

Draw the semicircle whose diameter is $$AC$$ (centre at the mid-point of $$AC$$, radius $$2.5\,\text{cm}$$), placing the arc on one side of $$AC$$.

By Thales' theorem, any point $$B$$ on this semicircle (except $$A$$ and $$C$$ themselves) gives $$\angle ABC = 90^\circ$$ — the angle in a semicircle is a right angle.

Step 3 — Construct the triangle

Choose any convenient point $$B$$ on the arc, then join $$\overline{AB}$$ and $$\overline{CB}$$. Triangle $$\triangle ABC$$ has

$$\angle B = 90^\circ \quad\text{and}\quad AC = 5\,\text{cm}.$$

Why infinitely many different triangles are possible

• The only fixed data are $$\angle B = 90^\circ$$ and $$AC = 5\,\text{cm}$$.
• The other two angles must satisfy $$\angle A + \angle C = 90^\circ$$, with both being positive and acute:

$$\angle A$$ can take any value strictly between $$0^\circ$$ and $$90^\circ$$. Each choice puts $$B$$ at a different point on the semicircle and gives a different triangle (with different lengths $$AB$$ and $$BC$$).

So there is no unique triangle — infinitely many non-congruent right-angled triangles satisfy the given conditions.

Background recall

• In an equilateral triangle all three sides are equal, hence each interior angle is $$60^{\circ}$$ because the angle–sum property gives
  \[60^{\circ}+60^{\circ}+60^{\circ}=180^{\circ}.\]
• In an isosceles triangle two sides (and therefore the two base angles) are equal; the third angle can be found from the angle–sum property $$a+a+b=180^{\circ}$$.

We now test the four cases through ruler-and-compass construction and logical checking.

(A) Equilateral right-angled triangle

Step 1 Assume we wish to construct such a triangle. Draw a right angle at point $$A$$, taking $$AB$$ as one arm and $$AC$$ as the other.

Step 2 To make the triangle equilateral we would have to set $$AB=AC=BC$$. But if $$\angle BAC=90^{\circ}$$, the other two angles must satisfy

$$\angle ABC+\angle ACB=180^{\circ}-90^{\circ}=90^{\circ}.$$

Because the sides are equal, $$\angle ABC=\angle ACB$$, so each would have to be $$45^{\circ}$$, contradicting the requirement that every angle be $$60^{\circ}$$.

Conclusion: construction impossible; an equilateral triangle can never be right-angled.

(B) Equilateral obtuse-angled triangle

If one angle were obtuse, say $$>90^{\circ}$$, the sum of the three angles would exceed $$180^{\circ}$$ (because the other two angles are at least $$60^{\circ}$$ each). Hence impossible by the angle–sum property.

Conclusion: construction impossible; an equilateral triangle can never be obtuse-angled either.

(C) Isosceles right-angled triangle — construction

1. Draw a segment $$AB$$ of any convenient length. Place point $$A$$ at the left end.
2. Construct a right angle at $$A$$ by drawing a ray $$AX$$ perpendicular to $$AB$$.
3. With centre $$A$$ and radius $$AB$$, cut the ray $$AX$$ at point $$C$$.
   This ensures $$AC=AB$$.
4. Join $$BC$$ to complete $$\triangle ABC$$.

We now have $$AB=AC$$ (isosceles) and $$\angle BAC=90^{\circ}$$ (right angle). The third angle automatically becomes $$45^{\circ}$$ because
$$45^{\circ}+45^{\circ}+90^{\circ}=180^{\circ}.$$

(D) Isosceles obtuse-angled triangle — construction

1. Draw a base segment $$PQ$$ of any length.
2. At $$P$$ construct an angle of, say, $$110^{\circ}$$ with one side along $$PQ$$, producing ray $$PR$$.
3. With centre $$P$$ and any convenient radius $$r$$, mark a point $$S$$ on $$PR$$ (thus $$PS=r$$).
4. With centre $$Q$$ and the same radius $$r$$, swing an arc to meet the circle already drawn about $$P$$ at point $$S$$. Because we used the same radius, $$QS=PS$$.
5. Join $$QS$$ to obtain $$\triangle PQS$$.

The triangle has $$PS=QS$$ (isosceles) and vertex angle $$\angle SPQ=110^{\circ}>90^{\circ}$$ (obtuse). The two base angles are therefore each $$\dfrac{180^{\circ}-110^{\circ}}{2}=35^{\circ}$$, satisfying the angle–sum property.

Overall result

• Equilateral triangle: neither right-angled nor obtuse-angled is possible.
• Isosceles triangle: both right-angled and obtuse-angled constructions are possible and steps are given above.

Step 1  Label the box
Draw a rectangular box and name its vertices as in the figure below (describe the figure to the class): face ABCD is the bottom, EFGH the top, the vertical edges are AE, BF, CG and DH. Let the spider be at vertex A and the point it wants to reach be the diagonally opposite vertex G.
Denote the three edge-lengths of the box by
$$\text{length}=l, \;\text{breadth}=b, \;\text{height}=h.$$

Step 2  Why can we unfold the box?
The spider is allowed to walk only on the surfaces. If we cut the box open along a few edges and flatten the required faces on the table, the spider’s walk on the surfaces turns into a straight line in the plane. In a plane the straight line joining two points is always the shortest path, so the shortest walk on the box will be the straight line that joins the two points in some net (unfolded layout) of the box.

Step 3  Try all possible nets that keep A and G on neighbouring faces
The starting point A lies on three faces that meet at A. To reach G the spider must move across two faces that share an edge through A, because G does not lie on any single face with A.
Three different pairs of faces are possible:

• Faces ABCD and ABFE (bottom & front)
• Faces ABCD and ADHE (bottom & left)
• Faces ABFE and ADHE (front & left)

When each pair is unfolded, A and G become opposite corners of a rectangle:

Step 4  Choose the smallest of the three diagonals
The spider must take that unfolding for which the diagonal is the least. Hence the length of the required shortest path is
\[\min\bigl\{\,d_1,\;d_2,\;d_3\bigr\}.\]

Special case – a cube of side
If the box is a cube with $$l=b=h=s,$$ every diagonal is
$$\sqrt{(2s)^2+s^2}=\sqrt{5s^2}=s\sqrt 5,$$
so the spider must crawl a distance $$s\sqrt 5.$$

Step 5  How to mark the path on a real box
Open the box along one of the three pairs of adjacent faces, flatten those two faces to form a rectangle, join the two opposite corners by a straight line with a ruler, and refold the box. That straight segment becomes the shortest path on the actual 3-D box. Ask your friends to unfold their boxes in a different way; the path that gives the smallest measured length should match the distance calculated above.

Therefore, the shortest walk is obtained by unfolding two suitable adjacent faces so that the spider and the opposite corner fall at opposite ends of one straight diagonal; its length equals the smallest of $$\sqrt{(l+b)^2+h^2},\;\sqrt{(l+h)^2+b^2},\;\sqrt{(b+h)^2+l^2}.$$