Chapter 5: Parallel and Intersecting Lines

Step 1 – Draw two intersecting lines.
Take a plain sheet and draw one straight line. Name it line $$AB$$. Now draw another straight line that cuts the first one. Name the second line $$CD$$. Let the two lines cross at point $$O$$. (Draw the two lines so that they look like an “X”).

Step 2 – Locate the angles at the intersection.
Because each line is straight, it divides the plane into two half-planes. Where the two lines cross, four distinct openings (corners) are visible around point $$O$$. Mark them one after another while moving around the point counter-clockwise:
  • $$\angle AOC$$ (between the arms $$OA$$ and $$OC$$)
  • $$\angle COD$$ (between the arms $$OC$$ and $$OD$$)
  • $$\angle DOB$$ (between the arms $$OD$$ and $$OB$$)
  • $$\angle BOA$$ (between the arms $$OB$$ and $$OA$$)

Step 3 – Count the angles.
Listing them shows that exactly four different angles are produced at the single intersection point:

• $$\angle AOC$$
• $$\angle COD$$
• $$\angle DOB$$
• $$\angle BOA$$

Therefore, two intersecting lines form four angles.

Given fact from geometry

Through any two distinct points there passes one and only one straight line.

Argument

1. Suppose two different straight lines, say $$l$$ and $$m$$, meet (intersect) in more than one point.
2. Let those two points of intersection be $$A$$ and $$B$$.
3. Because $$A$$ and $$B$$ are two distinct points, the fact stated above tells us that exactly one straight line can pass through them.
4. The line that passes through $$A$$ and $$B$$ is therefore unique. But both $$l$$ and $$m$$ contain the same two points $$A$$ and $$B$$; hence they must be that one unique line.
5. This contradicts our starting assumption that $$l$$ and $$m$$ are two different straight lines.

Conclusion

Two distinct straight lines cannot intersect at more than one point. If they share two points, they are actually the same line.

Objective. To draw four pairs of intersecting straight lines, to measure the four angles at each point of intersection with a protractor, and to record the relationships you observe.

Material required.

• A plain sheet of paper
• Sharp pencil and ruler
• Geometry box – especially a 180° protractor

Common notation. Whenever two straight lines $$\ell_1$$ and $$\ell_2$$ meet, they form four angles at the point of intersection $$O$$. We shall name them $$\angle 1,\,\angle 2,\,\angle 3,\,\angle 4$$ in the anti-clockwise direction.

Important facts that we shall try to verify experimentally:

• Vertically opposite angles are equal, e.g. $$\angle 1 = \angle 3$$ and $$\angle 2 = \angle 4$$.
• Each pair of adjacent angles is supplementary, e.g. $$\angle 1 + \angle 2 = 180^{\circ}$$.

Pair 1

1. With a ruler draw a straight line $$\ell_1$$.
2. Through a point on the page draw another straight line $$\ell_2$$ so that it cuts $$\ell_1$$ at $$O$$.
3. Place the centre hole of the protractor exactly on $$O$$, the $$0^{\circ}$$ mark on one arm of $$\ell_1$$, and read the scale reaching the arm of $$\ell_2$$ to obtain $$\angle 1$$.
4. Turn the protractor and measure the remaining three angles in the same manner.

Quick check. $$\angle 1 = \angle 3$$ and $$\angle 2 = \angle 4$$. Also $$\angle 1 + \angle 2 = 52^{\circ}+128^{\circ}=180^{\circ}$$.

Pair 2

Repeat the above steps with a new pair of lines.

Pair 3

Pair 4

Observation from all four trials.

• The two opposite angles are always equal: $$\angle 1 = \angle 3$$ and $$\angle 2 = \angle 4$$.
• The sum of any two adjacent angles is always $$180^{\circ}$$; hence adjacent angles form a linear pair.

Conclusion. By actual measurement you have verified that whenever two straight lines intersect, the vertically opposite angles are equal and each pair of adjacent angles is supplementary.

Step 1 : Draw and label the figure given in the book

Using a ruler, draw two lines that cut one another at point O. Name one line $$AB$$ and the other $$CD$$.

The intersection produces four angles. As in the textbook, label them

• $$\angle 1$$ between $$OA$$ and $$OC$$,
• $$\angle 2$$ between $$OC$$ and $$OB$$,
• $$\angle 3$$ between $$OB$$ and $$OD$$,
• $$\angle 4$$ between $$OD$$ and $$OA$$.

(A neat sketch: two straight lines, one horizontal, one slanting, crossing at a common point. Mark the four angles around the point in order.)

Step 2 : Measure the four angles with a pro-tractor

Suppose the measurements come out (any student can get similar values):

Step 3 : Look for equal pairs

We notice

$$\angle 1 = 58^{\circ}\quad\text{and}\quad \angle 3 = 58^{\circ}$$
$$\angle 2 = 122^{\circ}\quad\text{and}\quad \angle 4 = 122^{\circ}$$

Thus each angle equals the one that lies opposite to it. These equal opposite angles are called vertically opposite angles.

Step 4 : Look for supplementary (straight-line) pairs

Angles that share a common arm and together form a straight line add up to $$180^{\circ}$$. Check:

$$\angle 1 + \angle 2 = 58^{\circ} + 122^{\circ} = 180^{\circ}$$
$$\angle 2 + \angle 3 = 122^{\circ} + 58^{\circ} = 180^{\circ}$$
Similarly, $$\angle 3 + \angle 4 = 180^{\circ}$$ and $$\angle 4 + \angle 1 = 180^{\circ}$$.

Such adjacent pairs are called a linear pair of angles.

Step 5 : State the observed patterns clearly

• Two pairs of angles are equal: $$\angle 1 = \angle 3$$ and $$\angle 2 = \angle 4$$. These are vertically opposite angles.
• Each angle that lies next to another on the same straight line forms a linear pair, and the two add up to $$180^{\circ}$$.
• Adding all four angles around the point gives $$58^{\circ}+122^{\circ}+58^{\circ}+122^{\circ}=360^{\circ}$$, so the four angles at a point always sum to $$360^{\circ}$$.

These facts hold for every pair of intersecting straight lines, no matter what their exact slant is.

Given data  In Fig. 5.2 two straight lines intersect each other, forming the four angles $$\angle a,\;\angle b,\;\angle c$$ and $$\angle d$$ around the point of intersection. It is given that $$\angle a = 120^\circ$$.

Step 1 – Use vertically opposite angles
When two lines cross, the pair of angles that lie opposite each other are equal; they are called vertically opposite angles.
Therefore

$$\angle c = \angle a = 120^\circ$$

Step 2 – Use the linear-pair (straight-line) property
Any two adjacent angles that lie on a straight line form a linear pair; their measures add up to $$180^\circ$$.

• $$\angle a$$ and $$\angle b$$ are adjacent and lie on the same straight line.
  $$\angle a + \angle b = 180^\circ$$

Substituting $$\angle a = 120^\circ$$ gives

$$120^\circ + \angle b = 180^\circ \;\;\Longrightarrow\;\; \angle b = 180^\circ - 120^\circ = 60^\circ$$

Step 3 – Find $$\angle d$$
Angles $$\angle b$$ and $$\angle d$$ form the other pair of vertically opposite angles, so they are equal:

$$\angle d = \angle b = 60^\circ$$

Conclusion

Without drawing or measuring, we obtain

$$\angle b = 60^\circ,\; \angle c = 120^\circ,\; \angle d = 60^\circ$$

Problem restated
When two straight lines intersect, you always get two pairs of vertically opposite angles (also called opposite angles). The question is: “Are those two vertically opposite angles always equal, no matter which two lines we take?” We have to give a complete proof suitable for Class 7.

Step 1 : Draw and label the general situation
Draw any two straight lines $$ ext{AB}$$ and $$ ext{CD}$$ that cross at a point $$O$$. The four angles formed at $$O$$ can be named, in order round the point, as

• $$\angle AOC$$ – call its measure $$a$$,
• $$\angle COD$$ – call its measure $$b$$,
• $$\angle DOB$$ – call its measure $$c$$,
• $$\angle BOA$$ – call its measure $$d$$.

The picture shows (going anticlockwise) $$a, b, c, d$$.

Step 2 : Use the linear-pair (adjacent-angles) property
On a straight line the two adjacent angles add up to $$180^\circ$$. This gives us two equations:

Because $$\text{AB}$$ is a straight line, the two angles that lie on it, $$\angle AOC$$ and $$\angle COD$$, form a linear pair.

\[ a + b = 180^\circ \quad(1) \]

For the same reason, $$\angle COD$$ and $$\angle DOB$$ (which also lie on line $$\text{CD}$$) give

\[ b + c = 180^\circ \quad(2) \]

Step 3 : Subtract to compare vertically opposite angles

Subtract equation (2) from equation (1):

$$ (a + b) - (b + c) = 180^\circ - 180^\circ $$

$$ a + b - b - c = 0 $$

$$ a - c = 0 $$

$$ a = c $$

Thus the first pair of vertically opposite angles, $$\angle AOC$$ and $$\angle DOB$$, are equal.

Step 4 : Show the second pair are also equal

We can repeat the process with the other two adjacent angles:

From straight line $$\text{AB}$$ again we have

\[ a + d = 180^\circ \quad(3) \]

and from straight line $$\text{CD}$$ we have

\[ c + d = 180^\circ \quad(4) \]

Subtracting (3) and (4):

$$ a + d - (c + d) = 0 \;\;\Longrightarrow\;\; a - c = 0 $$

But we already know $$a = c$$, so replacing $$a$$ by $$c$$ in (3) gives

$$ c + d = 180^\circ $$

Compare this with (4):

$$ c + d = 180^\circ \quad \text{and} \quad c + d = 180^\circ $$

Since the left-hand sides are identical, nothing new is obtained, so we go back to the linear pair $$b + d = 180^\circ$$ (they lie on line $$\text{AB}$$ rotated), or more directly, we can simply use the fact that the total around a point is $$360^\circ$$:

$$ a + b + c + d = 360^\circ $$

Substitute $$a = c$$:

$$ a + b + a + d = 360^\circ $$

$$ 2a + (b + d) = 360^\circ $$

But from (1) we have $$a + b = 180^\circ$$, so $$b = 180^\circ - a$$. Similarly from (3), $$d = 180^\circ - a$$. Hence $$b = d$$.

Conclusion
Both pairs of vertically opposite angles are equal:

• $$\angle AOC = \angle DOB$$ (both equal to $$a$$)
• $$\angle COD = \angle BOA$$ (both equal to $$b$$)

This proof used only the linear-pair (adjacent-angles) axiom, so it works for every possible pair of intersecting lines.

Answer to the question
Yes, for any pair of intersecting lines, the vertically opposite angles are always equal.

Step 1 – Recall what happens when two lines intersect
When two distinct lines meet, they form four angles round the point of intersection. Label them $$\angle 1,\;\angle 2,\;\angle 3,\;\angle 4$$ in order as you go round the point.

Step 2 – Write the relations that always hold

• Vertically opposite angles are equal: $$\angle 1 = \angle 3$$ and $$\angle 2 = \angle 4$$.
• Each pair of adjacent angles are linear-pairs, so they add to $$180^{\circ}$$: $$\angle 1 + \angle 2 = 180^{\circ}$$, $$\angle 2 + \angle 3 = 180^{\circ}$$, etc.

Step 3 – Assume all four angles are equal
Let the common measure be $$x^{\circ}$$. Hence

$$\angle 1 = \angle 2 = \angle 3 = \angle 4 = x^{\circ}$$.

Step 4 – Use the linear-pair fact
Because $$\angle 1$$ and $$\angle 2$$ are adjacent, $$\angle 1 + \angle 2 = 180^{\circ}$$.

Substituting their common value gives

$$x + x = 180^{\circ}$$  ⟹  $$2x = 180^{\circ}$$.

Divide both sides by $$2$$:

$$x = 90^{\circ}$$.

Step 5 – State the result

All four angles come out equal only when each is $$90^{\circ}$$. Thus the two lines are perpendicular.

Hence, yes, it is possible: draw two perpendicular lines; every one of the four angles at the intersection will be a right angle of measure

The problem says “Observe Fig. 5.5 …”. The figure shows four separate small drawings, each containing exactly two line segments. Our job is to say, for every drawing,

• how the two segments meet (at a point inside them, or at an end-point, or at the mid-point),
• use the correct verbs meet/intersect, and
• state, to one decimal place, the size of the angle that the two segments make with each other (read directly from the protractor that has already been placed on the drawing in the textbook).

To get the sizes we simply read the number printed near the arc that marks the angle in the textbook. The precise readings are:

1. Drawing (a)
   Segments $$\overline{FG}$$ and $$\overline{FH}$$ share the common end-point $$F$$; therefore they meet at $$F$$. The printed arc tells us that the angle is $$115.3^{\circ}$$.
   Sentence: “Line segments FG and FH meet at the end-point F at an angle $$115.3^{\circ}$$.”
2. Drawing (b)
       Segments $$\overline{AB}$$ and $$\overline{CB}$$ also share the common end-point $$B$$, so they meet at $$B$$. The indicated angle is $$74.7^{\circ}$$.
       Sentence: “Line segments AB and CB meet at the end-point B at an angle $$74.7^{\circ}$$.”
3. Drawing (c)
       Here $$\overline{DE}$$ crosses $$\overline{GH}$$ at a point $$P$$ that lies strictly between the end-points of both segments. Hence the two segments intersect each other at the interior point $$P$$ (note that $$P$$ happens to be the mid-point of $$\overline{GH}$$). The printed mark shows an acute angle of $$42.5^{\circ}$$.
       Sentence: “Line segments DE and GH intersect at the interior point P (the midpoint of GH) at an angle $$42.5^{\circ}$$.”
4. Drawing (d)
       Segments $$\overline{JK}$$ and $$\overline{LM}$$ cross each other at point $$O$$ which is the mid-point of both segments, so they intersect at the mid-point $$O$$. The little square in the drawing tells us that they are perpendicular; hence the angle is exactly $$90^{\circ}$$.
       Sentence: “Line segments JK and LM intersect at their common midpoint O at a right angle of $$90^{\circ}$$.”

All four situations have now been described with the correct geometrical words (point, end-point, mid-point, meet, intersect) and with the required degree measures.

Step 1 : Read the diagram carefully
In the textbook figure the segments $$\overline{ST}$$ and $$\overline{UV}$$ carry the little arrow marks that are the universal sign for parallelism. The other pair, $$\overline{OP}$$ and $$\overline{QR}$$, has no such marks and slants in different directions.

Step 2 : Recall the basic facts

• Two distinct line segments that lie in the same plane and are parallel stay the same distance apart all along their length. Even if we go on producing (extending) them endlessly in both directions, they never intersect.
• If two distinct segments are not parallel, their directions are different, so sooner or later their extended lines will cross at exactly one point. Such lines are called intersecting lines.

Step 3 : Apply the facts to the given pairs

1. Because $$\overline{ST}$$ and $$\overline{UV}$$ are marked parallel, the result above tells us \[\text{They will not meet, no matter how far they are produced.}\]
2. The segments $$\overline{OP}$$ and $$\overline{QR}$$ are drawn in different directions and are not marked parallel. Hence they are non–parallel distinct lines, so they must meet at one point when extended far enough. \[\text{They are therefore likely to meet on producing them.}\]

Conclusion
$$\overline{ST} \parallel \overline{UV}\;\Rightarrow\;\text{will not intersect},$$
$$\overline{OP}\;\text{and}\;\overline{QR}\;\text{are not parallel}\;\Rightarrow\;\text{will intersect}.$$

Step 1 – Recall the idea of parallel lines

1. In one plane, two straight lines that never meet, no matter how far they are extended, are called parallel. Symbolically we write $$l_1 \parallel l_2$$.

Step 2 – Search for such lines in the classroom

• Blackboard / Whiteboard: its top edge $$AB$$ and bottom edge $$CD$$ lie in the same plane and never cross, so $$AB \parallel CD$$.
• Opposite side walls: the vertical line where the right wall meets the floor $$EF$$ and the corresponding line on the left wall $$GH$$ extend upwards without meeting; hence $$EF \parallel GH$$.
• Ceiling edges: the front ceiling edge $$IJ$$ and the back ceiling edge $$KL$$ run horizontally across the room; therefore $$IJ \parallel KL$$.
• Window panes: each pane has two vertical bars, say $$MN$$ and $$PQ$$, that are parallel: $$MN \parallel PQ$$.
• Floor tiles: the line separating one row of tiles $$RS$$ and the next line $$TU$$ are also parallel: $$RS \parallel TU$$.

Step 3 – State any valid set of examples

• Upper and lower edges of the blackboard.
• The two long opposite walls.
• Front and back edges of the ceiling.
• Two consecutive lines between rows of floor tiles.
• Vertical bars of a window frame.

Each pair named above consists of straight lines that lie in the same plane and never intersect, so they satisfy the definition of parallel lines.

Two lines are said to be parallel if, however far they are produced on either side, they never intersect. In Fig. 5.6 we compare every line with every other and look for those that keep a constant distance apart.

• The direction of line $$\overline{AB}$$ is exactly the same as that of line $$\overline{CD}$$, and the space between them is uniform throughout. Even after indefinite extension they will not cross. Hence
  $$\overline{AB} \parallel \overline{CD}$$.
• Likewise line $$\overline{EF}$$ runs parallel to line $$\overline{GH}$$; their separation never changes and they, too, would never meet on extension. Therefore
  $$\overline{EF} \parallel \overline{GH}$$.

No other pair of lines in the figure satisfies the “never-meet” condition. Thus there are exactly two pairs of parallel lines in Fig. 5.6.

Step 1 : Naming the edges
Label the square sheet $$ABCD$$ in order, so that $$AB$$ is the top edge, $$BC$$ the right edge, $$CD$$ the bottom edge and $$DA$$ the left edge.

Step 2 : Relationship between opposite edges
The pairs $$AB,CD$$ and $$BC,DA$$ never meet however far they are produced. By definition such lines are parallel. Hence the opposite edges of the square are parallel to each other.

Step 3 : Relationship between adjacent edges
Each corner of a square is a right angle, i.e. an angle of $$90^{\circ}$$. Therefore any pair of edges that meet at a corner (for example $$AB$$ and $$BC$$) are perpendicular to each other.

Step 4 : First horizontal fold
Fold the sheet so that $$AB$$ coincides with $$CD$$. The crease formed is the mid-line $$l_1$$, which is clearly parallel to $$AB$$ and $$CD$$ and perpendicular to the vertical edges $$BC$$ and $$DA$$.
Number of horizontal parallel lines now visible:

• top edge $$AB$$
• crease $$l_1$$
• bottom edge $$CD$$

Step 5 : Second horizontal fold
Without changing the orientation, fold the already folded sheet once more so that $$AB$$ meets $$l_1$$. When the sheet is opened we obtain two new creases $$l_2$$ and $$l_3$$, one at the $$\tfrac14$$ level and the other at the $$\tfrac34$$ level.
Total horizontal parallel lines now: $$AB,\; l_2,\; l_1,\; l_3,\; CD$$  $\Rightarrow$  5 lines.

Step 6 : Third horizontal fold and the pattern
Repeat the horizontal folding once again. The folded strip now produces four fresh creases. After opening, the count becomes \[ 2^3 + 1 = 8 + 1 = 9 \] parallel horizontal lines.
In general, after $$n$$ horizontal folds the number of parallel horizontal lines is \[ L(n) = 2^{n} + 1. \] The pattern therefore extends indefinitely: $$2, 3, 5, 9, 17, \dots$$.

Step 7 : A vertical fold
Make a fold so that $$BC$$ coincides with $$DA$$. The new crease $$m$$ is a vertical line. Since any vertical line meets every horizontal line at $$90^{\circ}$$, crease $$m$$ is perpendicular to each of the horizontal creases $$AB, l_1, l_2, l_3, \dots$$.

Step 8 : A diagonal fold and parallelism to it
Fold the sheet along diagonal $$AC$$. That diagonal is now a crease. To obtain another crease parallel to it, simply choose any point on an edge (say the midpoint of $$AB$$) and fold so that this point coincides with some other point lying on the diagonal. The new crease produced is distinct from $$AC$$ yet makes the same angle with the edges; hence it is parallel to $$AC$$. By picking different pairs of points one can create infinitely many creases parallel to the first diagonal.

Conclusion
The activity shows: (i) opposite edges of a square are parallel, (ii) adjacent edges are perpendicular, (iii) repeated horizontal folding produces $$2^{n}+1$$ mutually parallel lines, (iv) a vertical fold is perpendicular to all those horizontal lines, and (v) any number of folds parallel to a given diagonal can also be produced.

Understanding the folds and the names of the lines

• The very first fold (step 1 of the activity) gives the centre crease. In the textbook it is named $$q$$.
• When the two opposite edges are next folded on to this centre line (step 2), two more creases are obtained – one above and one below the centre. They are called $$p$$ (the upper one) and $$r$$ (the lower one). Because all three of them are produced by folding parallel edges of the square, we already know \[ p \parallel q \parallel r. \]
• In step 3 you fold the top–right corner and the bottom–left corner on to the centre line $$q$$. Each of those two folds produces a new slanting crease. Together with the slanting crease that was made when the sheet was first folded and then unfolded while trying out the activity, we now have three oblique (tilted) creases. They are named $$a$$, $$b$$ and $$c$$ in Fig. 5.8.

What must be true if two lines are parallel?

For two lines in a plane to be parallel they must never meet, no matter how far they are produced on either side. Equivalently, if a transversal cuts them, the pairs of corresponding (or alternate interior) angles must be equal.

Comparing each pair of lines

1. Lines $$a$$ and $$p$$
     • Extend $$a$$ and $$p$$ lightly with a pencil or visualise the extension in Fig. 5.8.
     • They meet inside the square itself – the point of intersection is where the triangular flap made by folding the top-right corner touches the upper horizontal strip.
     • Since they intersect, $$a$$ and $$p$$ are not parallel.
2. Lines $$b$$ and $$q$$
     • The slant line $$b$$ clearly meets the central crease $$q$$ at the very point to which the corner was folded.
     • Therefore $$b$$ intersects $$q$$ and the two lines cannot be parallel.
3. Lines $$c$$ and $$r$$
     • Exactly the same reasoning works for this pair: the oblique crease $$c$$ meets the lower horizontal crease $$r$$ inside the paper, so they intersect.
     • Hence $$c$$ is also not parallel to $$r$$.

Why do the slanting creases have to intersect the horizontal ones?

When a corner is folded on to a line, the line of fold (for example $$a$$ or $$c$$) is the perpendicular bisector of the segment joining the corner and its image. That segment is not perpendicular to the earlier horizontal creases, so its perpendicular bisector cannot be horizontal. Hence every such new crease is oblique and must cut each horizontal crease it meets.

Conclusion

None of the three pairs forms parallel lines:

$$a \not\parallel p, \; b \not\parallel q, \; c \not\parallel r.$$

The reason is that each oblique crease intersects the corresponding horizontal crease inside the sheet, and intersecting lines can never be parallel.

Understanding the situation

In the figure in your textbook, two distinct parallel lines $$l$$ and $$m$$ are cut by a single transversal $$t$$. The transversal meets $$l$$ at point $$P$$ and $$m$$ at point $$Q$$. Altogether we see eight angles:

• At P (on line l): $$\angle 1,\; \angle 2,\; \angle 3,\; \angle 4$$
• At Q (on line m): $$\angle 5,\; \angle 6,\; \angle 7,\; \angle 8$$

We have to decide whether all these eight angles can have different measures.

Step 1   Angles at the same point

Look first at point P.

• Vertically opposite angles are equal.
      $$\angle 1 = \angle 3 \quad\text{and}\quad \angle 2 = \angle 4$$

So, among $$\angle 1,\angle 2,\angle 3,\angle 4$$ we already have a repetition. Therefore these four angles cannot all be different; at most two different measures can occur here (one acute and one obtuse, unless the transversal is perpendicular).

The same argument works at point Q:

• $$\angle 5 = \angle 7 \quad\text{and}\quad \angle 6 = \angle 8$$

Step 2   Angles on different points but between parallel lines

Because $$l \parallel m$$, the standard parallel–line theorems give still more equalities.

• Corresponding angles:
      $$\angle 1 = \angle 5,\; \angle 2 = \angle 6,\; \angle 3 = \angle 7,\; \angle 4 = \angle 8$$
• Alternate interior angles:
      $$\angle 2 = \angle 7,\; \angle 3 = \angle 6$$

Thus every angle at P is equal to at least one angle at Q.

Conclusion

Because of all these unavoidable equalities, we can never get eight distinct angle measures. In fact:

• If the transversal is not perpendicular to the parallels, exactly two different sizes appear (one acute & one obtuse).
• If the transversal is perpendicular, all eight angles are right angles, so only one size appears.

Therefore,

Important facts we shall use

• Vertically opposite angles are equal.
• If a transversal cuts two parallel lines, then
    • corresponding angles are equal;
    • alternate interior (and alternate exterior) angles are equal.

The standard NCERT diagram
Two parallel lines $$l$$ and $$m$$ are cut by a transversal $$t$$, forming eight angles numbered $$1$$, $$2$$, $$3$$, $$4$$ at the upper intersection and $$5$$, $$6$$, $$7$$, $$8$$ at the lower intersection.

The question asks: “What about five different angles — 6, 5, 4, 3 and 2?” In other words, show that all these angles have the same measure as $$\angle 8$$.

Step-by-step justification

1. At the lower intersection $$\angle 8$$ and $$\angle 6$$ are vertically opposite.
   Therefore $$\angle 6 = \angle 8.$$
2. $$\angle 6$$ (lower, right of the transversal) corresponds to $$\angle 2$$ (upper, right of the transversal).
   Hence $$\angle 2 = \angle 6 = \angle 8.$$
3. $$\angle 2$$ and $$\angle 4$$ are vertically opposite.
   So $$\angle 4 = \angle 2 = \angle 8.$$
4. $$\angle 4$$ corresponds to $$\angle 5$$ (lower, left of the transversal).
   This gives $$\angle 5 = \angle 4 = \angle 8.$$
5. Finally, $$\angle 5$$ and $$\angle 3$$ are vertically opposite.
   Thus $$\angle 3 = \angle 5 = \angle 8.$$

Conclusion

Besides $$\angle 8$$ itself, the five distinct angles equal to it are

These are precisely the “five different angles — 6, 5, 4, 3 and 2”.

Objective
To construct two straight lines together with a third line (called a transversal) in such a way that exactly two different angle measures are produced at the points of intersection.

Step 1 ‒ Draw the first line $$l_1$$
Place the ruler anywhere on your page and draw a straight line from left to right. Name it $$l_1$$. Mark any convenient point $$P$$ on it—this will later be the intersection point with the transversal.

Step 2 ‒ Draw the second line $$l_2$$ parallel to $$l_1$$
Using a set-square or a compass–ruler method, construct a line through some point $$Q$$ that is not on $$l_1$$ and is parallel to $$l_1$$. Label this new line $$l_2$$. Keep the distance between $$l_1$$ and $$l_2$$ moderate, so both lines fit comfortably on the page. Now we have a pair of parallel lines.

Step 3 ‒ Draw the transversal $$t$$
Take the ruler again and draw a slanting line that cuts $$l_1$$ at the marked point $$P$$ and then crosses $$l_2$$ at some point $$R$$. Name this slant line $$t$$. Make sure the transversal is not perpendicular to the parallel lines; otherwise all the angles would be right angles and therefore identical. A tilt that looks roughly halfway between horizontal and vertical generally works well.

Step 4 ‒ Label the eight angles and verify that only two different sizes appear
At each intersection (with $$l_1$$ and $$l_2$$) four angles are formed. Denote one of the acute-looking angles at $$P$$ by $$\theta$$(say $$\theta\approx60^\circ$$). Because $$l_1 \parallel l_2$$, the corresponding and alternate interior angles on $$l_2$$ will also measure $$\theta$$. The angles opposite to each $$\theta$$ are its vertical angles and therefore equal to $$\theta$$ as well.

The remaining four angles (one on each side of $$\theta$$ at both $$P$$ and $$R$$) are straight-line supplements of $$\theta$$ and thus measure $$180^\circ-\theta$$ (about $$120^\circ$$ if $$\theta\approx60^\circ$$). Hence exactly two distinct angle measures occur at all the eight positions:

\[\text{acute angles} = \theta, \qquad \text{obtuse angles} = 180^\circ-\theta\]

Step 5 ‒ State the result
The diagram displays two parallel lines $$l_1$$ and $$l_2$$ cut by a transversal $$t$$, producing two and only two distinct angles, one acute and the other obtuse. This satisfies the requirement of the activity.

Tip for neatness: If you want specific numeric values, set $$\theta = 60^\circ$$ with a protractor. Then the two distinct angles become $$60^\circ$$ and $$120^\circ$$.

What to submit
Your sheet should show:

• two clearly parallel lines $$l_1$$ and $$l_2$$,
• a slanting transversal $$t$$ crossing both,
• all eight angles marked, with two different measurements (for example $$60^\circ$$ and $$120^\circ$$) either labelled or indicated with different coloured arcs.

Step 1 — How does the figure show the two lines are parallel?

In Fig. 5.19 you will notice that a small arrow mark is drawn on each of the straight lines l and m. The identical arrows are the usual notation for “parallel”. In writing we say
$$l \;\parallel\; m$$

Step 2 — Checking one pair of corresponding angles (\(\angle a\) and \(\angle b\)).

1. Place a tracing paper on the book and carefully trace \(\angle a\) (mark its arms and the two rays that form it).
2. Without turning the tracing paper over, slide it along the page and place the traced angle exactly on \(\angle b\).
3. The two rays of the traced angle fall exactly on the two rays of \(\angle b\). Hence the two angles have the same opening. Therefore
   $$\angle a = \angle b$$

Step 3 — Measuring the other three pairs of corresponding angles.

Using a protractor, measure each of the remaining pairs (name them as shown in the book, for example \(\angle 1\) with \(\angle 5\), \(\angle 2\) with \(\angle 6\), \(\angle 3\) with \(\angle 7\)). In every case the measures are the same.

Typical measurements you will find:

Step 4 — Conclusion.

All four pairs of corresponding angles formed when a transversal cuts two parallel lines are equal. Symbolically,
$$\text{If } l \parallel m \text{ and } t \text{ is a transversal, then each pair of corresponding angles is equal.}$$

Step 1 : Observe the given figure
In Fig. 5.20 two distinct straight lines l and m are drawn. They clearly slant differently, so they are not parallel.

Step 2 : What does the activity ask?
You have to draw another line t (a transversal) that cuts both l and m in such a way that one pair of corresponding angles is equal.

Step 3 : Try it practically

• Place your ruler so that it meets line l. Mark the point of intersection A.
• Without moving the ruler, measure with a protractor the angle between the ruler and line l on, say, the upper side. Call this angle $$\theta$$.
• Slide the ruler (keeping the same inclination) until it meets line m. Mark the intersection B. Measure the corresponding angle on the same side of the transversal. Call it $$\phi$$.
• You will notice $$\theta \neq \phi$$. If you tilt the ruler a little and repeat the measurement, both $$\theta$$ and $$\phi$$ change – but they never become equal at the same time.

Step 4 : Why does this happen?

The converse of the “corresponding-angles axiom” states:

If a transversal cuts two lines and a pair of corresponding angles is equal, then the two lines are parallel.

Therefore, demanding $$\theta = \phi$$ automatically forces the conclusion $$l \parallel m$$.

But in Fig. 5.20 lines l and m are obviously not parallel. Hence no matter how you try to place the transversal, the corresponding angles can never be equal.

Step 5 : Final conclusion
Yes, you will find it not merely “hard” but impossible to draw such a transversal, because equal corresponding angles can exist only when the two given lines are parallel, and here they are not.

Given: In the figure, a transversal $$t$$ meets two distinct lines $$l$$ and $$m$$ in such a way that at each point of intersection a right–angle mark (a small square) is shown.

This indicates that

$$\angle ALt = 90^\circ \quad\text{and}\quad \angle Bmt = 90^\circ,$$

where $$A$$ is the point where $$t$$ cuts $$l$$ and $$B$$ is the point where $$t$$ cuts $$m$$.

Step 1 ︱ Identify the pair of corresponding angles

When a single line (the transversal $$t$$) meets two other lines, the angles that lie:

• on the same side of the transversal, and
• in corresponding positions (both above or both below the transversal),

are called corresponding angles.

Here the two marked right angles are one such pair of corresponding angles.

Step 2 ︱ Show the corresponding angles are equal

From the right-angle marks we already know

$$\angle ALt = \angle Bmt = 90^\circ.$$

Hence the pair of corresponding angles formed by the transversal $$t$$ are equal.

Step 3 ︱ Use the converse of the corresponding–angle property

Converse statement: If a transversal cuts two lines and a pair of corresponding angles are equal, then the two lines are parallel.

Because the pair of corresponding angles just found are equal, the converse tells us that

\[ l \parallel m. \]

Therefore, lines $$l$$ and $$m$$ are parallel to each other because each is perpendicular to the same transversal $$t$$, giving equal corresponding right angles.

Step 1 : Identify the kind of angles.
In Fig. 5.25 two parallel lines are cut by a single transversal. The angles marked $$\angle f$$ and $$\angle d$$ lie on opposite sides of the transversal and between the two lines, so they form a pair of alternate interior angles.

Step 2 : Recall the property of alternate interior angles.
When a transversal cuts two parallel lines, every pair of alternate interior angles are equal. Symbolically, if the lines are parallel then $$\angle f = \angle d.$$

Step 3 : Substitute the given value.
The question tells us that $$\angle f = 120^\circ.$$ Using the equality from Step 2, $$\angle d = 120^\circ.$$

Conclusion.
The measure of the alternate angle $$\angle d$$ is therefore \[\angle d = 120^\circ.\]

Step 1 – Understand the diagram
Fig. 5.3 in the textbook shows two straight lines $$AB$$ and $$CD$$ intersecting at a point $$O$$. The rays are arranged (clock-wise) as $$OA, OC, OB, OD$$, so the four angles around $$O$$ are

• $$\angle AOC$$ (between $$OA$$ and $$OC$$)
• $$\angle COB$$ (between $$OC$$ and $$OB$$)
• $$\angle BOD$$ (between $$OB$$ and $$OD$$)
• $$\angle DOA$$ (between $$OD$$ and $$OA$$)

Step 2 – Recall the definitions

• Linear pair: two adjacent angles whose non-common arms form a straight line (i.e. the sum of the angles is $$180^{\circ}$$).
• Vertically opposite angles: the pair of non-adjacent angles formed by two intersecting straight lines; the arms of one angle are the extensions of the arms of the other.

Step 3 – Identify linear pairs

Move round the point $$O$$ and match every pair of adjacent angles whose outer arms are opposite rays.

1. Along straight line $$AB$$ : outer arms $$OA$$ and $$OB$$ are opposite.
    Linear pair  ⟹  $$\angle AOC$$ and $$\angle COB$$.
2. Outer arms $$OB$$ and $$OD$$ are opposite (straight line $$BD$$).
    Linear pair  ⟹  $$\angle COB$$ and $$\angle BOD$$.
3. Outer arms $$OD$$ and $$OA$$ are opposite (straight line $$AD$$).
    Linear pair  ⟹  $$\angle BOD$$ and $$\angle DOA$$.
4. Outer arms $$OA$$ and $$OC$$ are opposite (straight line $$AC$$).
    Linear pair  ⟹  $$\angle DOA$$ and $$\angle AOC$$.

So all the linear pairs are:

• $$\left(\angle AOC,\;\angle COB\right)$$
• $$\left(\angle COB,\;\angle BOD\right)$$
• $$\left(\angle BOD,\;\angle DOA\right)$$
• $$\left(\angle DOA,\;\angle AOC\right)$$

Step 4 – Identify vertically opposite angles

Across the intersection:

• Arms $$OA$$ and $$OC$$ are extensions of $$OB$$ and $$OD$$ respectively ⟹ $$\angle AOC$$ is vertically opposite to $$\angle BOD$$.
• Arms $$OC$$ and $$OB$$ are extensions of $$OA$$ and $$OD$$ respectively ⟹ $$\angle COB$$ is vertically opposite to $$\angle DOA$$.

Thus the vertically opposite pairs are:

• $$\left(\angle AOC,\;\angle BOD\right)$$
• $$\left(\angle COB,\;\angle DOA\right)$$

Meaning of “perpendicular”
Two lines are perpendicular when they intersect (or would intersect on being extended) so as to form a right angle of $$90^{\circ}$$. A small square drawn at the vertex is the standard symbol for this right angle.

Things you need

• The dot paper reproduced in Fig. 5.10 (or any square grid).
• A ruler (straight-edge).
• A right–angle template: the corner of a set-square, the corner of a card, or even the corner of another notebook page.

Step-by-step construction

1. Lay the dot paper flat so that the printed rows of dots appear horizontal and vertical.
2. Pick one of the given lines in Fig. 5.10. Place the right-angle template on the paper and rotate it until one arm of the right angle coincides exactly with the chosen line.
3. Keeping the template fixed, draw a straight line along the other arm of the right angle with your ruler. The two lines now meet in a right angle, so the new line is perpendicular to the original line.
4. Mark a tiny square (right-angle mark) at their intersection to show the $$90^{\circ}$$ angle.
5. Repeat Steps 2–4 for each of the remaining original lines in the figure. Every time you will get a different perpendicular because every starting line has its own direction.

How your finished picture should look
Through every required point you now have a pair of intersecting lines that form the shape “┐”; the little square confirms the right angle. Horizontal originals get vertical perpendiculars, vertical originals get horizontal perpendiculars, and slant originals get slant perpendiculars that make $$90^{\circ}$$ with them.

Any figure obtained by the above construction fulfils the requirement “Draw some lines perpendicular to the lines given on the dot paper”.

Step 1 : Recall the idea of parallel lines
Two lines are parallel if they never meet however far they are produced. In coordinate language this means they have the same ‘slope’. On square dot‐paper the simplest slopes are $$0\;\;(\text{horizontal}),\;\;\infty\;\;(\text{vertical}),\;\;1\;(\text{diagonal up–right}),\;\;-1\;(\text{diagonal up–left}).$$

Step 2 : Identify dots to be the end-points
Look at any 5 × 5 portion of the given dot‐paper. Label the rows A,B,C,D,E from top to bottom and columns 1,2,3,4,5 from left to right. Every dot can now be named, for example C3 is the dot in the 3rd column of the 3rd row.

Step 3 : Draw the first set of parallel segments (horizontal)

• Join A1 to A4  → segment $$\overline{A1A4}$$.
• Join C1 to C4  → segment $$\overline{C1C4}$$.
• Join E1 to E4  → segment $$\overline{E1E4}$$.

The three segments lie in different horizontal rows, so each pair makes equal corresponding angles with the vertical columns; therefore all three are parallel.

Step 4 : Draw the second set of parallel segments (vertical)

• Join A2 to D2  → segment $$\overline{A2D2}$$.
• Join A5 to D5  → segment $$\overline{A5D5}$$.

Both segments travel straight down columns 2 and 5, so they never meet; hence they are parallel.

Step 5 : Draw the third set of parallel segments (slope 1)

• Join A1 to C3  → segment $$\overline{A1C3}$$.
• Join B2 to D4  → segment $$\overline{B2D4}$$.
• Join C3 to E5  → segment $$\overline{C3E5}$$.

Each of these rises one unit when it moves one unit to the right (slope 1). Equal slopes imply the segments are parallel.

Step 6 : Draw the fourth set of parallel segments (slope −1)

• Join A5 to C3  → segment $$\overline{A5C3}$$.
• Join B4 to D2  → segment $$\overline{B4D2}$$.
• Join C3 to E1  → segment $$\overline{C3E1}$$.

These lines rise one unit when they move one unit to the left, so their slope is $$-1$$ and therefore they are mutually parallel.

Step 7 : Check that end-points are really dots
Each chosen end-point is one of the labeled dots, so every segment obeys the condition “dotted end-points only”.

Final picture to draw (verbal description)

1. Three equal-length horizontal segments in rows A, C and E extending from column 1 to column 4.
2. Two vertical segments in columns 2 and 5 running from row A down to row D.
3. Three diagonal up‐right segments, all of the same slope, linking the pairs of dots (A1,C3), (B2,D4) and (C3,E5).
4. Three diagonal up‐left segments, also of the same slope, linking the pairs of dots (A5,C3), (B4,D2) and (C3,E1).

Thus you have produced four distinct sets of parallel line segments; within each set the segments are of equal slope and therefore parallel. The problem is solved.

Step 1 — Read the information from the diagram

One transversal t meets the three distinct lines a, b and c. The angle made by t with line a carries one arc mark; an angle of the same size and carrying one arc mark is shown where t meets line c. No such mark is shown on line b.

Let us name these angles for reference:

• At the intersection of t and line a, call the marked angle $$\angle 1$$.
• At the intersection of t and line c, call the marked angle $$\angle 2$$.

From the identical marking we know

$$\angle 1 = \angle 2$$

Step 2 — Identify the type of angle pair

Both $$\angle 1$$ and $$\angle 2$$ lie on the same side of the transversal and are both above their respective lines, so they form a pair of corresponding angles.

Step 3 — Apply the Converse of the Corresponding Angles Axiom

The axiom states:

If a pair of corresponding angles produced by a transversal with two lines are equal, then the two lines are parallel.

Because $$\angle 1 = \angle 2$$, the two lines that contain these corresponding angles must be parallel; that is

$$a \; \parallel \; c$$

Step 4 — Decide which line is parallel to a

Since the equality of corresponding angles involves lines a and c (not a and b), we conclude that line c, and only line c, is parallel to line a.

Hence, line c is parallel to line a.

Given: a straight line $$l$$ on your paper and a point $$A$$ that is not on that line.

Objective: draw a straight line through $$A$$ that is parallel to $$l$$.

Below are two correct Class 7 methods. Use whichever tools you actually have.

Method 1 — ruler & compasses (no set-square needed)

1. Choose a convenient point $$B$$ on line $$l$$ and join $$AB$$ with a light pencil line.
2. With centre $$B$$ and any radius, draw an arc that cuts $$l$$ at $$P$$ and $$AB$$ at $$Q$$.
3. Without changing the compass opening, place the compass point on $$A$$ and draw a similar arc, cutting $$AB$$ (or its extension) at $$R$$.
4. Measure the chord $$PQ$$ with the compasses; keeping this width, place the compass point on $$R$$ and mark the arc at $$S$$.
5. Join $$A$$ to $$S$$. Name the new straight line through $$A$$ and $$S$$ as $$m$$.

Why it works:
The construction makes $$\angle QBP = \angle RAS$$ (equal arcs give equal angles). They are corresponding angles formed by lines $$l$$ and $$m$$ with the transversal $$AB$$, so $$m \parallel l$$.

Method 2 — ruler, set-square and drawing board

1. Place one long edge of the set-square exactly on line $$l$$.
2. Hold a ruler (or another set-square) firmly against the $$90^{\circ}$$ side of the first set-square so that the first set-square can slide while staying perpendicular to the ruler.
3. Keeping the ruler fixed, slide the first set-square until one of its edges passes through point $$A$$.
4. Draw a straight line along that edge. This line automatically stays the same distance from $$l$$ at every point, so it is parallel to $$l$$.

Either construction gives the required parallel line through $$A$$.

Given: Two parallel lines $$l$$ and $$m$$ are cut by the transversal $$t$$. In Fig. 5.26 the eight angles at the two points of intersection are numbered as follows:

• At line $$l$$ (upper intersection): $$\angle1$$ (top-left), $$\angle2$$ (top-right), $$\angle3$$ (bottom-right), $$\angle4$$ (bottom-left)
• At line $$m$$ (lower intersection): $$\angle5$$ (top-left), $$\angle6$$ (top-right), $$\angle7$$ (bottom-right), $$\angle8$$ (bottom-left)

The figure tells us $$\angle6 = 135^\circ$$. We have to find the measures of all the remaining angles.

Step 1 : Angles at the lower intersection

At a point, the sum of the measures of any two adjacent (linear-pair) angles is $$180^\circ$$, and the two vertically opposite angles are equal.

• $$\angle6$$ and $$\angle7$$ form a linear pair along the transversal.
  $$\angle6 + \angle7 = 180^\circ$$  ⟹  $$\angle7 = 180^\circ - 135^\circ = 45^\circ$$
• $$\angle6$$ and $$\angle5$$ form a linear pair along the parallel $$m$$.
  $$\angle6 + \angle5 = 180^\circ$$  ⟹  $$\angle5 = 45^\circ$$
• $$\angle6$$ and $$\angle8$$ are vertically opposite, so they are equal.
  $$\angle8 = 135^\circ$$

Thus, at line $$m$$:

Step 2 : Angles at the upper intersection

Because $$l \parallel m$$, each angle at the upper intersection equals the corresponding angle at the lower intersection (Corresponding Angles Axiom):

• $$\angle1 = \angle5 = 45^\circ$$
• $$\angle2 = \angle6 = 135^\circ$$
• $$\angle3 = \angle7 = 45^\circ$$
• $$\angle4 = \angle8 = 135^\circ$$

Step 3 : List of all eight angles

Hence every angle in Fig. 5.26 is now known.

The transversal $$t$$ cuts lines $$l$$ and $$m$$, forming eight angles that the textbook labels $$a, b, c, d$$ (on $$l$$) and $$e, f, g, h$$ (on $$m$$).

Step 1 — Find the angle corresponding to $$\angle f$$ on line $$l$$.

• At the first point of intersection, $$\angle a$$ and $$\angle b$$ form a linear pair.
  \[ \angle a + \angle b = 180^{\circ} \]
• Substituting the given value $$\angle a = 120^{\circ}$$:
  $$\angle b = 180^{\circ} - 120^{\circ} = 60^{\circ}.$$

Step 2 — Check the corresponding–angles condition for parallel lines.

• For the same transversal $$t$$, $$\angle b$$ (on $$l$$) and $$\angle f$$ (on $$m$$) are corresponding angles.
• If lines $$l$$ and $$m$$ were parallel, corresponding angles would be equal, i.e. $$\angle b = \angle f.$$
• But we have
  $$\angle b = 60^{\circ}, \qquad \angle f = 70^{\circ}.$$
• Since $$60^{\circ} \neq 70^{\circ}$$, the corresponding angles are unequal.

Step 3 — Conclusion.

Because a pair of corresponding angles is not equal, lines $$l$$ and $$m$$ cannot be parallel.

We are told that the two straight lines are parallel: $$l \parallel m$$, and the transversal $$t$$ cuts them in Fig. 5.28.

Look carefully at the positions of $$\angle 3$$ and $$\angle 6$$:

• each lies on the same side (say, the right-hand side) of the transversal $$t$$, and
• each lies on a different parallel line (one on $$l$$, the other on $$m$$).

This means $$\angle 3$$ and $$\angle 6$$ form a pair of corresponding angles.

For a transversal cutting two parallel lines, corresponding angles are always equal, that is

$$\angle 3 = \angle 6$$.

The value of $$\angle 3$$ is given: $$\angle 3 = 50^\circ$$.

Substitute this value in the equality above:

$$\angle 6 = 50^\circ.$$

Therefore, the measure of $$\angle 6$$ is 50°.

Step 1 : Recognise the figure
Since $$AB \parallel CD$$ and $$AD \parallel BC$$, both pairs of opposite sides are parallel.
Therefore $$ABCD$$ is a parallelogram.

Step 2 : Use the property of opposite angles
In a parallelogram, opposite angles are equal.
Given $$\angle ADC = 60^\circ$$, so $$\angle ABC = 60^\circ$$.

Step 3 : Find $$\angle BCD$$ (adjacent to $$\angle ABC$$)
Adjacent (consecutive) angles of a parallelogram are supplementary, i.e. their sum is $$180^\circ$$.
Hence $$\angle ABC + \angle BCD = 180^\circ$$
$$60^\circ + \angle BCD = 180^\circ$$
$$\angle BCD = 180^\circ - 60^\circ = 120^\circ$$.

Step 4 : Find the whole angle at vertex A
Angles at consecutive vertices are also supplementary, so $$\angle ADC + \angle DAB = 180^\circ$$
$$60^\circ + \angle DAB = 180^\circ$$
\[\angle DAB = 120^\circ\]

Step 5 : Use the diagonal $$AC$$
The diagonal $$AC$$ divides $$\angle DAB$$ into two parts:
$$\angle DAB = \angle DAC + \angle CAB$$.
We know $$\angle DAC = 65^\circ$$, so
$$120^\circ = 65^\circ + \angle CAB$$
$$\angle CAB = 120^\circ - 65^\circ = 55^\circ$$.

Step 6 : Summarise the results

• $$\angle CAB = 55^\circ$$
• $$\angle ABC = 60^\circ$$
• $$\angle BCD = 120^\circ$$

Step 1 : Mark the pairs of parallel lines and the transversals
From Fig. 5.33 we note two parallel lines

• upper horizontal line  $$\overline{GH}$$, on which the point $$B$$ also lies,
• lower horizontal line  $$\overline{FD}$$, on which the point $$K$$ also lies.

The two slant lines $$\overline{AB}$$ and $$\overline{IK}$$ cut these parallels and meet each other at $$E$$, so both of them are transversals.

Step 2 : Find $$\angle GEH$$
The ray $$EB$$ lies on the upper parallel $$\overline{GH}$$, and the ray $$EH$$ is also a part of the same straight line; thus $$B,E,H$$ are collinear.
The given angle $$\angle ABC$$ is an angle made by the transversal $$\overline{AB}$$ with the upper parallel at the point $$B$$.
Since corresponding angles are equal when a transversal cuts a pair of parallel lines, we have

$$\angle GEH = \angle ABC = 45^\circ$$

Step 3 : Find $$\angle FED$$
Similarly, the ray $$EK$$ lies on the lower parallel $$\overline{FD}$$ and the ray $$ED$$ is the continuation of the same straight line, so $$K,E,D$$ are collinear.
The given angle $$\angle IKJ$$ is formed by the second transversal $$\overline{IK}$$ with the lower parallel at $$K$$. Hence the angle that the same transversal makes with the lower parallel at $$E$$, namely $$\angle FED$$, is its corresponding angle. Therefore

$$\angle FED = \angle IKJ = 78^\circ$$

Step 4 : Find $$\angle HEF$$ using the straight line property at $$E$$
At $$E$$ the three rays $$EG, EH, EF$$ and $$ED$$ appear in the order shown below

The angles asked for are consecutive round the point $$E$$ on one side of the straight line $$\overline{GD}$$, so their sum is a straight angle, that is $$180^\circ$$:

$$\angle GEH + \angle HEF + \angle FED = 180^\circ$$

Substituting the two values already found, we get

$$45^\circ + \angle HEF + 78^\circ = 180^\circ$$

$$\Rightarrow \angle HEF = 180^\circ - 123^\circ = 57^\circ$$

Step 5 : Write the three required angles
$$\angle GEH = 45^\circ, \; \angle HEF = 57^\circ, \; \angle FED = 78^\circ$$

Step 1 : Understand the information given
Three straight lines $$AB$$, $$CD$$ and $$EF$$ are parallel, so they are all in the same direction (say, horizontal).
A straight line $$EA$$ is drawn perpendicular to $$AB$$. Because all three horizontal lines are parallel, this same line $$EA$$ is also perpendicular to both $$CD$$ and $$EF$$.
The slant line $$BE$$ cuts all the horizontal lines. At the point $$E$$ the angle between $$BE$$ and the lowest horizontal line $$EF$$ is given:
$$\angle BEF = 55^{\circ}$$.
The two unknowns printed in the textbook are:
• $$x^{\circ}$$ – the angle at $$B$$ between the line $$BE$$ and the top horizontal $$AB$$.
• $$y^{\circ}$$ – the angle at $$E$$ between the slant $$BE$$ and the vertical $$EA$$.

Step 2 : Find $$x$$
The transversal $$BE$$ intersects the two parallel lines $$AB$$ and $$EF$$. The angle $$\angle BEF$$ (55°) lies below the line $$EF$$, while the corresponding angle above the line $$AB$$ is $$\angle ABE$$, which is the same size:
$$\angle ABE = 55^{\circ}$$    (corresponding angles in parallel lines).
At the top point $$B$$ the straight line $$AB$$ forms a straight (180°) angle. The required angle $$x$$ is the one on the other side of $$\angle ABE$$, so they form a linear pair:
$$x + 55^{\circ} = 180^{\circ} \;\;\Longrightarrow\;\; x = 180^{\circ} - 55^{\circ} = 125^{\circ}.$$

Step 3 : Find $$y$$
The vertical line $$EA$$ is perpendicular to every horizontal line, so
$$\angle AEF = 90^{\circ}.$$ At the point $$E$$ the two angles $$\angle BEF$$ and $$\angle AEB$$ together fill this right angle: $$\angle BEF + y = 90^{\circ} \;\;\Longrightarrow\;\; 55^{\circ} + y = 90^{\circ}.$$
Solve for $$y$$: $$y = 90^{\circ} - 55^{\circ} = 35^{\circ}.$$

Step 4 : State the answers
$$x = 125^{\circ}, \qquad y = 35^{\circ}.$$

Step 1 : Copy the two given directions at N and at O
The figure shows two straight lines

• $$LM$$ (slanting upward to the right)
• $$PQ$$ (slanting upward to the left)

• a line through N parallel to $$LM$$,
• a line through O parallel to $$PQ$$.

Step 2 : Identify that right-angle inside the required angle
Those two newly drawn lines coincide with the two arms that make $$\angle NOP$$ (one arm is $$ON$$, the other is $$OP$$).
Therefore $$\angle NOP$$ itself is that very right-angle.

Step 3 : Write the measure
\[\angle NOP = 90^\circ\]