Chapter 5: I’m Up and Down, and Round and Round

First observe that the outlines in Fig. 5.1 are ideal geometrical figures that we encounter every day in natural as well as man-made objects. A quick comparison with common surroundings immediately tells us where each outline comes from.

• Figure (i) – Circle. The perfectly round outline is the same shape that we see in the Sun, the Moon, a coin, the cross–section of a tree-trunk or the rim of any wheel. In geometry we later describe it precisely as the set of all points that remain at a fixed distance (the radius) from one fixed point (the centre).
• Figure (ii) – Rectangle (or square). This four-sided right-angled shape is the face of a brick, a sheet of paper, a blackboard or the panes of a window. Early builders, when cutting stone blocks or wooden planks, were led naturally to this outline.
• Figure (iii) – Triangle. Roof-tops of houses, a road-side warning sign, the sloping face of a pyramid or even the cross-section of a mountain ridge all give the same three-sided silhouette. Such day-to-day sightings produced the abstract idea of a triangle.
• Figure (iv) – Regular hexagon. Bees have been making honey-combs with exactly this six-sided pattern for millions of years. Salt crystals or snow-flakes also display the same outline. These natural patterns suggested the study of regular hexagons.
• Figure (v) – Spiral curve. The shell of a snail, the curling of a fern tendril, the path of water in a whirl-pool or the windings of a watch-spring all trace an ever-widening spiral. By abstracting that outline, mathematicians defined and analysed plane spirals.

Thus every shape in Fig. 5.1 owes its origin to some familiar object or pattern that human beings observed around them. Abstract geometry was built only after first recognising these concrete outlines in nature and in the artefacts of daily life.

First, recall that a circle is the set of all points in a plane that are at a fixed distance (called the radius) from a fixed point (called the centre). Whenever a natural object has an outline that keeps the same distance from a central point, it looks circular to the eye.

Examples found in nature include:

• The apparent shape of the Sun and the full Moon when viewed from Earth.
• A cross-section of many fruits such as an orange or an apple sliced perpendicular to its axis.
• Water drops (seen from above) resting on a smooth surface, because surface tension pulls the perimeter into a circle.
• Soap bubbles when they settle on a flat surface, viewed from the top.
• The annual growth rings of a tree trunk, visible when the trunk is cut at right angles to its length.
• Lily pads or some broad circular leaves, whose outlines are nearly perfect circles.

All these shapes have boundaries that are (to a very good approximation) equidistant from a central point, so they resemble circles.

Goal. Locate the exact centre of the given circular sheet.

Idea recalled from Euclid. In any circle, the perpendicular bisector of a chord always passes through the centre. Conversely, the point where the perpendicular bisectors of two different chords meet must be the centre of the circle.

Step 1 — Choose the first chord.
Pick any two points A and B on the rim and join them to get chord $$AB$$.

Step 2 — Construct/fold its perpendicular bisector.
• If you have just a ruler and compass: find the midpoint M of $$AB$$ and draw the line through M that is perpendicular to $$AB$$.
• If you are only folding paper (simpler!): fold the paper so that point A exactly falls on B. The crease you obtain is automatically the line that is both the perpendicular to and the bisector of $$AB$$. Call this crease $$\boxed{l_1}$$.

Why does $$l_1$$ pass through the centre?
Let O be the true (unknown) centre. Radii $$OA$$ and $$OB$$ are equal, so $$\triangle OAB$$ is isosceles. In an isosceles triangle the segment joining the vertex to the midpoint of the base is perpendicular to the base; hence $$OM \perp AB$$. Therefore the true centre O lies on $$l_1$$.

Step 3 — Repeat with a second chord.
Pick another pair of rim-points C and D to get chord $$CD$$ (do not choose C or D on $$AB$$ so the two chords are different). Fold as before so that C falls on D; the new crease is the perpendicular bisector $$\boxed{l_2}$$ of $$CD$$.

By exactly the same reasoning, the true centre O also lies on $$l_2$$.

Step 4 — Locate the centre.
Mark the unique intersection point of the two creases $$l_1$$ and $$l_2$$. Call it O. Because O lies on both perpendicular bisectors, it is equidistant from A and B and from C and D, i.e. $$OA = OB = OC = OD$$. That common distance is the radius, so O is the required centre.

Conclusion. Amina’s suggestion was: “Draw (or fold) any two chords of the circular paper and then draw (or crease) their perpendicular bisectors; the point where the two bisectors intersect is the centre of the circle.”

Pre–requisite fact. A regular n–gon looks exactly the same after a rotation of $$\dfrac{360^{\circ}}{n}$$ about its centre. It also has as many reflection (mirror) axes as it has sides.

Below we apply this fact to a square (n = 4), a regular pentagon (n = 5) and a regular hexagon (n = 6).

1. Square (n = 4)

1. Rotational symmetry
   A full turn is $$360^{\circ}$$. The least positive angle at which a square matches its original outline is $$\dfrac{360^{\circ}}{4}=90^{\circ}$$. Hence the identical appearances occur at $$0^{\circ},\;90^{\circ},\;180^{\circ},\;270^{\circ}\;(=360^{\circ}).$$ So there are 4 rotational symmetries.
2. Reflection symmetry
   Draw the two diagonals and the two lines joining the mid-points of opposite sides; each is a mirror line. Therefore a square has 4 lines of symmetry.

2. Regular pentagon (n = 5)

1. Rotational symmetry
   Least turning angle: $$\dfrac{360^{\circ}}{5}=72^{\circ}.$$ Successive matching turns are $$0^{\circ},\;72^{\circ},\;144^{\circ},\;216^{\circ},\;288^{\circ}\;(=360^{\circ}).$$ Thus it has 5 rotational symmetries.
2. Reflection symmetry
   Each mirror line passes through one vertex and the midpoint of the opposite side, giving 5 lines of symmetry.

3. Regular hexagon (n = 6)

1. Rotational symmetry
   Least turning angle: $$\dfrac{360^{\circ}}{6}=60^{\circ}.$$ Identical positions occur at $$0^{\circ},\;60^{\circ},\;120^{\circ},\;180^{\circ},\;240^{\circ},\;300^{\circ}\;(=360^{\circ}).$$ Hence there are 6 rotational symmetries.
2. Reflection symmetry
   There are three mirror lines joining opposite vertices and three joining mid-points of opposite sides – altogether 6 lines of symmetry.

Therefore:

• Square – 4 rotational, 4 reflection lines.
• Regular pentagon – 5 rotational, 5 reflection lines.
• Regular hexagon – 6 rotational, 6 reflection lines.

Given A circle with radius $$r = 5\text{ units}$$.

To find (i) the length of its longest chord, (ii) whether a smallest (shortest non-zero) chord exists.

Step 1 · Recall the fact about chords and diameter
For any circle, the diameter is the chord that passes through the centre. A fundamental result proved in Euclid’s geometry is:

• The greater the perpendicular distance of a chord from the centre, the shorter is the chord, and
• The chord with zero distance from the centre (that is, the chord containing the centre itself) is the longest possible chord.

Hence, the diameter is always the longest chord of a circle.

Step 2 · Compute the diameter when $$r = 5\text{ units}$$

Step 3 · Investigate the “smallest” chord

Take any point $$P$$ on the circumference. Through $$P$$ draw a chord $$PQ$$ making a very small arc near $$P$$. By moving $$Q$$ closer and closer to $$P$$ along the circle we make the chord $$PQ$$ shorter and shorter. There is no positive lower bound on the length thus obtained; the length can be made as close to $$0$$ as we wish, although a chord of exact length $$0$$ would collapse to a single point and hence would not be a chord.

Therefore, while the circle has a longest chord, it has no smallest non-zero chord.

Conclusion

• The longest chord is the diameter, of length $$10\text{ units}$$.
• There is no smallest chord; chords can be made arbitrarily small in length.

Given : Two fixed points $$A$$ and $$B$$ in a plane.

To prove : The set (locus) of all points which are equidistant from $$A$$ and $$B$$ is the perpendicular bisector of the line segment $$AB$$.

We break the work into two parts:

1. Every point that is equidistant from $$A$$ and $$B$$ lies on the perpendicular bisector of $$AB$$.
2. Every point on the perpendicular bisector of $$AB$$ is equidistant from $$A$$ and $$B$$.

Construction for reference (make this in your notebook):
Draw the segment $$AB$$. Mark its midpoint $$M$$ by taking $$AM = MB$$. Through $$M$$ draw a straight line $$l$$ perpendicular to $$AB$$. Line $$l$$ is called the perpendicular bisector of $$AB$$.

Part 1  ("Only if")

Let $$P$$ be any point such that $$PA = PB$$.

• In $$\triangle PAB$$, we have two equal sides: $$PA = PB$$. Hence $$\triangle PAB$$ is isosceles with base $$AB$$.
• In an isosceles triangle the angles at the base are equal, so
  $$\angle PAB = \angle PBA$$.
• When the base angles are equal, the line segment from the vertex $$P$$ to the midpoint $$M$$ of the base $$AB$$ is perpendicular to the base (Class-IX theorems) and also bisects it.

Thus $$M$$ is the midpoint of $$AB$$ and $$PM \perp AB$$, which places $$P$$ on the perpendicular bisector $$l$$.

Therefore every point equidistant from $$A$$ and $$B$$ lies on the perpendicular bisector of $$AB$$.

Part 2  ("If")

Now take any point $$P$$ on the perpendicular bisector $$l$$.

• Since $$l$$ passes through $$M$$, we have $$AM = MB$$ (definition of a midpoint).
• Since $$l$$ is perpendicular to $$AB$$ at $$M$$, the triangles $$\triangle PMA$$ and $$\triangle PMB$$ are right triangles with the right angle at $$M$$.
• In these two right triangles we know:
     $$PM$$ is common,
     $$AM = MB$$  (midpoint condition),
     and each has a right angle at $$M$$.

Thus, by the Hypotenuse–Leg (HL) criterion (right-triangle version of SAS),
\[ \triangle PMA \cong \triangle PMB. \]

Corresponding parts of congruent triangles are equal, so
$$PA = PB$$.

Therefore every point on the perpendicular bisector of $$AB$$ is equidistant from $$A$$ and $$B$$.

Combining Parts 1 and 2 we see that

• The points equidistant from $$A$$ and $$B$$ lie only on the perpendicular bisector, and
• every point on the perpendicular bisector is equidistant from $$A$$ and $$B$$.

Hence the perpendicular bisector of $$AB$$ is precisely the locus of points equidistant from the two given points.

Given : Two distinct points A and B in a plane.

To prove : Infinitely many (an un-countable number of) circles can be drawn which pass through A and B.

Definition : A circle is the collection of all points in the plane that are at a fixed distance (called the radius) from a fixed point (called the centre).

Let the distance between the given points be $$AB = d$$.

For any circle through A and B, its centre O must satisfy the condition

$$OA = OB = r$$

because A and B are both points on that circle, so they must be at the same distance r from the centre O.

This means the centre O is equidistant from A and B. The set of all points that are equidistant from two given points is the perpendicular bisector of the segment AB.

Construct the perpendicular bisector ℓ of $$\overline{AB}$$. (If you actually draw the figure, mark the midpoint M of $$\overline{AB}$$, erect a perpendicular at M, and extend it in both directions; that entire line is ℓ.)

Every point O lying on ℓ automatically gives $$OA = OB$$, hence serves as a possible centre of a circle through A and B. Conversely, no point outside ℓ can be the centre, because then the distances to A and B would be unequal.

Now observe that a straight line contains infinitely many points. Therefore ℓ contains infinitely many choices for the centre O. For each such choice the radius is fixed as $$r = OA = OB$$, so we obtain a different circle (the radii vary from a value just greater than $$\tfrac{d}{2}$$ up to as large as we like).

Hence there is no upper bound on the number of distinct circles through A and B; there are infinitely many of them.

Conclusion : Through any two distinct points in a plane, infinitely many circles can be drawn.

Given. Two fixed, distinct points $$A$$ and $$B$$ in the plane.

We have to investigate all the circles that pass through both $$A$$ and $$B$$ and then find the least and the greatest possible radius among them.

1. Locus of the centres.

For a circle to pass through $$A$$ and $$B$$, its centre $$O$$ must be at equal distance from the two points, i. e. $$OA = OB$$. The set of all points that are equidistant from $$A$$ and $$B$$ is the perpendicular bisector of the segment $$AB$$ (a theorem proved earlier in the chapter). Hence

All the centres of the required circles lie on the perpendicular bisector of $$AB$$, and every point of that bisector is the centre of exactly one such circle (its radius is that common distance to $$A$$ and $$B$$).

2. Expressing the radius in terms of the centre’s position.

Let $$M$$ be the midpoint of $$AB$$ and let $$O$$ be an arbitrary point on the perpendicular bisector. Denote by $$d$$ the distance from $$O$$ to the line $$AB$$, i. e. $$d = OM$$.

In right $$\triangle OMA$$ we have

$$OA^2 = OM^2 + MA^2$$ (by Pythagoras).

But $$MA = \dfrac{AB}{2}$$, therefore

$$OA^2 = d^2 + \left( \dfrac{AB}{2} \right)^2.$$

The radius $$r$$ of the circle with centre $$O$$ is $$r = OA = OB$$, so

$$r = \sqrt{ d^2 + \left( \dfrac{AB}{2} \right)^2 }.$$

3. The smallest possible radius.

• The quantity $$\left( \dfrac{AB}{2} \right)^2$$ is fixed (because $$A$$ and $$B$$ are fixed).
• $$d^2 \ge 0$$, and it attains its minimum value $$0$$ when the centre $$O$$ coincides with $$M$$ (i. e. when $$O$$ lies on the segment $$AB$$).

Thus the minimum radius is obtained at $$d = 0$$:

\[ r_{\text{min}} = \sqrt{ 0 + \left( \dfrac{AB}{2} \right)^2 } = \dfrac{AB}{2}. \]

This is the circle that has $$AB$$ as a diameter (centre $$M$$).

4. The largest possible radius.

Because the perpendicular bisector extends indefinitely on both sides of $$AB$$, the distance $$d$$ can be made as large as we please. As $$d \to \infty$$, the expression $$r = \sqrt{ d^2 + \left( \dfrac{AB}{2} \right)^2 }$$ also grows without bound. Therefore

There is no greatest radius; we can obtain circles of arbitrarily large radii by choosing centres farther and farther away from the segment $$AB$$ along its perpendicular bisector.

5. Answer to the three parts.

1. Yes. Infinitely many circles of different radii can be drawn through $$A$$ and $$B$$ – one for every point on the perpendicular bisector of $$AB$$.
2. The smallest possible radius is $$\dfrac{AB}{2}$$ (the circle with $$AB$$ as diameter).
3. There is no largest radius; the radius can be made as large as desired, so it is said to be unbounded or tends to infinity.

Given : A fixed segment $$AB$$. Any circle that passes through both $$A$$ and $$B$$ must have its centre somewhere on the perpendicular bisector of $$AB$$ (because the centre is equidistant from $$A$$ and $$B$$).

Let

• $$AB$$ have midpoint $$M$$, so $$AM = MB$$.
• $$AM = MB = c$$ (half the length of $$AB$$).
• $$O$$ be an arbitrary point on the perpendicular bisector, at a distance $$d$$ from the segment (more precisely, $$OM = d$$, where $$OM\perp AB$$).

Diagram to draw: Segment $$AB$$ with midpoint $$M$$; the perpendicular bisector drawn through $$M$$. Mark a point $$O$$ on that bisector, at some height above $$AB$$. Join $$OA$$ and $$OB$$.

Because $$M$$ is the midpoint and $$OM\perp AB$$, triangle $$OAM$$ is right-angled at $$M$$. Apply the Pythagoras theorem in $$\triangle OAM$$:

$$OA^2 = AM^2 + OM^2$$

Substituting the names of the lengths,

$$OA^2 = c^2 + d^2 \quad\Longrightarrow\quad OA = \sqrt{c^2 + d^2}$$

The same calculation gives $$OB = \sqrt{c^2 + d^2}$$, so the circle’s radius is

Now notice:

• The half-length $$c = AM$$ is a fixed constant (since $$A$$ and $$B$$ do not move).
• The perpendicular distance $$d = OM$$ increases precisely when we “move away from the segment $$AB$$ along its perpendicular bisector”.

Because the expression $$\sqrt{c^2 + d^2}$$ increases whenever $$d$$ increases (square-root is an increasing function and $$c^2$$ is fixed), the radius $$r$$ gets larger the farther we go from the segment.

Conclusion : As you move away from the segment $$AB$$ along its perpendicular bisector, the radii of all possible circles through $$A$$ and $$B$$ increase.

Let the two fixed points be A and B and let M be the midpoint of the segment $$AB$$. The perpendicular bisector of $$AB$$ is the line through M perpendicular to $$AB$$.

Choose any point P on this perpendicular bisector. Join $$PA$$ and $$PB$$. Because P is on the perpendicular bisector we have

$$PA = PB$$ (points on the perpendicular bisector of a segment are equidistant from its end-points).

Thus the circle with centre P and radius $$PA$$ passes through both A and B.

To compare how “curved” the different circles look, we compare their radii. The larger the radius, the flatter (less curved) the circle appears in the neighbourhood of $$A$$ and $$B$$; the smaller the radius, the tighter (more curved) it looks.

Introduce the notation

• $$AB = 2d\;(d > 0)\;,$$ so $$AM = MB = d$$.
• The perpendicular distance of P from $$AB$$, i.e. $$PM$$, is called $$h\:(h \ge 0)$$.

In right-angled triangle $$\triangle PAM$$ (right angle at M) we apply Pythagoras’ theorem:

$$PA^2 = PM^2 + AM^2\;.$$

That is,

$$PA^2 = h^2 + d^2 \;\;\Longrightarrow\;\; PA = \sqrt{h^2 + d^2}\;.$$

The radius $$r$$ of the required circle is therefore

$$r = PA = \sqrt{h^2 + d^2}\;.$$

Observe what happens as we move the centre P along the perpendicular bisector:

• If we go farther away from the segment $$AB$$, the value of $$h$$ increases, so $$r = \sqrt{h^2 + d^2}$$ also increases.
• If we come closer to the segment $$AB$$, the value of $$h$$ decreases, so $$r$$ decreases.

Since curvature is inversely related to the radius (larger radius ⇒ smaller curvature), a larger $$r$$ makes the arc through A and B appear flatter, i.e. less curved. Conversely, a smaller $$r$$ makes it look more curved.

Hence, as you proceed along the perpendicular bisector away from the segment $$AB$$, the circle you draw through $$A$$ and $$B$$ becomes less curved.

Let $$A$$ and $$B$$ be two distinct points in the Euclidean plane.

We treat separately

1. Squares with $$A$$ and $$B$$ anywhere on the boundary (perimeter)
2. Squares with $$A$$ and $$B$$ as corners (vertices)

1. Squares having $$A$$ and $$B$$ on the boundary

The boundary of a square consists of four equal straight-line segments. In order that the two given points lie on that boundary, it is not necessary that they be vertices; they may lie anywhere on any of the four sides.

Choose the following construction.

• Draw the line $$\overline{AB}$$. Decide any real number $$s$$ with $$s\gt AB$$ to be the length of a side of a square.
• On $$\overline{AB}$$ mark a point $$P$$ so that $$AP=s$$. (There are two possibilities, extending $$A$$ towards $$B$$ or away from it.)
• At $$A$$ erect a perpendicular of length $$s$$; call its end $$D$$. Through $$D$$ draw a line parallel to $$\overline{AB}$$ and through $$P$$ draw a line parallel to $$AD$$. Their intersection is $$C$$. The quadrilateral $$ADCP$$ is a square of side $$s$$ that certainly contains $$A$$ on one side and, by choice of the mark $$P$$, contains $$B$$ somewhere on the same side or on the continuation of that side.

Because the number $$s$$ can be chosen in infinitely many ways (every real number bigger than $$AB$$ is admissible) and because we may choose the perpendicular to rise on either side of $$\overline{AB}$$, there are infinitely many different squares containing both $$A$$ and $$B$$ on their perimeters.

Hence, the answer to part (1) is: infinitely many squares.

2. Squares having $$A$$ and $$B$$ as vertices

All squares are rigid figures, so only the relative position of $$A$$ to $$B$$ matters. There are exactly two geometric possibilities.

1. Segment $$AB$$ is a side of the square.

• At $$A$$ draw a line through $$A$$ perpendicular to $$\overline{AB}$$ and mark off a length equal to $$AB$$ to reach a point $$D$$.
• Through $$B$$ draw a line parallel to $$AD$$ and through $$D$$ draw a line parallel to $$AB$$; their intersection is $$C$$.
• The quadrilateral $$ABCD$$ is a square because all sides are $$AB$$ in length and adjacent sides are perpendicular.
• Instead of erecting the perpendicular on the $${\text{upper}}$$ side of $$\overline{AB}$$ we could equally well erect it on the $${\text{lower}}$$ side. These two choices give two non-overlapping squares.

Therefore, when $$AB$$ is taken as a side, exactly two squares are possible.

2. Segment $$AB$$ is a diagonal of the square.

• Let $$M$$ be the midpoint of $$\overline{AB}$$. In any square the diagonals bisect one another at right angles. Hence the other diagonal must be the line through $$M$$ perpendicular to $$\overline{AB}$$.
• Call $$\vec{v}=\overrightarrow{MA}$$. A rotation of $$\vec{v}$$ through $$90^{\circ}$$ gives a new vector $$\vec{w}$$ with the same length $$|\vec{v}|$$. The remaining two vertices are $$C=M+\vec{w},\; D=M-\vec{w}.$$
• Using the opposite rotation (\(-90^{\circ}\)) merely exchanges the names of $$C$$ and $$D$$, producing the same square. Thus there is one and only one square for which $$AB$$ is a diagonal.

Combining both sub-cases,

• 2 squares with $$AB$$ as a side,
• 1 square with $$AB$$ as a diagonal,

gives a total of $$2+1=3$$ distinct squares.

Final count

Given data

• Side $$AB = 5\,\text{cm}$$
• $$\angle A = 70^{\circ}$$
• $$\angle B = 60^{\circ}$$

Because the three interior angles of a triangle add up to $$180^{\circ}$$, we first find $$\angle C$$:

$$\angle C = 180^{\circ} - (70^{\circ} + 60^{\circ}) = 50^{\circ}.$$

Every angle of the triangle is less than $$90^{\circ}$$, so $$\triangle ABC$$ is an acute triangle.

Construction of $$\triangle ABC$$

1. Draw a horizontal base line and mark two points on it, naming them A and B, such that the length $$AB = 5\,\text{cm}$$ (use a ruler).
2. At point A construct an angle of $$70^{\circ}$$ with $$AB$$ as one arm.
     • Place a protractor at A, mark the $$70^{\circ}$$ point, and draw a ray that passes through that mark. Call the ray AX.
3. At point B construct an angle of $$60^{\circ}$$ with $$BA$$ as one arm.
     • Place the protractor at B, mark the $$60^{\circ}$$ point, and draw ray BY.
4. The two rays AX and BY meet at some point. Label that intersection as C.
5. Join BC and AC to complete $$\triangle ABC$$.

Construction of the circumcircle of $$\triangle ABC$$

1. Perpendicular bisector of $$AB$$:
     • With centres A and B and any radius more than half of $$AB$$, draw two arcs above and below $$AB$$.
     • Join the intersection points of those arcs. This straight line is the perpendicular bisector of $$AB$$.
2. Perpendicular bisector of $$BC$$:
     • Repeat the same process for side $$BC$$ to obtain its perpendicular bisector.
3. The two perpendicular bisectors meet at a point. Call this point O. By Euclid’s result, all three perpendicular bisectors of a triangle are concurrent; the common point O is called the circumcentre of the triangle.
4. Measure the distance $$OA$$ with a compass. This length is the radius of the required circumcircle (because the circumcentre is equidistant from all three vertices).
5. With centre O and radius $$OA$$, draw a circle. The circle passes through A, B and C; hence it is the circumcircle of $$\triangle ABC$$.

Location of the circumcentre

Since $$\triangle ABC$$ is acute (all three interior angles are strictly less than $$90^{\circ}$$), the circumcentre O necessarily lies inside the triangle. This follows from the basic property proved in Euclid’s geometry that:

"The perpendicular bisectors of the three sides of an acute triangle meet at a point that lies in the interior of the triangle."

Thus, in this construction, the centre O of the circumcircle lies inside $$\triangle ABC$$.

Given data

• Side $$AB = 5\;\text{cm}$$
• Side $$AC = 4\;\text{cm}$$
• Included angle $$\angle A = 100^{\circ}$$ (obtuse)

Required — (i) construct $$\triangle ABC$$, (ii) draw its circumcircle, (iii) state whether the circum-centre lies inside or outside the triangle.

Construction steps

1. Draw the base.
   Draw a straight line and mark two points A and B on it such that $$AB = 5\;\text{cm}$$.
2. Construct the given angle at A.
   Using a protractor place its centre at A; mark a point on the ray making $$100^{\circ}$$ with $$AB$$ on the side opposite to $$B$$. Draw the ray $$AX$$ through that mark. Thus $$\angle BAX = 100^{\circ}$$.
3. Locate point C.
   With centre A and radius $$4\;\text{cm}$$ draw an arc to meet ray $$AX$$ at C. Now $$AC = 4\;\text{cm}$$. Join $$BC$$. $$\triangle ABC$$ is obtained with the required measurements.
4. Construct the perpendicular bisector of side AB.
   Keeping the same radius more than half of $$AB$$, draw arcs above and below $$AB$$ with centres A and B. The arcs intersect at two points; join them to obtain the perpendicular bisector $$l_1$$ of $$AB$$.
5. Construct the perpendicular bisector of side AC.
   Repeat the same procedure for $$AC$$ to obtain its perpendicular bisector $$l_2$$.
6. Locate the circum-centre O.
   Mark the intersection of $$l_1$$ and $$l_2$$ as O. By definition O is equidistant from A, B and C.
7. Draw the circumcircle.
   With centre O and radius $$OA$$ draw a circle. It will pass through points B and C automatically; this is the circumcircle of $$\triangle ABC$$.

Reasoning about the position of O

The triangle has an obtuse angle at $$A\,(100^{\circ}\!>90^{\circ})$$. A well-known Euclidean result states:

“In an acute triangle the circum-centre lies inside, in a right triangle on the hypotenuse, and in an obtuse triangle outside the triangle on the side opposite the obtuse angle.”

Hence, since $$\triangle ABC$$ is obtuse at A, the circum-centre O must lie outside the triangle, on the far side of the line $$BC$$ which is opposite the obtuse angle.

Verification

• Measure $$OB$$ and $$OC$$ with a compass; they equal $$OA$$, confirming the correctness of the circumcircle.
• Notice that point O indeed lies outside the boundaries of $$\triangle ABC$$.

Result

The circumcircle is drawn and its centre lies outside $$\triangle ABC$$, on the side opposite the obtuse angle at A.

Given data

• $$AB = 6\,\text{cm}$$
• $$BC = 7\,\text{cm}$$
• $$CA = 7\,\text{cm}$$ (so $$\triangle ABC$$ is isosceles with vertex $$C$$ and base $$AB$$)

We have to (i) construct $$\triangle ABC$$, (ii) draw its circumcircle, (iii) locate the circum-centre $$O$$ and (iv) measure $$OA,\;OB,\;OC$$.

Construction steps

1. Draw the base.
   Draw a straight line segment $$AB = 6\,\text{cm}$$ with the help of a ruler.
2. Locate vertex $$C$$.
   
   • With centre $$A$$ and radius $$7\,\text{cm}$$ draw an arc above the line $$AB$$.
   • With centre $$B$$ and the same radius $$7\,\text{cm}$$ draw another arc to cut the first arc at a point; mark the intersection as $$C$$.
   Because both radii are $$7\,\text{cm}$$, we have automatically ensured $$CA = BC = 7\,\text{cm}$$.
3. Complete the triangle.
   Join $$C$$ to $$A$$ and $$C$$ to $$B$$. $$\triangle ABC$$ is now drawn.
4. Construct two perpendicular bisectors.
   (Any two sides suffice.)
   
   • Side $$AB$$: With centres $$A$$ and $$B$$ and radius more than half of $$AB$$, draw arcs above and below $$AB$$ to intersect at two points; join these intersection points to obtain the perpendicular bisector of $$AB$$.
   • Side $$AC$$: Repeat the same procedure for $$AC$$ to get its perpendicular bisector.
5. Mark the circum-centre.
   The two bisectors meet at a point $$O$$. This point is equidistant from $$A,\,B,\,C$$ and is called the circum-centre of $$\triangle ABC$$.
6. Draw the circumcircle.
   With centre $$O$$ and radius $$OA$$ (distance from $$O$$ to any one vertex), draw a circle. The circle will pass through all three vertices $$A,\,B,\,C$$; it is the required circumcircle.
7. Measure the radii.
   Using a ruler or dividers measure $$OA,\;OB,\;OC$$. All three lengths are the same (a useful check on your construction); in an accurately drawn figure the length is found to be about $$3.9\,\text{cm}$$ (≈ 4 cm if rounded to the nearest millimetre).

Why the three lengths are equal

The point $$O$$ is the intersection of the perpendicular bisectors of the three sides of a triangle, and Euclid proved that such a point is equidistant from the three vertices. Hence $$OA = OB = OC$$ by definition; the measurement merely verifies this numerically.

Numerical confirmation (optional calculation)

For any triangle the circum-radius $$R$$ is given by $$R = \dfrac{abc}{4\Delta}$$, where $$a = BC$$, $$b = CA$$, $$c = AB$$ and $$\Delta$$ is the area.

This theoretical value matches the practical measurement (≈ 3.9 cm), confirming that $$OA = OB = OC$$.

Result

The circum-radius of the triangle is about $$3.9\,\text{cm}$$, so

$$OA = OB = OC \approx 3.9\,\text{cm}.$$

Given : Two fixed points $$A$$ and $$B$$ in a plane.
To find : The smallest (least possible) radius $$r_{\min}$$ of a circle that can be drawn so that the circle passes through both $$A$$ and $$B$$.

Step 1 – Recall a basic fact about circles
For any circle, all points that are equidistant from $$A$$ and $$B$$ lie on the perpendicular bisector of the segment $$AB$$. Therefore, the centre of every circle through $$A$$ and $$B$$ must be some point $$O$$ on that perpendicular bisector, and for every such centre

$$OA = OB = R$$   (the common distance is the radius).

Step 2 – Describe the set of admissible centres
Let $$M$$ be the mid-point of $$AB$$. The line through $$M$$ that is perpendicular to $$AB$$ is the perpendicular bisector. Any admissible centre $$O$$ must lie on this line. Write $$OM = d$$ for the distance of the centre from $$M$$. By right-triangle geometry, with $$\triangle OMA$$ right-angled at $$M$$, we have

$$OA^2 = OM^2 + MA^2$$.

Step 3 – Express the radius in terms of $$d$$
Because $$MA = \dfrac{AB}{2}$$, we obtain

$$R^2 = d^2 + \left(\dfrac{AB}{2}\right)^2$$   or   $$R = \sqrt{d^2 + \left(\dfrac{AB}{2}\right)^2}$$.

Step 4 – Minimise the radius
The quantity $$\left(\dfrac{AB}{2}\right)^2$$ is fixed (the two points are fixed). The only variable is $$d = OM$$, which is a non-negative length. The square-root expression is clearly smallest when $$d = 0$$, that is, when

$$O \equiv M$$   (the centre coincides with the mid-point).

Hence the least possible radius is obtained for the unique circle whose diameter is the segment $$AB$$ itself. Substituting $$d = 0$$ gives

\[ r_{\min} = \dfrac{AB}{2}. \]

Conclusion
The smallest circle that can be drawn through two fixed points $$A$$ and $$B$$ is the one whose diameter is $$AB$$, and its radius is exactly half the distance between the two points.

Given three distinct collinear points $$A,B,C$$ lying on the same straight line $$ ext{ℓ}$$ (take the order $$A,B,C$$ on $$ ext{ℓ}$$).

We discuss four connected questions step-by-step.

1. Can a point P be found so that $$PA = PB = PC$$ ?

• Because $$PA = PB$$, the point P must lie on the perpendicular bisector of $$AB$$ (call this line $$m_1$$).
• Because $$PB = PC$$, the same point P must also lie on the perpendicular bisector of $$BC$$ (call this line $$m_2$$).
• A single point P would therefore have to be the intersection $$m_1 \cap m_2$$.

So the question reduces to: “Do the two perpendicular bisectors $$m_1$$ and $$m_2$$ intersect?” The next section shows that they are parallel; hence they never meet and no such point P exists.

2. Why are the perpendicular bisectors of $$AB$$ and $$BC$$ parallel?

• Segment $$AB$$ lies on line $$\text{ℓ}$$, so its perpendicular bisector $$m_1$$ is perpendicular to $$\text{ℓ}$$:
  $$m_1 \perp \text{ℓ}.$$
• Segment $$BC$$ also lies on the same line $$\text{ℓ}$$, hence its perpendicular bisector $$m_2$$ is perpendicular to $$\text{ℓ}$$ as well:
  $$m_2 \perp \text{ℓ}.$$
• If two lines are perpendicular to the same line, they are parallel to each other. Therefore
  $$m_1 \parallel m_2.$$

Because the bisectors are parallel and have different mid-points (mid-point of $$AB$$ is not the same as that of $$BC$$), they are distinct parallel lines, so they never intersect.

To draw and verify: On a sheet draw a straight line, mark three points $$A,B,C$$ on it, construct the perpendicular bisector of $$AB$$ and that of $$BC$$ with a compass and straightedge. You will see two distinct parallel lines.

3. Can a circle pass through the three collinear points $$A,B,C$$?

The centre of such a circle would have to be equidistant from all three points; i.e. it would have to lie at $$m_1 \cap m_2$$, a point that does not exist. Hence

No circle can be drawn through three distinct collinear points.

4. Can a straight line intersect a given circle in three distinct points?

• Algebraically, the points of intersection of a line and a circle are obtained by solving one linear and one quadratic equation; at most two real solutions are possible.
• Geometrically, when you draw a line across a circle you enter and leave the circle exactly once, giving at most two common points (a tangent is the limiting case with exactly one).

Therefore a line can meet a circle in 0, 1 or 2 points, but never in three distinct points.

Conclusion

• No point P exists with $$PA = PB = PC$$ for three distinct collinear points.$$\qquad$$
• The perpendicular bisectors of $$AB$$ and $$BC$$ are distinct, parallel lines.$$\qquad$$
• No single circle can pass through three collinear points.$$\qquad$$
• It is impossible for one straight line to cut a circle in three distinct points.

Let the circumcircle of the given $$\triangle ABC$$ have centre $$O$$ and radius $$R$$, so that

$$OA = OB = OC = R.$$

Idea. A rotation about the point $$O$$ keeps every distance unchanged (it is a rigid motion) and therefore carries any figure to a congruent copy of itself. At the same time, because the centre is fixed, the image of every vertex still lies exactly $$R$$ units from $$O$$ – that is, it still lies on the same circle.

Construction of another congruent triangle.

1. Choose any angle $$\theta$$ with $$0^{\circ}<\theta<360^{\circ}$$ and $$\theta\neq 360^{\circ}k$$ for every integer $$k$$ (so that the image does not fall back on the original triangle).
2. Rotate the whole plane about $$O$$ through the angle $$\theta$$. Denote the images of $$A,\,B,\,C$$ by $$A',\,B',\,C'$$ respectively.

Why the new triangle is congruent.

The rotation is a distance–preserving transformation, hence

$$A'B' = AB, \; B'C' = BC, \; C'A' = CA.$$

Therefore, by the SSS criterion,

Why the new triangle shares the same circumcircle.

Because the centre $$O$$ does not move in the rotation,

$$OA' = OB' = OC' = R,$$

so the three new vertices still lie on the very same circle that contains $$A,\,B,\,C$$.

Distinctness. Since $$\theta$$ is not a full multiple of $$360^{\circ}$$, at least one of $$A',\,B',\,C'$$ differs from the corresponding original point, so $$\triangle A'B'C'$$ is a different triangle, not just a relabelling of $$\triangle ABC$$.

Hence a second triangle, congruent to the given one, lying on the same circumcircle has been obtained. Because the angle $$\theta$$ may be chosen in infinitely many ways, there are in fact infinitely many such triangles. (Reflections in any diameter of the circle provide still more examples.)

Conclusion. Yes — there are infinitely many triangles congruent to the given $$\triangle ABC$$ that can be inscribed in its circumcircle.

Given: A circle with centre $$O$$ and a chord $$AB$$. Joining $$O$$ to the ends of the chord forms $$\triangle OAB$$.

To prove: $$\triangle OAB$$ is isosceles, i.e. $$OA = OB$$.

Proof:

1. By definition, a radius of a circle is a line segment that joins the centre of the circle to any point on the circle.
2. Because $$A$$ and $$B$$ lie on the circle (they are the endpoints of chord $$AB$$), the segments $$OA$$ and $$OB$$ are both radii.
3. All radii of the same circle are equal.
   Reason: Euclid’s Common Notion 1 states, “Things which are equal to the same thing are equal to one another.” Every radius is equal to the fixed distance called the radius of the circle; thus $$OA$$ equals that distance and $$OB$$ equals the same distance, giving $$OA = OB$$.
4. Since two sides of $$\triangle OAB$$ are equal, the triangle satisfies the definition of an isosceles triangle.

Therefore, the triangle formed by any chord and the centre of its circle is isosceles.

Recall. In Question 1 we proved that the triangle formed by any chord of a circle together with its centre is isosceles, the two equal sides being the radii drawn to the two end-points of the chord.

Given. Let the circle have centre $$O$$ and radius $$r$$. Take any two chords $$AB$$ and $$CD$$ of this circle and form the two isosceles triangles $$\triangle OAB\quad\text{and}\quad\triangle OCD,$$ in which the equal sides are the radii $$OA, OB$$ and $$OC, OD$$ respectively; the bases are the chords $$AB$$ and $$CD$$. Assume that the two bases have equal length, i.e.

$$AB = CD.$$

To prove. $$\triangle OAB \cong \triangle OCD.$$

Proof.

1. The points $$A,\,B,\,C,\,D$$ all lie on the same circle of centre $$O$$, so every segment joining $$O$$ to any one of them is a radius. Hence

   $$OA = OB = OC = OD = r.$$

2. Compare $$\triangle OAB$$ with $$\triangle OCD$$:

   • $$OA = OC$$  (each equals $$r$$),
   • $$OB = OD$$  (each equals $$r$$),
   • $$AB = CD$$  (given – equal base lengths).

   Thus the three sides of $$\triangle OAB$$ are equal, in pairs, to the three corresponding sides of $$\triangle OCD$$.

3. By the SSS (Side-Side-Side) congruence criterion,

   \[\triangle OAB \cong \triangle OCD.\]

Conclusion. Any two isosceles triangles of the kind described in Question 1 (i.e. formed by a chord and the centre of the circle) are congruent whenever their bases — the two chords — are equal in length. □

Given figure (Fig. 5.12) :
Inside a circle with centre C, a chord AB is drawn. From the centre a line segment CM is drawn to the chord such that $$CM \perp AB$$. Hence $$\angle CMA = \angle CMB = 90^\circ$$.

To prove : $$AM = MB$$, i.e. CM bisects the chord AB.

Construction & data noted from the figure

• CA and CB are radii of the same circle, therefore $$CA = CB$$.
• CM is common in both \(\triangle CMA\) and \(\triangle CMB\).
• Each triangle has a right angle at M: $$\angle CMA = \angle CMB = 90^\circ$$ (because CM is perpendicular to AB).

Proof using RHS congruence

1. Consider \(\triangle CMA\) and \(\triangle CMB\).
2. Both triangles are right–angled at M.
3. Their hypotenuses are equal because both are radii: $$CA = CB$$.
4. They share the side $$CM$$.
5. Therefore, by the RHS (Right-angle–Hypotenuse–Side) congruence criterion, \[\triangle CMA \cong \triangle CMB.\]
6. Corresponding parts of congruent triangles are equal (CPCT), hence $$AM = MB.$$

Conclusion : The perpendicular drawn from the centre of a circle to a chord bisects that chord. Thus the converse of Theorem 4 is proved.

Given: Triangle $$ABC$$ is isosceles with $$AB = AC$$ and is inscribed in a circle whose centre is $$O$$.

To prove: The altitude from $$A$$ to $$BC$$ (let this altitude meet $$BC$$ at $$D$$) passes through $$O$$.

Construction: Draw altitude $$AD$$ so that $$AD \perp BC$$ and $$D$$ lies on $$BC$$.

Proof:

1. Consider triangles $$\triangle ABD$$ and $$\triangle ACD$$.
   • $$AB = AC$$  (given)
   • $$\angle BDA = \angle CDA = 90^{\circ}$$  (because $$AD \perp BC$$)
   • $$AD = AD$$  (common side)
   By the RHS congruence criterion, $$\triangle ABD \cong \triangle ACD$$.
2. Corresponding parts of congruent triangles are equal, therefore $$BD = DC$$.
   Hence $$D$$ is the midpoint of $$BC$$ and $$AD$$ is both a median and a perpendicular to $$BC$$.
   Thus $$AD$$ is the perpendicular bisector of chord $$BC$$.
3. Euclid’s theorem: the perpendicular bisector of any chord of a circle passes through the centre of the circle.
   Since $$AD$$ is the perpendicular bisector of the chord $$BC$$, the centre $$O$$ of the circle must lie on $$AD$$.
4. Therefore the altitude $$AD$$ from vertex $$A$$ to base $$BC$$ passes through the centre $$O$$ of the circumcircle.

Hence proved.

Step 1 : Draw and label the figure

Draw a circle with centre $$O$$ and radius $$5\,\mathrm{cm}$$. Draw two parallel chords on opposite sides of the centre—name them $$AB$$ and $$CD$$ such that $$AB = 6\,\mathrm{cm}$$ and $$CD = 8\,\mathrm{cm}$$. Mark the mid-points of the chords as $$M_1$$ on $$AB$$ and $$M_2$$ on $$CD$$. Join $$O M_1$$ and $$O M_2$$ (these joins are perpendicular to the respective chords).

Step 2 : Apply the perpendicular-bisector property

The line from the centre of a circle to a chord is perpendicular to the chord and bisects it. Therefore

$$A M_1 = \dfrac{AB}{2} = \dfrac{6}{2} = 3\,\mathrm{cm}$$   and   $$C M_2 = \dfrac{CD}{2} = \dfrac{8}{2} = 4\,\mathrm{cm}$$.

Step 3 : Distance of each chord from the centre

In right $$\triangle O A M_1$$, radius $$O A = 5\,\mathrm{cm}$$ and half-chord $$A M_1 = 3\,\mathrm{cm}$$.

By Pythagoras theorem:

$$O M_1^2 = O A^2 - A M_1^2 = 5^2 - 3^2 = 25 - 9 = 16$$ $$\Rightarrow\; O M_1 = 4\,\mathrm{cm}$$.

In right $$\triangle O C M_2$$, radius $$O C = 5\,\mathrm{cm}$$ and half-chord $$C M_2 = 4\,\mathrm{cm}$$.

$$O M_2^2 = 5^2 - 4^2 = 25 - 16 = 9$$ $$\Rightarrow\; O M_2 = 3\,\mathrm{cm}.$$

Step 4 : Distance between the mid-points

Because the chords are on opposite sides of the centre, the mid-points lie on the same straight line through $$O$$ but on opposite sides. Hence $$M_1 M_2 = O M_1 + O M_2 = 4\,\mathrm{cm} + 3\,\mathrm{cm} = 7\,\mathrm{cm}.$$

Step 5 : Result

The distance between the mid-points of the two chords is $$7\,\mathrm{cm}$$.

Given : In $$\triangle ABC$$ we have $$\angle B = \angle C$$.

To prove : $$AB = AC$$ (this is exactly Theorem 6: the sides opposite equal angles of a triangle are equal).

Construction : From vertex $$A$$ draw the perpendicular $$AD$$ to the base $$BC$$, so that $$AD \perp BC$$ and $$D$$ lies on $$BC$$.

After the construction we have two right triangles

• $$\triangle ABD$$ with right angle at $$D$$, and
• $$\triangle ACD$$ with right angle at $$D$$.

Step 1 : Establish similarity of the two right triangles

Because $$\angle B = \angle C$$ (given) and both the triangles are right–angled at $$D$$, we get

$$\angle ABD = \angle ACD\quad\text{and}\quad \angle ADB = \angle ADC = 90^{\circ}.$$

Thus, by the AA criterion,

$$\triangle ABD \sim \triangle ACD.$$

Step 2 : A proportionality obtained from similarity

Corresponding (hypotenuse) sides of the similar triangles give

$$\frac{AB}{AC} = \frac{BD}{DC}. \quad(1)$$

Step 3 : Invoke the Baudhāyana–Pythagoras theorem in each right triangle

Applying the theorem to $$\triangle ABD$$ and $$\triangle ACD$$, we have

$$AB^{2} = AD^{2} + BD^{2} \quad(2)$$
$$AC^{2} = AD^{2} + DC^{2}. \quad(3)$$

Step 4 : Use (1), (2) and (3) to compare the two sides

From (1) we get $$BD = \dfrac{AB}{AC}\,DC$$. Substitute this relation into (2):

$$AB^{2} = AD^{2} + \left( \dfrac{AB}{AC}\,DC \right)^{2}$$
$$\Rightarrow\; AB^{2} = AD^{2} + \dfrac{AB^{2}}{AC^{2}}\,DC^{2}. \quad(4)$$

Replace $$AD^{2}$$ obtained from (3): $$AD^{2} = AC^{2} - DC^{2}$$, and insert it into (4):

$$AB^{2} = (AC^{2} - DC^{2}) + \dfrac{AB^{2}}{AC^{2}}\,DC^{2}.$$

Collect the $$DC^{2}$$ terms on the right:

$$AB^{2} = AC^{2} + DC^{2}\left( \dfrac{AB^{2}}{AC^{2}} - 1 \right).$$

Move $$AC^{2}$$ to the left and factor:

$$AB^{2} - AC^{2} = DC^{2}\left( \dfrac{AB^{2}}{AC^{2}} - 1 \right).$$

Notice that the right-hand factor is exactly $$\dfrac{AB^{2} - AC^{2}}{AC^{2}}$$, so we get

$$AB^{2} - AC^{2} = DC^{2}\cdot \dfrac{AB^{2} - AC^{2}}{AC^{2}}.$$

If $$AB \neq AC$$, we can divide both sides by $$AB^{2} - AC^{2}$$ (which would be non-zero) and obtain

$$1 = \dfrac{DC^{2}}{AC^{2}},\;\text{i.e.}\; DC = AC.$$

But $$DC$$ is only a part of $$AC$$, so the equality is impossible unless the common factor we divided by was actually zero. Hence

$$AB^{2} - AC^{2} = 0 \;\Rightarrow\; AB = AC.$$

Step 5 : Conclusion

Because the sides opposite the equal angles $$\angle B$$ and $$\angle C$$ have turned out to be equal, the triangle is isosceles as claimed.

Thus, using the Baudhāyana–Pythagoras theorem we have established Theorem 6.

Hence proved.

Given : In Fig. 5.15 we have

• $$CE \perp AB$$ at $$E$$,
• $$CH \perp GH$$ at $$H$$,
• $$CE = CH$$.

To prove : $$AB = GF$$.

Proof

1. Since $$CE = CH$$, we can superpose the segment $$CE$$ on the segment $$CH$$ keeping the common end point $$C$$ fixed (Euclid’s Fourth Axiom: “Things which coincide with one another are equal to one another”). Under this superposition the end point $$E$$ will fall exactly on $$H$$.
2. During the above superposition the right angle at $$E$$ is carried rigidly with $$CE$$. Thus the straight line that was perpendicular to $$CE$$ at $$E$$, namely $$AB$$, now becomes the straight line drawn through $$H$$ that is perpendicular to $$CH$$.
3. Through a given point on a straight line there is one and only one perpendicular. Hence the new position of $$AB$$ must be that unique perpendicular to $$CH$$ through $$H$$ — i.e. the straight line $$GH$$ together with its extension $$GF$$. In other words, the whole segment $$AB$$ now lies exactly on the straight line $$GF$$.
4. Because the two segments coincide, Euclid’s Fourth Axiom again gives
   \[AB = GF\]

Thus, the required relation $$AB = GF$$ is established.

Question  (NCERT Ch. 5, Ex. 5.2, Q. 6)
If P is the mid-point of the hypotenuse AC of a right–angled △ABC (∠B = 90°), prove that P is equidistant from A, B and C.

Solution using the Baudhāyana–Pythagoras theorem

Step 1 · Draw and label
Draw right △ABC with ∠B = 90°. Mark its hypotenuse AC. Mark the mid-point of AC as P; thus $$AP = PC$$ by construction. Join P to the remaining vertex B.

Step 2 · Introduce algebraic names
Let $$AB = c, \; BC = d \;\text{ and }\; AC = h$$.
Because ∠B is a right angle, the Baudhāyana–Pythagoras theorem gives

Step 3 · Express the known equal lengths
Since P is the mid-point of AC,

$$AP = PC = \dfrac{h}{2}. \quad(1)$$

Step 4 · Find PB using Baudhāyana–Pythagoras again

Place the triangle on a pair of perpendicular number lines so that
B(0, 0), A(c, 0) and C(0, d). (This is only a convenient way to use the theorem; no new facts are assumed.)
Coordinates of P, the mid-point of AC, are therefore $$\bigl(\tfrac{c}{2},\,\tfrac{d}{2}\bigr).$$

The distance PB now follows directly from Baudhāyana–Pythagoras:

$$PB^{2}=\left(\tfrac{c}{2}-0\right)^{2}+\left(\tfrac{d}{2}-0\right)^{2}=\tfrac{c^{2}+d^{2}}{4}. \quad(2)$$

Step 5 · Compare the three squared distances

Using (1) and the value of h from Step 2:

$$PA^{2} = \left(\dfrac{h}{2}\right)^{2} = \tfrac{h^{2}}{4} = \tfrac{c^{2}+d^{2}}{4}. \quad(3)$$ Similarly, $$PC^{2}=\tfrac{c^{2}+d^{2}}{4}. \quad(4)$$ From (2), (3) and (4) we have

$$PB^{2}=PA^{2}=PC^{2}.$$

Step 6 · Conclude
Since all three squared distances are equal and distances are non-negative,

$$PB = PA = PC.$$ Thus the mid-point P of the hypotenuse is equidistant from the three vertices A, B and C, as required.

Result: $$PA = PB = PC$$

Given data

• Radius of the circle: $$r = 7\,\text{cm}$$
• Perpendicular distance from the centre to the chord: $$d = 6\,\text{cm}$$

Step 1. Draw and label the figure

Draw a circle with centre $$O$$ and radius $$7\,\text{cm}$$. Mark a chord $$AB$$ such that $$OC \perp AB$$ at point $$C$$ and $$OC = 6\,\text{cm}$$. (Because $$OC$$ is drawn perpendicular to the chord, it bisects the chord, so $$C$$ is the midpoint of $$AB$$.)

Step 2. Use the right-triangle relation

In right triangle $$\triangle OCA$$ we know the hypotenuse $$OA = r = 7\,\text{cm}$$ and one leg $$OC = d = 6\,\text{cm}$$. Let $$AC$$ be the other leg.

By Pythagoras’ theorem,

$$OA^{2} = OC^{2} + AC^{2}$$

Substitute the known values:

$$7^{2} = 6^{2} + AC^{2}$$

$$49 = 36 + AC^{2}$$

Subtract $$36$$ from both sides:

$$49 - 36 = AC^{2}$$

$$13 = AC^{2}$$

Therefore:

$$AC = \sqrt{13}\,\text{cm}$$

Step 3. Find the full chord length

Since $$C$$ is the midpoint of $$AB$$, the whole chord length is twice $$AC$$:

$$AB = 2 \times AC = 2 \times \sqrt{13} = 2\sqrt{13}\,\text{cm}$$

(Numerically, $$2\sqrt{13} \approx 7.21\,\text{cm}$$.)

Conclusion

The length of the chord is $$2\sqrt{13}\,\text{cm}$$.

Let the circle have centre O and radius $$r$$. Let $$AB$$ be a chord whose perpendicular distance from the centre is $$d$$.

1. Draw the perpendicular $$OL$$ from O to the chord $$AB$$, with $$L$$ lying on $$AB$$.
   Because a perpendicular drawn from the centre of a circle to a chord bisects the chord, $$L$$ is the midpoint of $$AB$$.

2. Denote the required length of the chord by $$AB = 2x$$.
   Hence the half-chord lengths are $$AL = LB = x$$.

3. Consider right $$\triangle OLA$$:

   • $$OA = r$$   (radius)
   • $$OL = d$$   (given perpendicular distance)
   • $$AL = x$$   (half-chord)

   By the Pythagoras theorem,

   $$OA^{2} = OL^{2} + AL^{2}$$

   Substituting,

   $$r^{2} = d^{2} + x^{2}$$

   Rearranging,

   $$x^{2} = r^{2} - d^{2}$$
   $$x = \sqrt{r^{2} - d^{2}} \quad (x>0)$$

4. The whole chord is twice its half:

   \[AB = 2x = 2\sqrt{r^{2}-d^{2}}\]

Thus, if the perpendicular distance of a chord from the centre is $$d$$ and the radius is $$r$$, the chord length is $$2\sqrt{r^{2}-d^{2}}$$.

Step 1 : Draw and name the required elements

Draw a circle with centre $$O$$. Let $$AB$$ and $$CD$$ be two chords such that the perpendiculars from $$O$$ to the chords meet them at $$M$$ and $$N$$ respectively. Thus

$$OM \perp AB,\;\;ON \perp CD$$

We are told that the distance of $$AB$$ from the centre is twice the distance of $$CD$$ from the centre, i.e.

$$OM = 2\,ON$$     ……(1)

Step 2 : Relate the length of a chord to its distance from the centre

Since $$OM$$ is perpendicular to $$AB$$, it bisects the chord. Hence

$$AM = MB = \dfrac{AB}{2}$$

In right $$\triangle OAM$$, by Pythagoras theorem,

$$OA^{2} = OM^{2} + AM^{2}$$

Because $$OA = r$$ (radius of the circle), we get

$$r^{2} = OM^{2} + \left(\dfrac{AB}{2}\right)^{2} \;\;\Longrightarrow\;\; AB = 2\sqrt{r^{2} - OM^{2}}$$     ……(2)

Similarly, in right $$\triangle OCN$$,

$$r^{2} = ON^{2} + \left(\dfrac{CD}{2}\right)^{2} \;\;\Longrightarrow\;\; CD = 2\sqrt{r^{2} - ON^{2}}$$     ……(3)

Step 3 : Use the given condition $$OM = 2ON$$

Let $$ON = x$$ ⇒ $$OM = 2x$$ (from (1)).
Substitute these in (2) and (3):

$$AB = 2\sqrt{r^{2} - (2x)^{2}} = 2\sqrt{r^{2} - 4x^{2}}$$     ……(4)

$$CD = 2\sqrt{r^{2} - x^{2}}$$     ……(5)

Step 4 : Check whether $$CD = 2AB$$ is necessarily true

Suppose we did have $$CD = 2AB$$. Replace $$CD$$ and $$AB$$ using (4) and (5):

$$2\sqrt{r^{2} - x^{2}} = 2\Bigl(2\sqrt{r^{2} - 4x^{2}}\Bigr)$$

Dividing by 2 and squaring both sides,

$$r^{2} - x^{2} = 4\bigl(r^{2} - 4x^{2}\bigr)$$

$$r^{2} - x^{2} = 4r^{2} - 16x^{2}$$

$$0 = 3r^{2} - 15x^{2}\;\;\Longrightarrow\;\; r^{2} = 5x^{2}$$

Thus $$CD = 2AB$$ can hold only when the radius happens to satisfy $$r = \sqrt{5}\,x$$. The question, however, gives no information about the value of the radius; it merely states the ratio of the two distances.

Step 5 : A quick counter-example

Take a circle of radius $$r = 5\text{ cm}$$.
Choose $$ON = 1\text{ cm}$$ ⇒ $$OM = 2\text{ cm}$$.

Then from (4) and (5):

$$AB = 2\sqrt{25 - 4} = 2\sqrt{21} \approx 9.17\text{ cm}$$
$$CD = 2\sqrt{25 - 1} = 2\sqrt{24} \approx 9.80\text{ cm}$$

The ratio $$\dfrac{CD}{AB} \approx 1.07\neq 2$$, proving that doubling the distance from the centre does not double the chord.

Conclusion

From the information $$OM = 2ON$$ alone we cannot conclude that $$CD = 2AB$$. The lengths of chords are related to the squares of the distances from the centre, not directly to the distances themselves.

Step 1 : Join the radii that intercept each arc
Draw the radii $$OA$$, $$OK$$ and $$OB$$. These three radii cut off the arc labelled $$AKB$$ on the circumference.
Similarly draw the radii $$OC$$, $$OL$$ and $$OD$$ to cut off the arc labelled $$CLD$$.

Step 2 : Measure the central angles
Using a protractor (or the angular scale of a geometry-box) measure the angles whose arms are the radii drawn in Step 1.

• The arc $$AKB$$ subtends the angle $$\angle AOB$$ at the centre.
   A careful measurement gives approximately
   $$\angle AOB \approx 110^\circ$$.
• The arc $$CLD$$ subtends the angle $$\angle COD$$ at the centre.
   A careful measurement gives approximately
   $$\angle COD \approx 250^\circ$$.

Step 3 : Classify each arc

• Because $$\angle AOB = 110^\circ < 180^\circ$$, the arc $$AKB$$ is a minor arc.
• Because $$\angle COD = 250^\circ > 180^\circ$$, the arc $$CLD$$ is a major arc.

(Any slight variation in the measured values is immaterial; the first angle is clearly less than $$180^\circ$$ whereas the second is clearly more than $$180^\circ$$, so the same conclusion is reached.)

Step 1 — Identify the triangle
Points $$A$$ and $$B$$ lie on the circle and $$O$$ is the centre, so the three points form triangle $$\triangle OAB$$.
Because $$OA$$ and $$OB$$ are radii,
$$OA = OB = 12\,\text{cm}.$$

Step 2 — State the known angle
The central angle is given as
$$\angle AOB = 60^{\circ}.$$

Step 3 — Use the Cosine Rule
In any triangle with sides $$a, b, c$$ opposite angles $$A, B, C$$ respectively, the Cosine Rule says
$$c^{2} = a^{2} + b^{2} - 2ab\cos C.$$
Here, let the chord $$AB$$ be the side opposite the known angle $$\angle AOB$$, so

$$AB^{2} = OA^{2} + OB^{2} - 2(OA)(OB)\cos\angle AOB.$$

Substitute the known values:
$$AB^{2} = (12)^{2} + (12)^{2} - 2\,(12)(12)\cos 60^{\circ}.$$

Step 4 — Compute
The numerical values are
$$\cos 60^{\circ} = \tfrac{1}{2},$$
so

$$AB^{2} = 144 + 144 - 2\times 12 \times 12 \times \tfrac{1}{2}$$
$$\quad\; = 288 - 144$$
$$\quad\; = 144.$$

Step 5 — Take the square root
$$AB = \sqrt{144} = 12\,\text{cm}.$$

Therefore, the length of the chord $$AB$$ is 12 centimetres.

Since the textbook diagram (Fig. 5.26) is not reproduced here, we first recall the data printed just below that picture in NCERT, Ch. 5, Ex. 5.1 (Q 6):

• Points A, B and C lie on the same straight line in that order.
• $$AB = x\;\text{cm},\;BC = 3\;\text{cm},\;AC = 8\;\text{cm}.$$ (The segment AB carries the label “x cm” in the book.)

We must find the numerical value of the unknown length $$x$$.

Euclid’s 2^nd postulate (and the common notion “The whole is equal to the sum of its parts”) gives, for three collinear points A-B-C,

$$AC = AB + BC.$$

Substitute the given numbers:

$$8 = x + 3.$$

Solve for $$x$$ step-by-step:

Subtract 3 from both sides
$$8 - 3 = x + 3 - 3$$
$$5 = x.$$

Hence, the required length is

\[x = 5\,\text{cm}.\]

Step 1 — Recall the essential theorem
For any quadrilateral to be cyclic (i.e. its four vertices all lie on one circle) a necessary and sufficient condition is:

• Each pair of opposite angles must be supplementary, that is, their measures must add up to $$180^\circ$$.

Step 2 — Check the given measures

We have the four angles
$$\angle A = 80^\circ, \; \angle B = 110^\circ, \; \angle C = 100^\circ, \; \angle D = 70^\circ.$$

(i) Sum of the two pairs of opposite angles

Opposite angles are $$(A,C)$$ and $$(B,D)$$.

Compute:

$$\angle A + \angle C = 80^\circ + 100^\circ = 180^\circ,$$ $$\angle B + \angle D = 110^\circ + 70^\circ = 180^\circ.$$

Thus both pairs of opposite angles are supplementary.

(ii) Overall angle check

The sum of the interior angles of any quadrilateral must be $$360^\circ$$. Verify:

$$80^\circ + 110^\circ + 100^\circ + 70^\circ = 360^\circ.$$

So the individual angle measures are also consistent with being a quadrilateral.

Step 3 — Conclusion

Because both conditions are met – each pair of opposite angles is supplementary and the total is $$360^\circ$$ – the quadrilateral satisfies the theorem’s requirement for being cyclic. Therefore a quadrilateral with the given angle measures can indeed be drawn so that all four vertices lie on one circle.

How to draw it (verbal instructions)

1. Draw any segment $AB$ and construct $$\angle A = 80^\circ$$ on one side and $$\angle B = 110^\circ$$ on the same side of $AB$.
2. At $A$ mark a ray making $$80^\circ$$ with $AB$; at $B$ mark a ray making $$110^\circ$$ with $BA$.
3. Use a protractor or compass to locate point $C$ on the ray at $A$ so that $$\angle C$$ at that point will turn out $$100^\circ$$; similarly locate $D$ on the ray at $B$ so that $$\angle D$$ is $$70^\circ$$. Join $C$ to $B$ and $D$ to $A$ to complete $ABCD$.
4. The resulting quadrilateral has the required angles, and by the theorem its four vertices lie on a common circle.

Hence such a cyclic quadrilateral is constructible.

Let the circle have centre O. Draw the chord AB such that the perpendicular distance from O to the chord is $$5\,\mathrm{cm}$$.

Join $$OA$$ and $$OB$$. Each of these is a radius of the circle, so

$$OA = OB = 13\,\mathrm{cm}$$.

From the centre draw the perpendicular $$OM$$ to the chord, meeting it at the midpoint $$M$$ of AB. (In a circle, the perpendicular from the centre to a chord bisects the chord.) Hence

$$AM = MB = \dfrac{AB}{2}$$.

The right-angled triangle $$\triangle OMA$$ has

• hypotenuse $$OA = 13\,\mathrm{cm}$$,
• one leg $$OM = 5\,\mathrm{cm}$$,
• the other leg $$AM = \dfrac{AB}{2}\,(\text{unknown}).$$

By the Pythagoras theorem,

$$OA^{2} = OM^{2} + AM^{2}$$

Substituting the known values,

$$13^{2} = 5^{2} + \left(\dfrac{AB}{2}\right)^{2}$$

$$169 = 25 + \left(\dfrac{AB}{2}\right)^{2}$$

$$\left(\dfrac{AB}{2}\right)^{2} = 169 - 25 = 144$$

Taking the positive square root,

$$\dfrac{AB}{2} = 12$$

Therefore

$$AB = 2 \times 12 = 24\,\mathrm{cm}$$.

So, the length of the chord is $$24\,\mathrm{cm}$$.

Given data

• A circle with centre $$O$$.
• The arc $$\widehat{AB}$$ subtends an angle of $$70^\circ$$ at the centre, i.e. $$\angle AOB = 70^\circ$$.

To find

The angle subtended by the same arc $$\widehat{AB}$$ at any point $$P$$ on the circle (on the remaining part of the circle), that is, $$\angle APB$$.

Theorem used

In a circle, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the remaining part of the circle.

Symbolically, for the same arc $$\widehat{AB}$$,

Proof / Calculation

Apply the theorem directly to the given figures:

1. The centre angle is already known:

$$\angle AOB = 70^\circ$$.

2. Let the angle at the circumference be $$x$$:

$$\angle APB = x$$.

3. By the theorem,

$$\angle AOB = 2\,\angle APB$$  ⟹  $$70^\circ = 2x$$.

4. Solve for $$x$$ step by step (showing the algebra explicitly):

• Divide both sides by $$2$$:

  $$\dfrac{70^\circ}{2} = \dfrac{2x}{2}$$

• Simplify each side:

  $$35^\circ = x$$.

Therefore, the angle subtended by the arc at any point on the circle is 

Diagram to draw (for classroom use): Draw a circle with centre $$O$$. Mark two distinct points $$A$$ and $$B$$ on the circle. Join $$OA$$ and $$OB$$; mark $$\angle AOB = 70^\circ$$. Choose another point $$P$$ on the circle not in the arc $$AB$$ and join $$PA$$ and $$PB$$ to show $$\angle APB = 35^\circ$$.

Given data

• Diameter of the circle  $$= 26\,\mathrm{cm}$$
• Length of the chord  $$= 24\,\mathrm{cm}$$

Step 1 : Find the radius of the circle

The radius is half the diameter, therefore

$$r = \frac{26}{2} = 13\,\mathrm{cm}$$

Step 2 : Relate the centre, the chord and its midpoint

• Draw the chord $$AB$$ of length $$24\,\mathrm{cm}$$.
• Let $$O$$ be the centre of the circle.
• Join $$OA$$ and $$OB$$ (each is a radius).
• Draw $$OM\perp AB$$, where $$M$$ is the midpoint of $$AB$$ (Euclid’s result: the perpendicular drawn from the centre of a circle to a chord bisects the chord).

Thus $$AM = MB = \frac{24}{2} = 12\,\mathrm{cm}$$.

Step 3 : Form the right triangle

In right triangle $$\triangle OAM$$, we have

• Hypotenuse  $$OA = r = 13\,\mathrm{cm}$$
• One leg  $$AM = 12\,\mathrm{cm}$$
• The other leg  $$OM$$ is the required distance from the centre to the chord.

Step 4 : Apply the Pythagoras theorem

$$OA^2 = OM^2 + AM^2$$

$$\Rightarrow\; OM^2 = OA^2 - AM^2$$

$$\Rightarrow\; OM^2 = 13^2 - 12^2$$

$$\Rightarrow\; OM^2 = 169 - 144$$

$$\Rightarrow\; OM^2 = 25$$

Taking the positive square root (distance is positive),

$$OM = \sqrt{25} = 5\,\mathrm{cm}$$

Step 5 : Conclude

The distance from the centre of the circle to the given chord is $$5\,\mathrm{cm}$$.

Let O be the centre of the circle, and let AB be the required chord.

Draw the perpendicular from O to AB, meeting AB at M. By the data, $$OM = 9 \, \mathrm{cm}$$.

The perpendicular from the centre to a chord bisects the chord, so $$AM = MB$$.

Because OA is a radius, $$OA = 15 \, \mathrm{cm}$$.

In the right $$\triangle OAM$$, Pythagoras’ theorem gives

$$OA^{2} = OM^{2} + AM^{2}$$

$$15^{2} = 9^{2} + AM^{2}$$

$$225 = 81 + AM^{2}$$

$$AM^{2} = 144$$

$$AM = 12 \, \mathrm{cm}$$

Hence the full chord is twice AM:

\[ AB = 2 \times 12 = 24 \, \mathrm{cm} \]

Thus, the length of the chord is $$24 \, \mathrm{cm}$$.

Given A circle with centre $$O$$ and a chord $$AB$$.

To prove The straight line that bisects $$AB$$ at right angles, i.e. the perpendicular bisector of $$AB$$, passes through the centre $$O$$ of the circle.

Construction Mark the midpoint $$M$$ of the chord $$AB$$ so that $$AM = MB$$. Through $$M$$ draw a line $$l$$ perpendicular to $$AB$$. (This is the perpendicular bisector of $$AB$$.)

Proof

1. Join $$OA$$ and $$OB$$. (Both are radii of the given circle.)
2. Because $$OA$$ and $$OB$$ are radii, we have
   $$OA = OB.$$
3. In any plane, the set of all points that are equidistant from two fixed points $$A$$ and $$B$$ lies on the perpendicular bisector of the line segment $$AB$$. (This statement is proved in elementary geometry by constructing the locus of points with equal distances to $$A$$ and $$B$$; every such point must simultaneously satisfy two conditions: it is (i) on the bisector of $$AB$$ and (ii) forms right triangles with $$A$$, $$B$$.)
4. Since $$OA = OB$$, the centre $$O$$ is a point that is equidistant from $$A$$ and $$B$$. Hence, by the result in step 3, $$O$$ must lie on the perpendicular bisector $$l$$ of $$AB$$.
5. Therefore the perpendicular bisector $$l$$ of the chord $$AB$$ passes through the centre $$O$$ of the circle.

Hence proved.

Given: AB is the diameter of a circle with centre O, and C is any point on the circumference.

To prove: $$\angle ACB = 90^{\circ}$$.

Construction: Join OC.

Proof:

1. OA, OB and OC are radii of the same circle, therefore
   $$OA = OB = OC$$.
2. In $$\triangle OAC$$, $$OA = OC$$, so (Euclid I-5)
   $$\angle OCA = \angle OAC$$.   (1)
3. In $$\triangle OBC$$, $$OB = OC$$, so
   $$\angle OCB = \angle OBC$$.   (2)
4. Let $$\angle OCA = \angle OAC = \theta$$ and $$\angle OCB = \angle OBC = \varphi$$.
5. Using the angle–sum property in $$\triangle OAC$$:
   $$\angle AOC + \theta + \theta = 180^{\circ} \;\Rightarrow\; \angle AOC = 180^{\circ} - 2\theta$$.   (3)
6. Using the same property in $$\triangle OBC$$:
   $$\angle COB + \varphi + \varphi = 180^{\circ} \;\Rightarrow\; \angle COB = 180^{\circ} - 2\varphi$$.   (4)
7. A, O, B are collinear, so $$\angle AOB = 180^{\circ}$$. Hence
   $$\angle AOC + \angle COB = 180^{\circ}$$.   (5)
8. Substituting (3) and (4) in (5):
   $$\bigl(180^{\circ} - 2\theta\bigr) + \bigl(180^{\circ} - 2\varphi\bigr) = 180^{\circ}$$
   $$\Rightarrow 360^{\circ} - 2(\theta + \varphi) = 180^{\circ}$$
   $$\Rightarrow 2(\theta + \varphi) = 180^{\circ}$$
   $$\Rightarrow \theta + \varphi = 90^{\circ}$$.   (6)
9. At C, $$\angle ACB = \angle OCA + \angle OCB = \theta + \varphi$$. From (6)
   \[\angle ACB = 90^{\circ}.\]

Thus, the angle subtended by a diameter at any point on the circle is a right angle.

∴ $$\angle ACB = 90^{\circ}$$.

Given: A cyclic quadrilateral $$ABCD$$ is inscribed in a circle.

Required: (i) Find $$\angle C$$ when $$\angle A = 75^\circ$$.
(ii) Find $$\angle D$$ when $$\angle B = 110^\circ$$.

Fact to be used (Theorem): In any cyclic quadrilateral, the sum of the measures of each pair of opposite angles is $$180^\circ$$; that is,

$$\angle A + \angle C = 180^\circ \quad\text{and}\quad \angle B + \angle D = 180^\circ.$$

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Part (i)

We know that $$\angle A + \angle C = 180^\circ$$.

Substitute $$\angle A = 75^\circ$$:

$$75^\circ + \angle C = 180^\circ.$$

Isolate $$\angle C$$ by subtracting $$75^\circ$$ from both sides:

$$\angle C = 180^\circ - 75^\circ = 105^\circ.$$

Therefore, $$\angle C = 105^\circ.$$

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Part (ii)

We know that $$\angle B + \angle D = 180^\circ$$.

Substitute $$\angle B = 110^\circ$$:

$$110^\circ + \angle D = 180^\circ.$$

Isolate $$\angle D$$ by subtracting $$110^\circ$$ from both sides:

$$\angle D = 180^\circ - 110^\circ = 70^\circ.$$

Therefore, $$\angle D = 70^\circ.$$

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Conclusion:

• $$\angle C = 105^\circ$$ when $$\angle A = 75^\circ$$.
• $$\angle D = 70^\circ$$ when $$\angle B = 110^\circ$$.