1.

Given: Points $A,B,C$ lie on a circle with centre $O$. $\mathrm{\angle }BOC={30}^{\circ }$ and $\mathrm{\angle }AOB={60}^{\circ }$. $D$ is a point on the circle other than the arc $ABC$.
Find: $\mathrm{\angle }ADC$

Solution:
The central angle subtending arc $AC$ equals $\mathrm{\angle }AOC=\mathrm{\angle }AOB+\mathrm{\angle }BOC={60}^{\circ }+{30}^{\circ }={90}^{\circ }$. An inscribed angle subtending the same arc is half the central angle. So

$$\mathrm{\angle }ADC=\frac{1}{2}\mathrm{\angle }AOC=\frac{1}{2}\cdot {90}^{\circ }={45}^{\circ }\mathrm{.}$$

Answer: $\boxed{{\displaystyle {45}^{\circ }}}$

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2.

Problem: A chord of a circle equals the radius. Find the angle subtended by that chord at (a) a point on the minor arc, and (b) a point on the major arc.

Solution:
Let the circle have radius $R$. If a chord has length $R$ and subtends central angle $\theta$, then

$$\text{chord length}=2R\sin \frac{\theta }{2}=R\Rightarrow \sin \frac{\theta }{2}=\frac{1}{2}\mathrm{.}$$

So $\frac{\theta }{2}={30}^{\circ }$ (taking the acute value), hence $\theta ={60}^{\circ }$.

• At a point on the minor arc the inscribed angle equals half the central angle: $\frac{1}{2}\theta ={30}^{\circ }$.

• At a point on the major arc the inscribed angle subtends the reflex arc ${360}^{\circ }-\theta$, so its measure is $\frac{1}{2}({360}^{\circ }-\theta )={180}^{\circ }-\frac{\theta }{2}={180}^{\circ }-{30}^{\circ }={150}^{\circ }$

Answer: Minor arc: $\boxed{{\displaystyle {30}^{\circ }}}$. Major arc: $\boxed{{\displaystyle {150}^{\circ }}}$

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3.

Given (Fig. 9.24): $P,Q,R$ lie on a circle with centre $O$ and $\mathrm{\angle }PQR={100}^{\circ }$.
Find: $\mathrm{\angle }OPR$.

Solution:
The inscribed angle $\mathrm{\angle }PQR$ subtends arc $PR$. So the central angle subtending the same arc (the reflex or full measure) is $2\times {100}^{\circ }={200}^{\circ }$. The interior central angle between radii $OP$ and $OR$ is the smaller one, i.e. ${360}^{\circ }-{200}^{\circ }={160}^{\circ }$. In isosceles triangle $\mathrm{△}OPR$ (since $OP=OR$), the base angles at $P$ and $R$ are equal and each is

$$\frac{{180}^{\circ }-{160}^{\circ }}{2}=\frac{{20}^{\circ }}{2}={10}^{\circ }\mathrm{.}$$

Thus $\mathrm{\angle }OPR={10}^{\circ }$

Answer: $\boxed{{\displaystyle {10}^{\circ }}}$

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4.

Given (Fig. 9.25): $A,B,C,D$ are on a circle and $\mathrm{\angle }ABC={69}^{\circ },\mathrm{\angle }ACB={31}^{\circ }$
Find: $\mathrm{\angle }BDC$

Solution:
In $\mathrm{△}ABC$ the angle at $A$ is

$$\mathrm{\angle }BAC={180}^{\circ }-{69}^{\circ }-{31}^{\circ }={80}^{\circ }\mathrm{.}$$

Both $\mathrm{\angle }BAC$ and $\mathrm{\angle }BDC$ are inscribed angles that subtend the same arc $BC$. Angles in the same segment are equal. Hence

$$\mathrm{\angle }BDC=\mathrm{\angle }BAC={80}^{\circ }\mathrm{.}$$

Answer: $\boxed{{\displaystyle {80}^{\circ }}}$

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5.

Given (Fig. 9.26): $A,B,C,D$ are on a circle. Chords $AC$ and $BD$ meet at $E$ inside the circle. $\mathrm{\angle }BEC={130}^{\circ }$ and $\mathrm{\angle }ECD={20}^{\circ }$.
Find: $\mathrm{\angle }BAC$.

Solution:
Note $E$ lies on chord $AC$, so ray $CE$ lies along chord $CA$. Thus $\mathrm{\angle }ECD=\mathrm{\angle }ACD$. Since $\mathrm{\angle }ACD$ is an inscribed angle subtending arc $AD$,

$$\text{arc }AD=2\mathrm{\angle }ACD=2\times {20}^{\circ }={40}^{\circ }\mathrm{.}$$

For two chords intersecting inside a circle the angle between them is half the sum of the measures of the arcs intercepted by the angle and its vertical angle. Thus

$$\mathrm{\angle }BEC=\frac{1}{2}(\text{arc }BC+\text{arc }AD)\mathrm{.}$$

Given $\mathrm{\angle }BEC={130}^{\circ }$ and $\text{arc }AD={40}^{\circ }$, we get

$${130}^{\circ }=\frac{1}{2}(\text{arc }BC+{40}^{\circ })\Rightarrow \text{arc }BC={260}^{\circ }-{40}^{\circ }={220}^{\circ }\mathrm{}$$

The inscribed angle $\mathrm{\angle }BAC$ subtends arc $BC$, so

$$\mathrm{\angle }BAC=\frac{1}{2}\text{arc }BC=\frac{1}{2}\cdot {220}^{\circ }={110}^{\circ }\mathrm{}$$

Answer: $\boxed{{\displaystyle {110}^{\circ }}}$

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6.

Given: $ABCD$ is cyclic; diagonals meet at $E$. $\mathrm{\angle }DBC={70}^{\circ }$ and $\mathrm{\angle }BAC={30}^{\circ }$.
(a) Find $\mathrm{\angle }BCD$.
(b) If $AB=BC$, find $\mathrm{\angle }ECD$.

Solution (a):
$\mathrm{\angle }BAC={30}^{\circ }$ is an inscribed angle subtending arc $BC$, so arc $BC={60}^{\circ }$. $\mathrm{\angle }DBC={70}^{\circ }$is an angle at $B$ subtending arc $DC$, so arc $DC={140}^{\circ }$. Assume vertices are in order $A\text{\hspace{0.17em}⁣}-\text{\hspace{0.17em}⁣}B\text{\hspace{0.17em}⁣}-\text{\hspace{0.17em}⁣}C\text{\hspace{0.17em}⁣}-\text{\hspace{0.17em}⁣}D$ around the circle; then the arc $BD$ that does not contain $C$ equals arc $BA+AD={360}^{\circ }-(\text{arc }BC+\text{arc }CD)={360}^{\circ }-({60}^{\circ }+{140}^{\circ })={160}^{\circ }$. The angle $\mathrm{\angle }BCD$ is an inscribed angle that subtends arc $BD$ not containing $C$, so

$$\mathrm{\angle }BCD=\frac{1}{2}\cdot {160}^{\circ }={80}^{\circ }\mathrm{}$$

Answer (a): $\boxed{{\displaystyle {80}^{\circ }}}$

Solution (b): If $AB=BC$, then chord $AB$ equals chord $BC$ so their arcs are equal: arc $AB=$ arc $BC={60}^{\circ }$. The remaining arc $AD$ (between $A$ and $D$) equals ${360}^{\circ }-(\text{arc }AB+\text{arc }BC+\text{arc }CD)={360}^{\circ }-(60+60+140)={100}^{\circ }$Angle $\mathrm{\angle }ECD$ has vertex at $C$ and ray $CE$ lies along $CA$ (since $E$ is intersection of diagonals), so $\mathrm{\angle }ECD=\mathrm{\angle }ACD$, an inscribed angle subtending arc $AD$. Thus

$$\mathrm{\angle }ECD=\frac{1}{2}\cdot \text{arc }AD=\frac{1}{2}\cdot {100}^{\circ }={50}^{\circ }\mathrm{.}$$

Answer (b): $\boxed{{\displaystyle {50}^{\circ }}}$

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7.

Problem: If the diagonals of a cyclic quadrilateral are diameters of the circumcircle, prove the quadrilateral is a rectangle.

Solution:
Let the cyclic quadrilateral be $ABCD$ and suppose its diagonals $AC$ and $BD$ are diameters of the circle. Any angle subtended by a diameter is a right angle (angle in a semicircle). Hence

$$\mathrm{\angle }ABC(\text{subtends diameter }AC)={90}^{\circ },$$

$$\mathrm{\angle }CDA(\text{subtends diameter }AC)={90}^{\circ },$$

and similarly the other two angles subtend the other diameter so all interior angles are ${90}^{\circ }$. A quadrilateral with all angles right is a rectangle.

Answer: It is a rectangle.

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8.

Problem: If the non-parallel sides of a trapezium are equal, prove the trapezium is cyclic.

Solution:
Let trapezium $ABCD$ have $AB\parallel CD$ and non-parallel sides $AD=BC$. By standard isosceles trapezium arguments (use congruence of appropriate triangles formed by extending a side or constructing a suitable parallel), one shows $\mathrm{\angle }A=\mathrm{\angle }B$ and $\mathrm{\angle }C=\mathrm{\angle }D$. But since $AB\parallel CD$, consecutive interior angles on a transversal satisfy $\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$. Hence

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

Thus a pair of opposite angles sum to ${180}^{\circ }$, so the trapezium is cyclic (a quadrilateral is cyclic iff a pair of opposite angles sum to ${180}^{\circ }$).

Answer: The trapezium is cyclic.

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9.

Given (Fig. 9.27): Two circles meet at $B$ and $C$. Through $B$ draw a line meeting the first circle again at $A$ and the second at $D$. Another line through $B$ meets the first circle again at $P$ and the second at $Q$.
Prove: $\mathrm{\angle }ACP=\mathrm{\angle }QCD$.

Solution (clear synthetic argument):
Label the two lines through $B$ as ${\mathrm{\ell }}_1$ (containing $A,B,D$) and ${\mathrm{\ell }}_2$ (containing $P,B,Q$). Note:

• Points $A,B,P,C$ lie on the first circle, so $\mathrm{\angle }ACP$ (angle at $C$ formed by $CA$ and $CP$) is an inscribed angle subtending chord $AP$. By the “angles in the same segment” property,

$$\mathrm{\angle }ACP=\mathrm{\angle }ABP\text{(both subtend chord }AP\text{ on the first circle).}$$

• Points $Q,B,D,C$ lie on the second circle, so $\mathrm{\angle }QCD$ (angle at $C$ formed by $CQ$ and $CD$) is also an inscribed angle subtending chord $QD$. Thus

$$\mathrm{\angle }QCD=\mathrm{\angle }QBD\text{(both subtend chord }QD\text{ on the second circle).}$$

Now observe that $\mathrm{\angle }ABP$ and $\mathrm{\angle }QBD$ are vertically opposite or supplementary pairs produced by the two straight lines ${\mathrm{\ell }}_1$ and ${\mathrm{\ell }}_2$ through $B$. In fact, $BA$ and $BD$ are the same straight line (opposite directions of ${\mathrm{\ell }}_1$), and $BP$ and $BQ$are the same straight line (opposite directions of ${\mathrm{\ell }}_2$), so the angle between $BA$ and $BP$equals the angle between $DB$ and $BQ$. Hence

$$\mathrm{\angle }ABP=\mathrm{\angle }QBD\mathrm{.}$$

Combining the equalities gives $\mathrm{\angle }ACP=\mathrm{\angle }QCD$, as required.

Answer: $\boxed{{\displaystyle \mathrm{\angle }ACP=\mathrm{\angle }QCD}}$

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10.

Problem: If circles are drawn with two sides of a triangle as diameters, prove that the point(s) of intersection of these circles lie on the third side.

Solution:
Let triangle be $\mathrm{△}ABC$. Draw the circle with diameter $AB$ and the circle with diameter $AC$. Any point $P$common to both circles (other than $A$) satisfies

$$\mathrm{\angle }APB={90}^{\circ }\text{(since }AB\text{ is diameter)}\text{and}$$

$$\mathrm{\angle }APC={90}^{\circ }\text{(since }AC\text{ is diameter).}$$

So $\mathrm{\angle }BPC={180}^{\circ }-(\mathrm{\angle }APB+\mathrm{\angle }APC)={180}^{\circ }-{90}^{\circ }-{90}^{\circ }=0^{\circ }$, which means points $B,P,C$ are collinear. Thus intersection point(s) $P$ lie on line $BC$.

Answer: The intersection point(s) lie on the third side $BC$.

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11.

Problem: $ABC$ and $ADC$ are right triangles with common hypotenuse $AC$. Prove that $\mathrm{\angle }CAD=\mathrm{\angle }CBD$.

Solution:
Because both $\mathrm{△}ABC$ and $\mathrm{△}ADC$ are right triangles with hypotenuse $AC$, all four points $A,B,C,D$ lie on the circle with diameter $AC$ (angle in a semicircle is ${90}^{\circ }$). Thus quadrilateral $ABCD$ is cyclic. The angles $\mathrm{\angle }CAD$ and $\mathrm{\angle }CBD$ subtend the same arc $CD$, so by the “angles in the same segment” theorem,

$$\mathrm{\angle }CAD=\mathrm{\angle }CBD\mathrm{}$$

Answer: $\boxed{{\displaystyle \mathrm{\angle }CAD=\mathrm{\angle }CBD}}$

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12.

Problem: Prove that a cyclic parallelogram is a rectangle.

Solution:
Let $ABCD$ be a parallelogram which is cyclic. In a parallelogram opposite angles are equal: $\mathrm{\angle }A=\mathrm{\angle }C$ and $\mathrm{\angle }B=\mathrm{\angle }D$. In a cyclic quadrilateral opposite angles are supplementary: $\mathrm{\angle }A+\mathrm{\angle }C={180}^{\circ }$. Combining these two facts gives $2\mathrm{\angle }A={180}^{\circ }\Rightarrow \mathrm{\angle }A={90}^{\circ }$. Hence each interior angle is ${90}^{\circ }$; therefore the parallelogram is a rectangle.

Answer: A cyclic parallelogram is a rectangle.

Q1

Problem. Two circles of radii $5$ cm and $3$ cm intersect at two points and the distance between their centres is $4$ cm. Find the length of the common chord.

Solution.
Let the centres be $O_1$ (radius $R_1=5$) and $O_2$ (radius $R_2=3$), and the distance $O_1{O}_2=d=4$. Let the common chord be $XY$. Let the perpendicular from $O_1$ to chord $XY$ meet it at $M$. Let $O_{1}M=x$. Then for the two right triangles:

$$\{\begin{array}{l}{x}^2+{\left(\frac{XY}{2}\right)}^2=R_1^2=25\\ (d-x{)}^2+{\left(\frac{XY}{2}\right)}^2=R_2^2=9\end{array}$$

Subtract the second from the first:

$$25-9=x^2-(d-x{)}^2=2dx-d^2\mathrm{}$$

So

$$16=2\cdot 4\cdot x-16\Rightarrow 16=8x-16$$
$$\Rightarrow 8x=32\Rightarrow x=4$$

Then half–chord length $\frac{XY}{2}=\sqrt{25-x^2}=\sqrt{25-16}=\sqrt{9}=3$
Hence the common chord length $XY=2\times 3=\boxed{{\displaystyle 6\text{ cm}}}\mathrm{<span\; class="mord"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist"><span\; class="boxpad"><span\; class="mspace"> </span><span\; class="mord\; text">cm</span></span></span><span\; class="vlist-s"></span></span></span></span><span\; class="mord"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

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Q2

Problem. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to the corresponding segments of the other chord.

Solution.
Let chords $AB$ and $CD$ be equal and intersect at $E$. Put $AE=x,EB=s-x$ where $s=AB$. Put $CE=y,ED=s-y$ (since $CD=AB=s$). By the intersecting-chords theorem (power of a point):

$$AE\cdot EB=CE\cdot ED\Rightarrow x(s-x)=y(s-y)\mathrm{}$$

Rearrange:

$$(x-y)(s-(x+y))=0.$$

The factor $s-(x+y)$ would be zero only if $x+y=s$, i.e. $AE+CE=AB$, which would place $C$ on segment $AB$ (not possible for two distinct chords intersecting inside). Hence $x-y=0$, so $x=y$. Therefore

$$AE=CE\text{and similarly}EB=ED\mathrm{}$$

So corresponding segments are equal.

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Q3

Problem. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the two chords.

Solution.
With the notation of Q2, let $O$ be the circle’s centre and $AE=CE$ and $BE=DE$ (from Q2). Consider triangles $\mathrm{△}OAE$ and $\mathrm{△}OCE$. They have

$$OA=OC(\text{radii}),AE=CE,OE=OE\mathrm{}$$

So $\mathrm{△}OAE\cong \mathrm{△}OCE$ (SSS). Corresponding angles at $E$ are equal:

$$\mathrm{\angle }OEA=\mathrm{\angle }OEC\mathrm{.}$$

Thus the line $OE$ makes equal angles with chords $AB$ and $CD$(Similarly $\mathrm{\angle }OEB=\mathrm{\angle }OED$.)

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Q4

Problem. If a line intersects two concentric circles (same centre $O$) at $A,B,C,D$ (in that order along the line), prove that $AB=CD$. (See Fig. 9.12.)

Solution.
Let the line meet the two concentric circles so that along the line, from left to right, the points are $A$ (outer circle), $B$(inner circle), $O$ (centre), $C$ (inner circle), $D$ (outer circle). Denote outer radius $R$ and inner radius $r$. Then

$$AB=OB-OA=(\text{distance }O\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}-\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}B)-(\text{distance }O\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}-\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}A)\mathrm{.}$$

But because of symmetry about $O$ the distances satisfy $OA=OD=R$ and $OB=OC=r$ (on appropriate signed scale). Hence

$$AB=OA-OB=R-r,$$

and

$$CD=OD-OC=R-r\mathrm{.}$$

Therefore $AB=CD$. (Geometrically: the two segments from outer to inner circle on opposite sides are equal by symmetry about $O$.)

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Q5

Problem. Three girls Reshma, Salma and Mandip stand on a circle of radius $5$ m. Reshma–Salma distance $=6$, Salma– Mandip distance $=6$. Find the distance between Reshma and Mandip.

Solution.
All three lie on the same circle radius $R=5$. Chord length $L$ corresponding to a central angle $\theta$ satisfies

$$L=2R\sin \frac{\theta }{2}\mathrm{}$$

Given chord $=6$, so

$$6=2\cdot 5\sin \frac{\theta }{2}\Rightarrow \sin \frac{\theta }{2}=\frac{6}{10}=\mathrm{0.6.}$$

Thus $\cos \frac{\theta }{2}=\sqrt{1-{0.6}^2}=\sqrt{0.64}=0.8$Then

$$\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=2\cdot 0.6\cdot 0.8=\mathrm{0.96.}$$

The triangle formed by the three girls has two equal sides (6,6) so the central angles subtending those equal chords are equal (call each $\theta$). The remaining central angle is ${360}^{\circ }-2\theta$. The chord between Reshma and Mandip corresponds to central angle ${360}^{\circ }-2\theta$; its length is

$$\text{RM}=2R\sin \frac{{360}^{\circ }-2\theta }{2}=2R\sin ({180}^{\circ }-\theta )=2R\sin \theta \mathrm{.}$$

So

$$\text{RM}=2\cdot 5\cdot 0.96=10\cdot 0.96=\boxed{{\displaystyle 9.6\text{ m}}}\mathrm{}$$

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Q6

Problem. A circular park has radius $20$. Three boys sit at equal distances on its boundary (i.e., they are vertices of an inscribed equilateral triangle). Find the length of the string of each toy telephone (distance between any two boys).

Solution.
Three equal points on a circle divide the circumference into three equal arcs; each central angle is ${120}^{\circ }$. The chord length for central angle ${120}^{\circ }$ is

$$L=2R\sin \frac{{120}^{\circ }}{2}=2R\sin {60}^{\circ }=2R\cdot \frac{\sqrt{3}}{2}=R\sqrt{3}\mathrm{}$$

With $R=20$, string length $=20\sqrt{3}\text{ m}\approx 34.64\text{ m}\mathrm{}$
So exact answer: $\boxed{{\displaystyle 20\sqrt{3}\text{ m}}}$