Exercise-9.3, Class 9th, Maths, Chapter 9, NCERT

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.