Solution

It can be observed that

$\angle A O C=\angle A O B+\angle B O C$

$=60^{\circ}+30^{\circ}$

$=90^{\circ}$

We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.

$ \angle ADC=\frac{1}{2} \angle AOC=\frac{1}{2} \times 90^{\circ}=45^{\circ} $

Solution

In $\triangle OAB$,

$A B=O A=O B=$ radius

$\angle \triangle OAB$ is an equilateral triangle.

Therefore, each interior angle of this triangle will be of $60^{\circ}$. $\angle \angle AOB=60^{\circ}$

$\angle ACB=\frac{1}{2} \angle AOB=\frac{1}{2}(60^{\circ})=30^{\circ}$

In cyclic quadrilateral $A C B D$,

$\angle ACB+\angle ADB=180^{\circ}$ (Opposite angle in cyclic quadrilateral)

$\angle \angle ADB=180^{\circ}-30^{\circ}=150^{\circ}$

Therefore, angle subtended by this chord at a point on the major arc and the minor arc are $30^{\circ}$ and $150^{\circ}$ respectively.

Solution

Consider PR as a chord of the circle.

Take any point $S$ on the major arc of the circle.

PQRS is a cyclic quadrilateral.

$\angle PQR+PSR=180^{\circ}$ (Opposite angles of a cyclic quadrilateral)

$\angle P S SR=180^{\circ}-100^{\circ}=80^{\circ}$

We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. $\angle \angle POR=2 \angle PSR=$ $2(80^{\circ})=160^{\circ}$

In $\triangle POR$,

$OP=OR$ (Radii of the same circle)

$\angle$ OPPR $=$ ORP (Angles opposite to equal sides of a triangle)

$\mathrm{OPR}+\mathrm{ORP}+\mathrm{POR}=180^{\circ}$ (Angle sum property of a triangle)

$\angle \mathrm{OPR}+160^{\circ}=180^{\circ} 2$

$\angle \mathrm{OPR}=180^{\circ}-160^{\circ}=20^{\circ} 2$

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

Solution

$\angle CDE+DCE=\angle CEB$ (Exterior angle)

$\angle C D E+20^{\circ}=130^{\circ}$

$\angle CDE=110^{\circ}$

For chord CD,

$<\quad<$

$\angle CAD=70^{\circ}$

$ \begin{aligned} & \angle \quad \angle \quad B C D+B A=180^{\circ} \text{ (Opposite angles of a } By \\ & +100^{\circ}=180^{\circ} \quad BCD \\ & \angle BCD=80^{\circ} \\ & \text{ In } \triangle ABC \text{, } \\ & AB=BC \text{ (Given) } \\ & \angle \angle BCA=\angle CAB \text{ (Angles opposite to equal sides of a triangle) } \\ & \angle \angle BCA=30^{\circ} \end{aligned} $

$ \angle \quad \angle \quad BCD+BAD=180^{\circ} \text{ (Opposite angles of a cyclic quadrilateral) } \angle $

However, $\angle BAC=\angle CDE$ (Angles in the same segment of a circle)

$\angle \angle BAC=110^{\circ}$

Solution

$BAD=BAC+CAD=30^{\circ}+70^{\circ}=100^{\circ}$

CD + BAD $=180^{\circ}$ (Opposite angles of a cyclic quadrilateral) $\angle$

We have, $B^{\prime} C D=80^{\circ}$

$ \begin{aligned} & \angle B^{\prime} C A+A C D=80^{\circ} \\ & \circ+A C D=80^{\circ} 30 \quad A C D \\ & \angle=\angle 50^{\circ} \\ & \angle \angle \\ & E C D=50^{\circ} \end{aligned} $

Solution

Let $A B C D$ be a cyclic quadrilateral having diagonals $B D$ and $A C$, intersecting each other at point $O$.

$ \angle BAD=\frac{1}{2} \angle BOD=\frac{180^{\circ}}{2}=90^{\circ} $

$ \begin{aligned} & \angle BCD+\angle BAD=180^{\circ} \text{ (Cyclic quadrilateral) (Consider } BD \text{ as a chord) } \\ & \angle BCD=180^{\circ}-90^{\circ}=90^{\circ} \\ & \angle ADC=\frac{1}{2} \angle AOC=\frac{1}{2}(180^{\circ})=90^{\circ} \\ & \angle ADC+\angle ABC=180^{\circ}(\text{ Cyclic quadrilateral })^{\circ} \\ & +\angle ABC=180^{\circ} 90 \end{aligned} $

$\angle A B C=90^{\circ}$

(Considering $AC$ as a chord)

Each interior angle of a cyclic quadrilateral is of $90^{\circ}$. Hence, it is a rectangle.

Solution

Consider a trapezium $A B C D$ with $A B|| C D$ and $B C=A D$.

Draw $AM \angle CD$ and $BN \angle CD$.

In $\triangle A M D$ and $\triangle B N C$,

$A D=B C$ (Given)

$\angle AMD=\angle BNC(.$ By construction, each is $.90^{\circ})$

$AM=BM$ (Perpendicular distance between two parallel lines is same)

$\angle \triangle AMD \triangle BNC$ (RHS congruence rule)

$\angle A D C=B C \mathscr{C}(C P C T) \ldots$

${ }^{\angle} B A D$ and $A D C$ are on the same side of transversal AD.

$\angle BAD+\stackrel{\angle ADC}{ }=180^{\circ}$.

$\angle \quad \angle$

$B A D+B C D=180^{\circ}[$ Using equation (1)]

This equation shows that the opposite angles are supplementary.

Therefore, $A B C D$ is a cyclic quadrilateral.

Solution

Join chords AP and DQ. $\quad \angle \angle$

For chord AP,

$\angle PBA=\angle ACP$ (Angles in the same segment) …

For chord DQ,

$\angle DBQ=\Phi CD$ (Angles in the same segment) … (2) ABD and

PBQ are line segments intersecting at $B$.

$\angle PBA=$ DBQ (Vertically opposite angles) … (

From equations ( 1 ), ( 2 ), and ( 3 ), we obtain $\angle A C P$

$=\angle QCD$

Solution

Consider a $\triangle ABC$.

Two circles are drawn while taking $A B$ and $A C$ as the diameter.

Let they intersect each other at $D$ and let $D$ not lie on $B C$.

Join AD.

$\angle ADB=90^{\circ}$ (Angle subtended by semi-circle)

$\angle ADC=90^{\circ}$ (Angle subtended by semi-circle)

$\angle$

$ BDC=\angle ADB+\angle ADC=90^{\circ}+90^{\circ}=180^{\circ} $

Therefore, BDC is a straight line and hence, our assumption was wrong.

Thus, Point $D$ lies on third side $B C$ of $\triangle A B C$.

Solution

In $\triangle ABC$,

$«<$

$\angle 90^{\circ}+\angle BCA+\angle CAB=180^{\circ}$

$\angle \angle BCA+\angle CAB=90^{\circ} \ldots$ (1) $ABC+BCA+CAB=180^{\circ}$ (Angle sum property of a triangle)

In $\triangle ADC$,

$\angle C D A+A C D+D A C=180^{\circ}$ (Angle sum property of a triangle)

$\angle 90^{\circ}+A^{\prime} C D+D^{\prime} C=180^{\circ}$

${ }^{\angle} A C D+D A C=90^{\circ}$

Adding equations (1) and (2), we obtain

$ BCA+CAB+ACD+DAC=180^{\circ} $

$\angle(BCA+\stackrel{\angle}{\angle} \angle ACD)+(CAB+DAC)=180^{\circ}$

$$ \begin{equation*} BCD+DAB=180^{\circ} \tag{3} \end{equation*} $$

However, it is given that

$\angle B+D=90^{\circ}+90^{\circ}=180^{\circ} \ldots$

From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral $A B C D$ is $180^{\circ}$. Therefore, it is a cyclic quadrilateral.

Consider chord CD.

$\angle CAD=\angle CBD$ (Angles in the same segment)

Solution

Let $A B C D$ be a cyclic parallelogram. $\angle<$ A $+C=180^{\circ}$ (Opposite angles of a cyclic quadrilateral) … (1)

$\angle A=\angle C$ and $\angle B=\angle D$ We know that opposite angles of a parallelogram are equal. $\angle$

From equation (1), $\angle A+\angle C=180^{\circ}$

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

$\angle 2 \angle A=180^{\circ}$

$\angle \angle A=90^{\circ}$

Parallelogram $ABCD$ has one of its interior angles as $90^{\circ}$. Therefore, it is a rectangle.