NCERT Solutions Class 9 Maths Chapter 9 - Circles

In your daily life experiences, you might have noticed some objects having a round shape as in wheels of cars or other vehicles, bangles, the surfaces of clocks, certain coins like the one and five rupee coins or buttons of shirts. The mentioned shapes are examples of circles and that is why we shall focus on the topic of circles covered in the Class 9 maths syllabus chapter 9, which covers definition and properties of a circle. This chapter also presents useful theorems one of which states that a perpendicular from the centre of a circle to a chord will divide the chord into two equal parts while the other states that the chords of a circle which are equal in length lie at an equal distance from the centre of the circle. 

In this article, you will get the NCERT Solutions Class 9 Maths Chapter 9 Circle PDF which is made by the best of ALLEN's experts in depth for your ease in solving the questions. In addition to that, these exercises that are necessary for the completion of Class 9 examinations will also help you develop your logical and structural reasoning skills and geometric problem solving skills as well.

1.0Download NCERT Solutions Class 9 Maths Chapter 9 Circles

Download NCERT Solutions for Class 9 Maths Chapter 9 Circles in a free and easy-to-access PDF format. These solutions provide step-by-step explanations for all exercises, including key concepts like the properties of circles, tangents, and angles subtended by chords.

2.0NCERT Solutions Class 9 Maths Chapter 9 - Circles: All Exercises

3.0What Will Students Learn in Chapter 9 - Circles?

• Circles have a definition and fundamental concepts that constitute them.
• Key terms like radius, diameter, chord, arc, and circumference must be understood.
• Important theorems, such as:
• A jump from the center to any chord fails to lose its focus and accomplishes two things, namely, it is perpendicular and it bisects the chord.
• Two chords of equal length subtended from the center of a circle will be at the same distance from the center.
• What approach to take when these particular components, chords, arcs, and even tangents are applied in the context of a problem.
• Use of theorems to establish and justify relationships between figures contained in circles.

4.0NCERT Questions with Solutions for Class 9 Maths Chapter 9 - Detailed Solutions

Exercise : 9.1

• Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. Sol. Given : Two congruent circles $\mathrm{C}(0,\mathrm{r})$ and $\mathrm{C}\left({\mathrm{O}}^{\prime },\mathrm{r}\right)$ which have chords AB and CD respectively such that $\mathrm{AB}=\mathrm{CD}$.
  
  To prove : $\angle \mathrm{AOB}=\angle {\mathrm{CO}}^{\prime }\mathrm{D}$ Proof : From $△\mathrm{AOB}$ and $△{\mathrm{CO}}^{\prime }\mathrm{D}$, we have $\mathrm{AB}=\mathrm{CD}$ [Given] $\mathrm{OA}={\mathrm{O}}^{\prime }\mathrm{C}$ [Each equal to r] OB = O'D [Each equal to r] $\therefore △\mathrm{AOB}\cong △{\mathrm{CO}}^{\prime }\mathrm{D}$ [By SSS-congruence] $\Rightarrow \angle \mathrm{AOB}=\angle {\mathrm{CO}}^{\prime }\mathrm{D}[$ C.P.C.T. $]$
• Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. Sol. Given : Two congruent circle $\mathrm{C}(\mathrm{O},\mathrm{r})$ and $\mathrm{C}\left({\mathrm{O}}^{\prime },\mathrm{r}\right)$ which have chords AB and CD respectively, such that $\angle \mathrm{AOB}=\angle {\mathrm{CO}}^{\prime }\mathrm{D}$
  
  To prove : $AB=CD$ Proof : In $△\mathrm{AOB}$ and $△{\mathrm{CO}}^{\prime }\mathrm{D}$, we have :

Exercise : 9.2

• 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 common chord. Sol. We know that if two circles intersect each other at two points, then the line joining their centres is the perpendicular bisector of their common chord.
  
  In $△{\mathrm{PO}}^{\prime }0$ ${\mathrm{OP}}^2={\mathrm{OO}}^{\prime 2}+{\mathrm{PO}}^{\prime 2}$ $5^2=4^2+{\mathrm{PO}}^{\prime 2}$ $\Rightarrow {\mathrm{PO}}^{\prime }=3\mathrm{cm}$ $\therefore$ Length of the common chord $\Rightarrow \mathrm{PQ}=20^{\prime }\mathrm{P}=2\times 3=6\mathrm{cm}$
• If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. Sol. 0 is the centre of the circle. Chords AB and CD of the circle are equal. $P$ is the point of intersection of AB and CD . Join OP , draw $\mathrm{OL}\perp \mathrm{AB}$ and $\mathrm{OM}\perp \mathrm{OD}$. Here, we find $OL=OM$ $(\because \mathrm{AB}=\mathrm{CD})$ In $△\mathrm{OLP}$ and $△\mathrm{OMP}$,
  
  $\mathrm{OL}=\mathrm{OM}$ (By 1) $\mathrm{OP}=\mathrm{OP}$ (Common hypotenuse) $\angle \mathrm{OLP}=\angle \mathrm{OMP}$ (Each $=90^{\circ }$ ) Then we have $\Delta \mathrm{OLP}\cong \Delta \mathrm{OMP}$ (RHS congruence) By CPCT, PL = PM Now, $\mathrm{AL}=\mathrm{BL}=1/2\mathrm{AB}$ $\begin{array}{rl}& CM=DM=1/2CD\\ \Rightarrow & AL=CM(\because AB=CD)\end{array}$ and $\mathrm{BL}=\mathrm{DM}$ Subtracting (2) from (3), $\begin{array}{rl}& \mathrm{AL}-\mathrm{PL}=\mathrm{CM}-\mathrm{PM}\\ \Rightarrow & \mathrm{AP}=\mathrm{CP}\end{array}$ Adding (2) and (4), $\mathrm{PL}+\mathrm{BL}=\mathrm{PM}+\mathrm{DM}\Rightarrow \mathrm{PB}=\mathrm{PD}$
• 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 chords. Sol.
  
  Ois the centre of the circle. Chords AB and CD of the circle are equal. $P$ is the point of intersection of AB and CD . Join OP , draw $\mathrm{OL}\perp \mathrm{AB}$ and $\mathrm{OM}\perp \mathrm{CD}$. Here, we find. $\mathrm{OL}=\mathrm{OM}(\because \mathrm{AB}=\mathrm{CD})$ In $△\mathrm{OLP}$ and $△\mathrm{OMP}$, OL $=\mathrm{OM}$ (By 1) $\mathrm{OP}=\mathrm{OP}$ (Common hypotenuse) $\angle \mathrm{OLP}=\angle \mathrm{OMP}$ (Each $=90^{\circ }$ ) Then we have $\Delta \mathrm{OLP}\cong △\mathrm{OMP}$ (RHS congruence) By CPCT, $\angle \mathrm{OPL}=\angle \mathrm{OPM}$
• If a line intersects two concentric circles (circles with the same centre) with centre O at $\mathrm{A},\mathrm{B},\mathrm{C}$ and D , prove that $\mathrm{AB}=\mathrm{CD}$ (see fig).
  
  Sol. Given : Two circles with the common centre 0 . A line " $\ell$ " intersects the outer circle at A and D and the inner circle at B and C .
  
  To prove: $AB=CD$ Construction: Draw $\mathrm{OM}\perp \ell$. Proof: $\mathrm{OM}\perp \ell$ [Construction] For the outer circle, $\therefore AM=MD$ [Perpendicular from the centre bisects the chord] For the inner circle, $\begin{array}{ll}& \mathrm{OM}\perp \ell \text{ [Construction] }\\ \therefore & \mathrm{BM}=\mathrm{MC}\end{array}$ [Perpendicular from the centre to the chord bisects the chord] Subtracting (2) from (1), we have $\Rightarrow \mathrm{AM}-\mathrm{BM}=\mathrm{MD}-\mathrm{MC}$ $\Rightarrow AB=CD$
• Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip? Sol. We draw $\mathrm{SN}\perp \mathrm{RM}$ as shown in figure. Since, $△\mathrm{RSM}$ is an isosceles triangle hence SN bisects RM and also SN (produced) passes through the centre 0 .
  
  Put RN $=\mathrm{x}$ Then $\mathrm{ar}(△\mathrm{ORS})=\frac{1}{2}\times \mathrm{OS}\times \mathrm{RN}$ $=\frac{1}{2}\times 5\times \mathrm{x}(\because \mathrm{OS}=\mathrm{OR}=5\mathrm{m})$ i.e., $\mathrm{ar}(\Delta \mathrm{ORS})=\frac{5}{2}\mathrm{x}$ Now, draw $\mathrm{OP}\perp \mathrm{RS},\mathrm{P}$ is mid-point of RS . $\Rightarrow \mathrm{PR}=\mathrm{PS}=3\mathrm{m}$ $\Rightarrow {\mathrm{OP}}^2=(5{)}^2-(3{)}^2=16$ $\Rightarrow \mathrm{OP}=4\mathrm{m}$ Here, $\mathrm{ar}(△\mathrm{ORS})=\frac{1}{2}\times \mathrm{RS}\times \mathrm{OP}=\frac{1}{2}\times 6\times 4$ i.e., $\mathrm{ar}(△\mathrm{ORS})=12{\mathrm{m}}^2$ From (1) and (2), $\frac{5}{2}x=12\Rightarrow x=4.8\mathrm{m}$ $\Rightarrow \mathrm{RM}=2\mathrm{x}=2\times 4.8\mathrm{m}$ $\Rightarrow \mathrm{RM}=9.6\mathrm{m}$ Thus, distance between Reshma and Mandip is 9.6 m .
• A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. Sol. Let Ankur, Syed and David are sitting at A, $S$ and $D$ respectively such that $AS=SD=AD$ i.e., $△\mathrm{ASD}$ is an equilateral triangle. Let the length of each side of the equilateral triangle be $2x$ metres.
  
  Draw $\mathrm{AM}\perp \mathrm{SD}$. Since, $△\mathrm{ASD}$ is an equilateral triangle, $\therefore$ So, altitude AM divides SD into two equal parts. Also, $\mathrm{OM}\perp \mathrm{SD}$. $\Rightarrow$ OM bisects SP also. So, we can say AM passes through 0 . $\Rightarrow \mathrm{SM}=\frac{1}{2}\mathrm{SD}=\frac{1}{2}(2\mathrm{x})=\mathrm{x}$ Now, in $△\mathrm{ASM}$, we have ${\mathrm{AM}}^2+{\mathrm{SM}}^2={\mathrm{AS}}^2$ $\Rightarrow {\mathrm{AM}}^2={\mathrm{AS}}^2-{\mathrm{SM}}^2$ $=(2\mathrm{x}{)}^2-{\mathrm{x}}^2=4{\mathrm{x}}^2-{\mathrm{x}}^2=3{\mathrm{x}}^2$ $\Rightarrow \mathrm{AM}=\sqrt{3}\mathrm{x}$ Now, $\mathrm{OM}=\mathrm{AM}-\mathrm{OA}=(\sqrt{3}\mathrm{x}-20)\mathrm{m}$ $\Rightarrow (OS=OA=20\mathrm{cm})$ In $△\mathrm{OSM}$ $\Rightarrow (20{)}^2=x^2+(\sqrt{3}x-20{)}^2$ $\Rightarrow 400=x^2+3x^2-40\sqrt{3}x+400$ $\Rightarrow 4x^2=40\sqrt{3}x$ $\Rightarrow 4\mathrm{x}=40\sqrt{3}$ $\Rightarrow x=10\sqrt{3}m$ Now, SD $=2\mathrm{x}=2\times 10\sqrt{3}\mathrm{m}=20\sqrt{3}\mathrm{m}$ Thus, the length of the string of each phone $=20\sqrt{3}\mathrm{m}$

Exercise : 9.3

• In Fig. A, B and C are three points on a circle with centre 0 such that $\angle \mathrm{BOC}=30^{\circ }$ and $\angle \mathrm{AOB}=60^{\circ }$. If D is a point on the circle other than the arc ABC , find $\angle \mathrm{ADC}$.
  
  Sol. $\angle \mathrm{ADC}=\frac{1}{2}\angle \mathrm{AOC}=\frac{1}{2}\left(60^{\circ }+30^{\circ }\right)=45^{\circ }$ [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]
• A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc. Sol.
  
  $\because \mathrm{OA}=\mathrm{OB}=\mathrm{AB}$ [Given] $\therefore △\mathrm{OAB}$ is equilateral $\therefore \angle \mathrm{AOB}=60^{\circ }$ and $\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}$ $=\frac{1}{2}\times 60=30^{\circ }$ $\because \mathrm{ADBC}$ is a cyclic quadrilateral. $\therefore \angle \mathrm{ADB}+\angle \mathrm{ACB}=180^{\circ }$ [The sum of either pair of opposite angles of a cyclic quadrilateral is $180^{\circ }$ ] $\Rightarrow \angle \mathrm{ADB}+30^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{ADB}=180^{\circ }-30^{\circ }$ $\Rightarrow \angle \mathrm{ADB}=150^{\circ }$
• In figure, $\angle \mathrm{PQR}=100^{\circ }$, where $\mathrm{P},\mathrm{Q}$ and R are points on a circle with centre 0 . Find $\angle \mathrm{OPR}$.
  
  Sol.
  
  Take a point $S$ in the major arc. Join PS and RS. $\because \mathrm{PQRS}$ is a cyclic quadrilateral. $\therefore \angle \mathrm{PQR}+\angle \mathrm{PSR}=180^{\circ }$ [The sum of either pair of opposite angles of a cyclic quadrilateral is $180^{\circ }$ ] $\Rightarrow 100^{\circ }+\angle \mathrm{PSR}=180^{\circ }$ $\Rightarrow \angle \mathrm{PSR}=180^{\circ }-100^{\circ }$ $\Rightarrow \angle \mathrm{PSR}=80^{\circ }$ Now $\angle \mathrm{POR}=2\angle \mathrm{PSR}$ [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] $\angle \mathrm{POR}=2\times 80^{\circ }=160^{\circ }$ [Using (1)] In $△\mathrm{OPR}$, $\because \mathrm{OP}=\mathrm{OR}$ [radii of a circle] $\therefore \angle \mathrm{OPR}=\angle \mathrm{ORP}$ [Angles opposite to equal sides of a triangle are same] In $△\mathrm{OPR}$, $\angle \mathrm{OPR}+\angle \mathrm{ORP}+\angle \mathrm{POR}=180^{\circ }$ [Sum of all the angles of a triangle is $180^{\circ }$ ] $\Rightarrow \angle \mathrm{OPR}+\angle \mathrm{OPR}+160^{\circ }=180^{\circ }$ [Using (2) and (3)] $\Rightarrow 2\angle \mathrm{OPR}+160^{\circ }=180^{\circ }$ $\Rightarrow 2\angle \mathrm{OPR}=180^{\circ }-160^{\circ }=20^{\circ }$ $\Rightarrow \angle \mathrm{OPR}=10^{\circ }$
• In fig. $\angle \mathrm{ABC}=69^{\circ },\angle \mathrm{ACB}=31^{\circ }$, find $\angle \mathrm{BDC}$.
  
  Sol. $\angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{BAC}=180^{\circ }$ (By angle sum property) $\Rightarrow 69^{\circ }+31^{\circ }+\angle \mathrm{BAC}=180^{\circ }$ $\Rightarrow \angle \mathrm{BAC}=180^{\circ }-100^{\circ }=80^{\circ }$ Since, angles in the same segment are equal $\begin{array}{rl}\angle \mathrm{BDC}& =\angle \mathrm{BAC}\\ \Rightarrow \angle \mathrm{BDC}& =80^{\circ }.\end{array}$
• In figure, $\mathrm{A},\mathrm{B},\mathrm{C}$ and D are four points on a circle. AC and BD intersect at a point E such that $\angle \mathrm{BEC}=130^{\circ }$ and $\angle \mathrm{ECD}=20^{\circ }$. Find $\angle \mathrm{BAC}$.
  
  Sol. $\angle \mathrm{CED}+\angle \mathrm{BEC}=180^{\circ }$ [Linear Pair] $\begin{array}{rl}\Rightarrow & \angle \mathrm{CED}+130^{\circ }=180^{\circ }\\ \Rightarrow & \angle \mathrm{CED}=180^{\circ }-130^{\circ }=50^{\circ }\\ & \angle \mathrm{ECD}=20^{\circ }\end{array}$ In $△\mathrm{CED}$, $\angle \mathrm{CED}+\angle \mathrm{ECD}+\angle \mathrm{CDE}=180^{\circ }$ [Sum of all the angles of a triangle is $180^{\circ }$ ] $\Rightarrow 50^{\circ }+20^{\circ }+\angle \mathrm{CDE}=180^{\circ }$ [Using (i) and (ii)] $\Rightarrow 70^{\circ }+\angle \mathrm{CDE}=180^{\circ }$ $\Rightarrow \angle \mathrm{CDE}=180^{\circ }-70^{\circ }$ $\Rightarrow \angle \mathrm{CDE}=110^{\circ }$ Now $\angle \mathrm{BAC}=\angle \mathrm{CDE}=110^{\circ }$ [Angle in the same segment of a circle are equal]
• ABCD is a cyclic quadrilateral whose diagonals intersect at a point E . If $\angle \mathrm{DBC}=$ $70^{\circ },\angle \mathrm{BAC}=30^{\circ }$, find $\angle \mathrm{BCD}$. Further, if AB $=\mathrm{BC}$, find $\angle \mathrm{ECD}$. Sol.
  
  Since angles in the same segment of a circle are equal $\therefore \angle \mathrm{BAC}=\angle \mathrm{BDC}$ $\Rightarrow \angle \mathrm{BDC}=30^{\circ }$ Also $\angle \mathrm{DBC}=70^{\circ }$ (Given) $\therefore$ In $△\mathrm{BCD}$, we have $\Rightarrow \angle \mathrm{BCD}+\angle \mathrm{DBC}+\angle \mathrm{CDB}=180^{\circ }$ [sum of angles of a triangle is $180^{\circ }$ ] $\Rightarrow \angle \mathrm{BCD}+70^{\circ }+30^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{BCD}=80^{\circ }$ Now, in $△\mathrm{ABC},\mathrm{AB}=\mathrm{BC}$ (Given) $\therefore \angle \mathrm{BCA}=\angle \mathrm{BAC}$ (angles opp. to equal sides of a triangle are equal) $\Rightarrow \angle \mathrm{BCA}=30^{\circ }$ $\left[\angle \mathrm{BAC}=30^{\circ }\right]$ Now, $\angle \mathrm{BCA}+\angle \mathrm{ECD}=\angle \mathrm{BCD}$ $\Rightarrow 30^{\circ }+\angle \mathrm{ECD}=80^{\circ }$ $\Rightarrow \angle \mathrm{ECD}=50^{\circ }$
• If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle. Sol.
  
  Since, AC and BD are diameters. $\Rightarrow \mathrm{AC}=\mathrm{BD}$ [all diameters of a circle are equal] Also, $\angle \mathrm{BAD}=90^{\circ }$ [angle formed in a semicircle is $90^{\circ }$ ] Similarly, $\angle \mathrm{ABC}=90^{\circ },\angle \mathrm{BCD}=90^{\circ }$ and $\angle \mathrm{CDA}=90^{\circ }$. Now, in right $△\mathrm{ABC}$ and $△\mathrm{BAD}$, we have $\mathrm{AC}=\mathrm{BD}$ (from (1)) $\mathrm{AB}=\mathrm{BA}$ (common) $\angle \mathrm{ABC}=\angle \mathrm{BAD}$ (each equal to $90^{\circ }$ ) $\therefore △\mathrm{ABC}\cong △\mathrm{BAD}$ (By RHS congruence) $\Rightarrow \mathrm{BC}=\mathrm{AD}$ (CPCT) Similarly, $\mathrm{AB}=\mathrm{DC}$ Thus, ABCD is a rectangle.
• If the non-parallel sides of a trapezium are equal, prove that it is a cyclic. Sol. Given : ABCD is a trapezium whose two non-parallel sides AD and BC are equal. To Prove : Trapezium ABCD is a cyclic. Construction : Draw BE||AD Proof: $\because \mathrm{AB}\Vert \mathrm{DE}$ [Given] $\mathrm{AD}\Vert$ BE [By construction] $\therefore$ Quadrilateral ABED is a parallelogram.
  
  $\therefore \angle \mathrm{BAD}=\angle \mathrm{BED}$ [Opp. $\angle$ s of a||gm] and $\mathrm{AD}=\mathrm{BE}$ [Opp. sides of a || gm] But $\mathrm{AD}=\mathrm{BC}$ [Given] From (ii) and (iii) $\mathrm{BE}=\mathrm{BC}$ $\therefore \angle \mathrm{BEC}=\angle \mathrm{BCE}$ [Angle opposite to equal sides] $\angle \mathrm{BEC}+\angle \mathrm{BED}=180^{\circ }$ [Linear pair] $\Rightarrow \angle \mathrm{BCE}+\angle \mathrm{BAD}=180^{\circ }$ [From (iv) and (i)] $\Rightarrow$ Trapezium ABCD is cyclic. $[\because$ If a pair of opposite angles of a quadrilateral is $180^{\circ }$, then the quadrilateral is cyclic]
• Two circles intersect at two points B and C. Through B, two line segment ABD and PBQ are drawn to intersect the circles at $\mathrm{A},\mathrm{D}$ and $\mathrm{P},\mathrm{Q}$ respectively. Prove that $\angle \mathrm{ACP}=\angle \mathrm{QCD}$.
  
  Sol. Given : Two circles intersect at two points B and C . Through B , two line segments ABD and PBQ are drawn to intersect the circles at $\mathrm{A},\mathrm{D}$ and $\mathrm{P},\mathrm{Q}$ respectively. To Prove : $\angle \mathrm{ACP}=\angle \mathrm{QCD}$ Proof: $\angle \mathrm{ACP}=\angle \mathrm{ABP}$ [Angles in the same segment of a circle are equal] $\angle \mathrm{QCD}=\angle \mathrm{QBD}$ [Angles in the same segment of a circle are equal] $\angle \mathrm{ABP}=\angle \mathrm{QBD}$ [Vertically Opposite Angles] From (i), (ii) and (iii), $\angle \mathrm{ACP}=\angle \mathrm{QCD}$
• If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side. Sol. We have $△\mathrm{ABC}$, and two circles described with diameter as AB and AC respectively. They intersect at a point D , other than A . Let us join A and D.
  
  AB is a diameter $\therefore \angle \mathrm{ADB}$ is an angle formed in a semicircle. $\Rightarrow \angle \mathrm{ADB}=90^{\circ }$ Similarly, $\angle \mathrm{ADC}=90^{\circ }$ adding (1) and (2) $\angle \mathrm{ADB}+\angle \mathrm{ADC}=180^{\circ }$ i.e., $\mathrm{B},\mathrm{D}$ and C are collinear points BC is a straight line. Thus, D lies on BC .
• ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle \mathrm{CAD}=\angle \mathrm{CBD}$. Sol. AC is a hypotenuse $\angle \mathrm{ADC}=90^{\circ }=\angle \mathrm{ABC}$
  
  $\therefore$ Both the triangles are in the same semicircle. $\Rightarrow \mathrm{A},\mathrm{B},\mathrm{C}$ and D are concyclic. Join BD DC is chord $\therefore \angle \mathrm{CAD}$ and $\angle \mathrm{CBD}$ are formed on the same segment $\therefore \angle \mathrm{CAD}=\angle \mathrm{CBD}$
• Prove that a cyclic parallelogram is a rectangle. Sol.
  
  We have a cyclic parallelogram ABCD. $\therefore$ Sum of its opposite angles is $180^{\circ }$ $\therefore \angle \mathrm{A}+\angle \mathrm{C}=180^{\circ }$ But $\angle \mathrm{A}=\angle \mathrm{C}$ From (1) and (2), we have $\angle \mathrm{A}=\angle \mathrm{C}=90^{\circ }$ Similarly, $\angle \mathrm{B}=\angle \mathrm{D}=90^{\circ }$ $\Rightarrow$ Each angle of the parallelogram ABCD is $90^{\circ }$ Thus, ABCD is a rectangle.