NCERT Solutions for Class 9 Maths Chapter 10 Circles

Q1 ) Recall that two circles are congruent, if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centers.

Answer :

Given: MN and PO are two equal chords of two congruent circles with centre at O and O'.

To prove: \(\angle{MON}\) and \(\angle{PO'Q}\)

Proof: In \(\triangle{MON}\) and \(\triangle{PO'Q}\), we have,

MO = PO' ...(Radii of congruent circles)
NO = QO' ...(Radii of congruent circles)
and MN = PQ ...(Given)
By SSS criterion, we get,

\(\triangle{MON}\) \(\displaystyle \cong\) \(\triangle{PO'Q}\)
Hence, \(\angle{MON}\) = \(\angle{PO'Q}\) ...(By CPCT)

Q2 ) Prove that, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer :

Given:
MW and PQ are two chords of congruent circles such that angles subtended by these chords at the centres O and O' of the circles are equal i.e., \(\angle{MON}\) = \(\angle{PO'Q}\)



To prove: MN = PQ

Proof:
In \(\triangle{MON}\) and \(\triangle{PO'Q}\), we have,

MO = PO' ...(Radii of congruent circles)
NO = QO' ...(Radii of congruent circles)
and \(\angle{MON}\) = \(\angle{PO'Q}\) ...(Given)
By SAS criterion, we get,

\(\triangle{MON}\) \(\displaystyle \cong\) \(\triangle{PO'Q}\)

Hence, MN = PQ ...(By CPCT)

Q1 ) Draw different pairs of circles. How many points does each pair have in common?
what is the maximum number of common points?

Answer :

Different pairs of circles are as follows:

i) No points in common (externally)	(ii) One point in common (internally)
iii) No points in common (internally)	iv) One point in common (externally)
v) Two points in common

From figures, it is obvious that these pairs many have 0 or 1 or 2 points in common.

Hence, a pair of circles cannot intersect each other at more than two points.

Q2 ) Suppose you are given a circle. Give a construction to find its centre.

Answer :

Steps of construction:

1. Taking three points P, Q and R on the circle.
2. Join PQ and QR.
3. Draw MQ and NS, respectively the perpendicular bisectors of PQ and RQ, which intersect each other at O.

Hence, O is the centre of the circle.

Q3 ) If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Answer :

Given:
Two circles with centres O and O' intersect at two points M and N so that MN is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO’ intersect MN at P.

To prove: OO' is the perpendicular bisector of MN.

Construction:
Draw line segments OM, ON, O'M and O'N.



Proof:
In \(\triangle{OMO'}\) and \(\triangle{ONO'}\), we have,

OM = ON ...(Radii of congruent circles)
O'M = O'N ...(Radii of congruent circles)
and OO' = O'O ...(Common)
By SSS criterion, we get,
\(\triangle{OMO'}\) \(\displaystyle \cong\) \(\triangle{ONO'}\)

Hence, \(\angle{MOO'}\) = \(\angle{NOO'}\) ...(By CPCT)
\(\therefore \) \(\angle{MOP}\) = \(\angle{NOP}\) ...(i)(\(\because \) \(\angle{MOO'}\) = \(\angle{MOP}\), \(\angle{NOO'}\) = \(\angle{NOP}\))

In \(\triangle{MOP}\) and \(\triangle{NOP}\), we have,
OM = ON ...(Radii of congruent circles)
OM = OM ...(Common)
and \(\angle{MOP}\) = \(\angle{NOP}\) ...(from eq. (i))
By SAS criterion, we get,
\(\triangle{MOP}\) \(\displaystyle \cong\) \(\triangle{NOP}\)

Hence, MP = NP ...(By CPCT)
And \(\angle{MPO}\) = \(\angle{NPO}\) ...(ii)

But \(\angle{MPO}\) + \(\angle{NPO}\) = \(180^\circ\)
(Since, MPN is a straight line)
\(\therefore \) 2 \(\angle{MPO}\) = \(90^\circ\) ...(from (ii))
\(\therefore \) \(\angle{MPO}\) = \(90^\circ\)

We have,
MP = NP and \(\angle{MPO}\) = \(\angle{NPO}\) = \(90^\circ\)

Hence, it is proved that OO' is the perpendicular bisector of MN.

Q1 ) 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.

Answer :

Let O and O' be the centres of the circles of radii 5 cm and 3 cm, respectively.

Let AB be their common chord.

Given: OA = 5cm, O'A = 3cm and OO' = 4cm



In \(\triangle{OO'A}\), we have,
\({AO'}^2 + {OO'}^2 = {3}^2 + {4}^2 = 9 + 16 = 25\)
\(\Rightarrow \) \({OA}^2 = {5}^2 = {AO'}^2 + {OO'}^2\)

Therefore, OO'A is a right angled traingle and right angled at O'

Now, Area of \(\triangle{OO'A}\) = \(\frac{1}{2} \) × O'A × OO'
= \(\frac{1}{2} \) × 3 × 4

Area of \(\triangle{OO'A}\) = 6 sq.cm ...(i)
Also, Area of \(\triangle{OO'A}\) = \(\frac{1}{2} \) × OO' × AM
= \(\frac{1}{2} \) × 4 × AM
Area of \(\triangle{OO'A}\) = 2 × AM ...(ii)

From eq. (i) and (ii), we get,
2 × AM = 6

Therefore, AM = 3 cm.

Since, when two circles intersect at two points, then their centres lie on the perpendicular bisector of the common chord.

Thus, AB= 2 × AM = 2 × 3 = 6cm.

Therefore, the length of the common chord is 6cm.

Q2 ) 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.

Answer :

Given:
MN and AB are two chords of a circle with centre O, AB and MN intersect at P and MN = AB.

To prove: MP = PB and PN = AP

Construction: Draw \(OD\perp{MN}\) and \(OC\perp{AB}\).Join OP



Proof:
DM = DN = (\(\frac{1}{2} \) ) MN
(Since, Perpendicular from centre bisects the chord)
Similarly,
AC = CB = (\(\frac{1}{2} \) ) AB
(Since, Perpendicular from centre bisects the chord)
\(\therefore \) we get,
MD = BC and DN = AC ... (i)(Since, MN = AB)

In \(\triangle{ODP}\) and \(\triangle{OPC}\),
OD = OC
(Equal chords of a circle are equidistant from the centre)
\(\angle{ODP}\) = \(\angle{OCP}\)
and OP = OP ...(Common side)
\(\therefore \) by RHS criterion of congruence,
\(\triangle{ODP}\) \(\displaystyle \cong\) \(\triangle{OPC}\)
Also, DP = PC ...(By CPCT) ...(ii)

On adding Equations (i) and (ii), we get,

MD + DP = BC + PC
\(\Rightarrow \) MP = PB
Also, On subtracting Equation (ii) and (i),we get,

DN + DP = AC + PC
\(\Rightarrow \) PN = AP

Hence, MP = PB and PN = AP are proved.

Q3 ) 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.

Answer :

Given:
RQ and MN are chords of a circle with centre O. MN and RQ intersect at P and MN = RQ.

To prove: \(\angle{OPC}\) = \(\angle{OPB}\)

Construction: Draw \(OC\perp{RQ}\) and \(OB\perp{MN}\). Join OP.

Proof:
In \(\triangle{OCP}\) and \(\triangle{OBP}\),
OC = OB
(Equal chords of a circle are equidistant from the centre)
\(\angle{OCP}\) = \(\angle{OBP}\)
(Each angle is \(90^\circ\))
and OP = OP ...(Common side)

Therefore, by RHS criterion of congruence,
\(\triangle{OCP}\) \(\displaystyle \cong\) \(\triangle{OBP}\)
Also, \(\angle{OPC}\) = \(\angle{OPB}\) ...(By CPCT)

Q4 ) If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D. prove that AB = CD (see figure).

Answer :

Let OP be the perpendicular from O on linel . Since, the perpendicular from the centre of a circle to a chord bisects the chords.



Now, BC is the chord of the smaller circle and \(OP\perp{BC}\).

Therefore, BC = PC ...(i)
Since, AD is a chord of the larger circle and \(OP\perp{AD}\).

Therefore, AP = PD ...(ii)

On subtracting Eq. (i) from Eq.(ii), we get,
AP - BP = PD - PC

Thus, AB = CD
Hence, it is proved.

Q5 ) Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, and 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?

Answer :

Let O be the centre of the circle and Reshma, Salma and Mandip are represented by the points R, S and M, respectively. Let RP = x m.



Area of \(\triangle{ORS}\) = \(\frac{1}{2} \) × x × 5 = \(\frac{5x}{2} \) ...(i)
(\(\because \) in \(\triangle{ORM}\), RM is a chord
\(\therefore \) \(OP\perp{RM}\))

In right angled triangle \(\triangle{RON}\),
\({OR}^2 = {RN}^2 + {NO}^2\)
\(\Rightarrow \) \({5}^2 = {3}^2 + {NO}^2\)
\(\Rightarrow \) \({NO}^2 = 25 - 9 = 16\)
\(\therefore \)NO = 4cm
So, Area of \(\triangle{ORS}\) = \(\frac{1}{2} \) × RS × ON = \(\frac{1}{2} \) × 6 × 4 = 12 ...(ii)

From Equations (i) and (ii), we get,

\(\Rightarrow \) \(\frac{5x}{2} \) = 12
\(\therefore \) x = \( \frac{24}{5} \) .

Since, P is the mid- Point of RM,

Thus, RM = 2RP = 2 × \(\frac{24}{5} \)
\(\Rightarrow \) RM = \(\frac{48}{5} \) = 9.6m

Hence, the distance between Reshma and Mandip is 9.6m.

Q6 ) 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.

Answer :

Let Ankur, Syed and David standing on the point P, Q and R.

Let PQ = QR = PR = X



\(\therefore \) DPQR is an equilateral traingle.

Drawn altitudes PC, QD and RN from vertices to the sides of a traingle and intersect these altitudes at the centre of a circle M.

As PQR is an equilateral, therefore these altitudes bisects their sides.

In \(\triangle{PQC}\),
\({PQ}^2 = {PC}^2 + {QC}^2 \)
(By Pythagoras theorem)
\(\Rightarrow \) \({x}^2 = {PC}^2 + ( \frac{x}{2} )^2 \)
\(\Rightarrow \)\({PC}^2 = {x}^2 - (\frac{x}{2} )^2\)
\(\Rightarrow \) \({PC}^2 = {x}^2 - \frac{x^2}{4} = 3\frac{x^2}{4} \)
(Since, QC = \( \frac{1}{2} \) QR = \(\frac{x}{2} \) )
\(\therefore \) PC = \(\frac{\sqrt{3}x}{2} \)
Now, MC = PC - PM = \(\frac{\sqrt{3}x}{2} \) - 20
(Since, PM = radius)

Now, in \(\triangle{QCM}\),
\({QM}^2 = {QC}^2 + {MC}^2 \)
(By Pythagoras theorem)
\(\Rightarrow \) \( 20^2 = (\frac{x}{2} )^2 + (\frac{\sqrt{3}x}{2} - 20)^2 \)
(Since, QM = radius)
\(\Rightarrow \) \(400 = \frac{x^2}{4} +( \frac{\sqrt{3}x}{2})^2 - 20\sqrt{3}x + 400\)
\(\Rightarrow \) \(0 = {x}^2 - 20\sqrt{3}x\)
\(\Rightarrow \) \({x}^2 = 20\sqrt{3}x\)
\(\therefore \) \(x = 20\sqrt{3}\)

Hence, PQ = QR = PR = \(20\sqrt{3}\)m.

Q1 ) In figure A, B and C are three points on a circle with centre O such that \(\angle{BOC}\) = \(30^\circ\) and \(\angle{AOB}\) = \(60^\circ\). If D is a point on the circle other than the arc ABC, find \(\angle{ADC}\).
Let PQ = QR = PR = X

Answer :

Here, \(\angle{AOC}\) = \(\angle{AOB}\) + \(\angle{BOC}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\)

\(\therefore \) Arc ABC makes \(90^\circ\) at the centre of the circle.

Thus, \(\angle{ADC}\) = (\(\frac{1}{2} \)) \(\angle{AOC}\)
(Since, The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle)
\(\Rightarrow\) = (\(\frac{1}{2} \) ) \(90^\circ\) = \(45^\circ\)

Hence, \(\angle{ADC}\) = \(45^\circ\).

Q2 ) 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.

Answer :

Let BC be chord, which is equal to the radius. Join OB and OC.



Given:
BC = OB = OC
\(\therefore, \) \(\triangle{OBC}\) is an equilateral traingle.
\(\therefore \) \(\angle{BOC}\) = \(60^\circ\)
But, we know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any part of the circle.

\(\therefore \) \(\angle{BAC}\) = (\(\frac{1}{2} \) ) \(\angle{BOC}\)
\(\Rightarrow \) \(\angle{BAC}\) = (\(\frac{1}{2} \) ) \(60^\circ\) = \(30^\circ\)

Also, here, ABMC is a cyclic quadrilateral.

\(\therefore \) \(\angle{BAC}\) + \(\angle{BMC}\) = \(180^\circ\)
(Since, in a cyclic quadrilateral the sum of opposite angles is \(180^\circ\)).
\(\therefore \) \(\angle{BMC}\) = \(180^\circ\) - \(30^\circ\) = \(150^\circ\)

Hence, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc is \(150^\circ\).

Q3 ) In figure, \(\angle{PQR}\) = \(100^\circ\) ,where P, Q and R are points on a circle with centre O. Find \(\angle{OPR}\).

Answer :

Since, the angle subtended by the centre is double the angle subtended by circumference, we get,

\(\angle{PQR}\) = 2 \(\angle{PQR}\) = \(200^\circ\)
(given: \(\angle{PQR}\) = \(100^\circ\))

\(\therefore \) in \(\triangle{OPR}\),
\(\angle{POR}\) = \(360^\circ\) - \(200^\circ\) = \(160^\circ\) ...(i)

Again, in \(\triangle{OPR}\),
OP = OR ...(Radii of the circle)
\(\angle{OPR}\) = \(\angle{ORP}\)
(By property of isosceles traingle)

In \(\triangle{POR}\),
\(\angle{OPR}\) + \(\angle{ORP}\) + \(\angle{POR}\) = \(180^\circ\) ...(ii)
(Sum of internal angles of triangle)

\(\therefore \) from eq.(i) and (ii), we get,

\(\angle{OPR}\) + \(\angle{OPR}\) + \(160^\circ\) = \(180^\circ\)
\(\Rightarrow \) 2 \(\angle{OPR}\) = \(180^\circ\) - \(160^\circ\) = \(20^\circ\)
Therefore, \(\angle{OPR}\) = \(10^\circ\).

Q4 ) In figure, \(\angle{ABC}\) = \(69^\circ\), \(\angle{ACB}\) = \(31^\circ\). Find \(\angle{BDC}\).

Answer :

As we know that, the angles in the same segment are equal.
We get,

\(\angle{BDC}\) = \(\angle{BAC}\) ...(i)

Now, in \(\triangle{ABC}\),

\(\angle{BAC}\) + \(\angle{ABC}\) + \(\angle{ACB}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{BAC}\) + \(69^\circ\) + \(31^\circ\) = \(180^\circ\)
\(\Rightarrow \)\(\angle{ABC}\) + \(100^\circ\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{ABC}\) = \(180^\circ\) - \(100^\circ\)
\(\therefore \) \(\angle{BAC}\) = \(80^\circ\)
\(\therefore \) \(\angle{BDC}\) = \(80^\circ\) ...(from (i))

Q5 ) In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that \(\angle{BEC}\) = \(130^\circ\) and \(\angle{ECD}\) = \(20^\circ\). Find \(\angle{BAC}\).

Answer :

In \(\triangle{ABC}\),

\(\angle{AED}\) = \(180^\circ\) -\(130^\circ\) = \(50^\circ\)
(linear pair of angles)

\(\therefore \) \(\angle{CED}\) = \(\angle{AED}\) = \(50^\circ\)
(Vertically opposite angles)

Also, \(\angle{ABD}\) = \(\angle{ACD}\)
(Since, the angles in the same segment are equal)
\(\Rightarrow \) \(\angle{ABE}\) = \(\angle{ECD}\)
\(\therefore \) \(\angle{ABE}\) = \(180^\circ\) ...(ii)

Now, in \(\triangle{CDE}\),
\(\angle{BAC}\) + \(20^\circ\) + \(50^\circ\) = \(180^\circ\)
(from eq.(i) and (ii))

\(\Rightarrow \) \(\angle{BAC}\) + \(70^\circ\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{BAC}\) = \(180^\circ\) - \(70^\circ\)
\(\therefore \) \(\angle{BAC}\) = \(110^\circ\)

Q6 ) ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \(\angle{DBC}\) = \(70^\circ\), \(\angle{BAC}\) = \(30^\circ\), Find \(\angle{BCD}\). Further, if AB = BC, find \(\angle{EDC}\).

Answer :

As, Angles in the same segment are equal, we get,

\(\angle{BDC}\) = \(\angle{BAC}\)
\(\therefore \) \(\angle{BDC}\) = \(30^\circ\).

In \(\triangle{DBC}\), we have,

\(\angle{BDC}\) + \(\angle{DBC}\) + \(\angle{BCD}\) = \(180^\circ\)
\(\Rightarrow \) \(30^\circ\) + \(70^\circ\) + \(\angle{BCD}\) = \(180^\circ\)
\(\Rightarrow \) \(100^\circ\) + \(\angle{BCD}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{BCD}\) = \(180^\circ\) - \(100^\circ\)
\(\therefore \) \(\angle{BCD}\) = \(80^\circ\)

If AB= BC, \(\angle{BCA}\) = \(\angle{BAC}\) = \(30^\circ\)
(Since, Angles opposite to equal sides in a traingle are equal)

Now,
\(\angle{ECD}\) = \(\angle{BCD}\) - \(\angle{BCA}\) = \(80^\circ\) - \(30^\circ\)
(Given \(\angle{BCD}\) = \(80^\circ\) and \(\angle{BCA}\) = \(30^\circ\))

\(\Rightarrow \) \(\angle{ECD}\) = \(30^\circ\)
\(\therefore \) \(\angle{EDC}\) = \(30^\circ\)

Q7 ) If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer :

Given:
Diagonals NP and QM of a cyclic quadrilateral are diameters of the circle through the vertices M, P, Q and N of the quadrilateral NQPM.



To prove: Quadrilateral NQPM is a rectangle.

Proof:
ON = OP = OQ = OM ...(Radii of circle)

Now, ON = OP = (\(\frac{1}{2} \) ) NP
and OM = OQ = (\(\frac{1}{2} \) ) MQ
\(\therefore \) NP = MQ

Hence, the diagonals MPQN are equal and bisect each other.

Hence, it is proved that quadrilateral NQPM is a rectangle.

Q8 ) If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer :

Given:
Non-parallel sides PS and QR of a trapezium PQRS are equal.

Construction:
Draw \(SM\perp{PQ}\) and \(RN\perp{PQ}\).



To prove: ABCD is cyclic trapezium.

Proof:
In \(\triangle{SMP}\) and \(\triangle{RNQ}\), we have,

SP = RQ ...(Given)
SM = RN
(Distance between two parallel twolines is always equal)

and \(\angle{SMP}\) = \(\angle{RNQ}\)
(Each \(90^\circ\))
By RHS criterion, we get,

\(\triangle{SMP}\) \(\displaystyle \cong\) \(\triangle{RNQ}\)

Hence, \(\angle{P}\) = \(\angle{Q}\) ...(By CPCT)

Also, \(\angle{PSM}\) = \(\angle{QRN}\)

Now, adding \(90^\circ\) on both sides, we get,

\(\therefore \) \(\angle{PSM}\) + \(90^\circ\) = \(\angle{QRN}\) + \(90^\circ\)
\(\therefore \) \(\angle{MSR}\) + \(\angle{PSM}\) = \(\angle{NRS}\) + \(\angle{QRN}\)
(Since, \(\angle{MSR}\) = \(\angle{NRS}\) = \(90^\circ\))

\(\therefore \) \(\angle{PSR}\) = \(\angle{QRS}\)
\(\Rightarrow \) \(\angle{S}\) = \(\angle{R}\)
\(\therefore \) \(\angle{P}\) = \(\angle{Q}\) and \(\angle{S}\) = \(\angle{R}\)

\(\because \) Sum of the angles of a quadrilateral is \(360^\circ\)
We get,

\(\angle{P}\) + \(\angle{Q}\) + \(\angle{S}\) + \(\angle{R}\) = \(360^\circ\)

From eq. (i), we get,

2[\(\angle{Q}\) + \(\angle{S}\)] = \(360^\circ\)
\(\therefore \) [\(\angle{Q}\) + \(\angle{S}\)] = \(180^\circ\)

Hence, it is proved that PQRS is a cyclic trapezium.

Q9 ) Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see figure). Prove that \(\angle{ACP}\) = \(\angle{QCD}\).

Answer :

Given:
Two circles intersect at two points B and C. Through B two line segment ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively.

To prove: \(\angle{ACP}\) = \(\angle{QCD}\)

Proof:
In circle I,
\(\angle{ACP}\) = \(\angle{ABP}\) ...(i)
(Angles in the same segment)

In circle II,
\(\angle{QCD}\) = \(\angle{QBD}\) ...(ii)
(Angles in the same segment)

Also, \(\angle{ABP}\) = \(\angle{QBD}\)
(Vertically opposite angles)

From Equation (i) and (ii), we get,

\(\angle{ACP}\) = \(\angle{QCD}\)
Hence, proved.

Q10 ) 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.

Answer :

Given:
Two circles are drawn with sides AC and AB of \(\triangle{ABC}\) as diameters. Both circles intersect each other at D.

To prove: D lies on BC.

Construction: Join AD.



Proof:
Since, AC and AB are the diameters of the two circles,
\(\angle{ADB}\) = \(90^\circ\) ...(i)
(Angles in a semi-circle)

And \(\angle{ADC}\) = \(90^\circ\) ...(ii)
(Angles in a semi-circle)

On adding Equation (i) and (ii), we get,

\(\angle{ADB}\) + \(\angle{ADC}\) = \(90^\circ\) + \(90^\circ\) = \(180^\circ\)

Hence, we can say that, BCD is a straight line.

So, D lies on BC.

Q11 ) ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that \(\angle{CAD}\) = \(\angle{CBD}\).

Answer :

Draw a circle with AC as diameter passing through B and D. Join BD.



\(\triangle{ADC}\) and \(\triangle{ABC}\) are right angled triangles with common hypotenuse.

\(\angle{CBD}\) = \(\angle{CAD}\)
(Since, angles in the same segment are equal)

Hence, proved.

Q12 ) Prove that a cyclic parallelogram is a rectangle.

Answer :

Given:
PQRS is a parallelogram inscribed in a circle.

To prove: PQRS is a rectangle.



Proof:
Since, PQRS is a cyclic quadrilateral.

Thus, \(\angle{P}\) + \(\angle{R}\) = \(180^\circ\) ...(i)
(Since, Sum of opposite angles in a cyclic quadrilateral is \(180^\circ\))

But, \(\angle{P}\) = \(\angle{R}\) ...(ii)
(Since, in a parallelogram, opposite angles are equal)

from eq. (i) and (ii), we get,

\(\angle{P}\) = \(\angle{R}\) = \(90^\circ\)

Similarly,
\(\angle{Q}\) = \(\angle{S}\) = \(90^\circ\)

Thus, Each angle of PQRS is \(90^\circ\).

Hence, it is proved that PQRS is a rectangle.