# ICSE Solutions for Selina Concise Chapter 17 Circles Class 10 Maths

**Exercise 17(A) **

**1. In the given figure, O is the center of the circle. ∠OAB and ∠OCB are 30 ^{o} and 40^{o} respectively. Find ∠AOC Show your steps of working.**

**Solution**

Firstly, let’s join AC.

And, let ∠OAC = ∠OCA = x **[Angles opposite to equal sides are equal]**

So, ∠AOC = 180° – 2x

Also,

∠BAC = 30° + x

∠BCA = 40° + x

Now, in ∆ABC

∠ABC = 180° – ∠BAC – ∠BCA **[Angles sum property of a triangle]**

= 180° – (30° + x) – (40^{o} + x)

= 110° – 2x

And, ∠AOC = 2∠ABC

**[Angle at the center is double the angle at the circumference subtend by the same chord]**

180° – 2x = 2(110° – 2x)

⇒ 2x = 40°

⇒ x = 20°

Thus, ∠AOC = 180° – 2×20° = 140°

^{}

**2. In the given figure, ∠BAD = 65****°****, ∠ABD = 70****°****, ∠BDC = 45°**

**(i) Prove that AC is a diameter of the circle.**

**(ii) Find ∠ACB.**

**Solution**

**(i)** In ∆ABD,

∠DAB + ∠ABD + ∠ADB = 180°

⇒ 65° + 70° + ∠ADB = 180°

⇒ 135° + ∠ADB = 180°

⇒ ∠ADB = 180° – 135° = 45°

Now,

∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°

As ∠ADC is the angle of semi-circle for AC as the diameter of the circle.

**(ii)** ∠ACB = ∠ADB **[Angles in the same segment of a circle]**

Hence, ∠ACB = 45°

^{}

**3. Given O is the centre of the circle and ∠AOB = 70****°****.**

**Calculate the value of:**

**(i) ∠OCA,**

**(ii) ∠OAC.**

**Solution**

Here, ∠AOB = 2∠ACB

**[Angle at the center is double the angle at the circumference subtend by the same chord]**

∠ACB = 70°/2 = 35°

Now, OC = OA **[Radii of same circle]**

Thus,

∠OCA = ∠OAC = 35°

^{}

**4. In each of the following figures, O is the centre of the circle. Find the values of a, b and c.**

**Solution**

Here, b = ½ ×130°

**[Angle at the center is double the angle at the circumference subtend by the same chord]**

Thus, b = 65°

Now,

a + b = 180°^{ }**[Opposite angles of a cyclic quadrilateral are supplementary]**

⇒ a = 180° – 65° = 115°

Here, c = ½ × Reflex (112°)

**[Angle at the center is double the angle at the circumference subtend by the same chord]**

Thus, c = ½ ×(360^{o} – 112^{o}) = 124°

a = 115°, b = 65°, c = 124°

^{}

**5. In each of the following figures, O is the center of the circle. Find the values of a, b, c and d.**

**Solution**

Here, ∠BAD = 90° **[Angle in a semi-circle]**

So, ∠BDA = 90° – 35° = 55°

And,

a = ∠ACB = ∠BDA = 55°

**[Angles subtended by the same chord on the circle are equal]**

Here, ∠DAC = ∠CBD = 25°

**[Angles subtended by the same chord on the circle are equal]**

And, we have

120° = b + 25°

**[Exterior angle property of a triangle]**

b = 95°

∠AOB = 2∠AOB = 2×50° = 100°

**[Angle at the center is double the angle at the circumference subtend by the same chord]**

Also, OA = OB

∠OBA = ∠OAB = c

c = (180°– 100°)/2 = 40°

We have, ∠APB = 90^{o} **[Angle in a semicircle]**

∠BAP = 90° – 45° = 45°

Now, d = ∠BCP = ∠BAP = 45°

**[Angles subtended by the same chord on the circle are equal]**

a = 55°, b = 95°, c= 40°, d = 45°

**6. In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O _{1} and O_{2} are the centers of two circles.**

**Solution**

It’s seen that,

∠DBA = ∠CBA = 90° **[Angle in a semi-circle is a right angle]**

So, adding both

∠DBA + ∠CBA = 180°

Thus, DBC is a straight line i.e. D, B and C form a straight line.

**7. In the figure, given below, find:**

**(i) ∠BCD,**

**(ii) ∠ADC,**

**(iii) ∠ABC.**

**Show steps of your working.**

**Solution**

From the given fig, it’s seen that

In cyclic quadrilateral ABCD, DC || AB

And given, ∠DAB = 105°

**(i) **So, ∠BCD = 180° – 105° = 75°

**[Sum of opposite angles in a cyclic quadrilateral is 180 ^{o}]**

**(ii)** Now,

∠ADC and ∠DAB are corresponding angles.

So,

∠ADC + ∠DAB = 180°

∠ADC = 180° – 105°

Thus, ∠ADC = 75°

**(iii)** We know that, the sum of angles in a quadrilateral is 360°

So,

∠ADC + ∠DAB +∠BCD + ∠ABC = 360°

75°^{ }+ 105° + 75° + ∠ABC = 360°

∠ABC = 360°^{ }– 255°

Thus, ∠ABC = 105°

^{}

**8. In the figure, given below, O is the centre of the circle. If ∠AOB = 140****° and ∠OAC = 50°;**

**find:**

**(i) ∠ACB,**

**(ii) ∠OBC,**

**(iii) ∠OAB,**

**(iv) ∠CBA.**

**Solution**

Given, ∠AOB = 140° and ∠OAC = 50°

**(i)** Now,

∠ACB = ½ Reflex (∠AOB) = ½ (360° – 140°) = 110°

**[Angle at the center is double the angle at the circumference subtend by the same chord]**

**(ii)** In quadrilateral OBCA,

∠OBC + ∠ACB + ∠OCA + ∠AOB = 360° **[Angle sum property of a quadrilateral]**

∠OBC + 110° + 50° + 140° = 360°

Thus, ∠OBC = 360° – 300° = 60°

**(iii)** In ∆AOB, we have

OA = OB **(radii)**

So, ∠OBA = ∠OAB

Hence, by angle sum property of a triangle

∠OBA + ∠OAB + ∠AOB = 180°

⇒ 2∠OBA + 140° = 180°

⇒ 2∠OBA = 40°

⇒ ∠OBA = 20°

**(iv)** We already found, ∠OBC = 60°

And, ∠OBC = ∠CBA + ∠OBA

⇒ 60^{o} = ∠CBA + 20°

∴ ∠CBA = 40°

^{}

**9. Calculate:**

**(i) ∠CDB,**

**(ii) ∠ABC,**

**(iii) ∠ACB.**

**Solution**

Here, we have

∠CDB = ∠BAC = 49°

∠ABC = ∠ADC = 43°

**[Angles subtended by the same chord on the circle are equal]**

Now, by angle sum property of a triangle we have

∠ACB = 180° – 49° – 43° = 88°

^{}

**10. In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75 ^{o}; ∠ABD = 58^{o} and ∠ADC = 77^{o}.**

**Find:**

**(i) ∠BDC,**

**(ii) ∠BCD,**

**(iii) ∠BCA.**

**Solution**

**(i)** By angle sum property of triangle ABD,

∠ADB = 180^{o} – 75^{o} – 58^{o} = 47^{o}

Thus, ∠BDC = ∠ADC – ∠ADB = 77^{o} – 47^{o} = 30^{o}

**(ii)** ∠BAD + ∠BCD = 180^{o}

**[Sum of opposite angles of a cyclic quadrilateral is 180 ^{o}]**

Thus, ∠BCD = 180^{o} – 75^{o} = 105^{o}

**(iii)** ∠BCA = ∠ADB = 47^{o}

**[Angles subtended by the same chord on the circle are equal]**

**11. In the figure given below, O is the centre of the circle and triangle ABC is equilateral.**

**Find:**

**(i) ∠ADB, **

**(ii) ∠AEB**

**Solution**

**(i) **As, it’s seen that ∠ACB and ∠ADB are in the same segment,

So, ∠ADB = ∠ACB = 60^{o}

**(ii)** Now, join OA and OB.

And, we have

∠AEB = ½ Reflex (∠AOB) = ½ (360^{o }– 120^{o}) = 120^{o}

**[Angle at the center is double the angle at the circumference subtend by the same chord]**

**12. Given: ∠CAB = 75 ^{o} and ∠CBA = 50^{o}. Find the value of ∠DAB + ∠ABD.**

**Solution**

Given, ∠CAB = 75^{o} and ∠CBA = 50^{o}

In ∆ABC, by angle sum property we have

∠ACB = 180^{o} – (∠CBA + ∠CAB)

= 180^{o} – (50^{o} + 75^{o}) = 180^{o} – 125^{o}

= 55^{o}

And,

∠ADB = ∠ACB = 55^{o}

**[Angles subtended by the same chord on the circle are equal]**

Now, taking ∆ABD

∠DAB + ∠ABD + ∠ADB = 180^{o}

⇒ ∠DAB + ∠ABD + 55^{o} = 180^{o}

⇒ ∠DAB + ∠ABD = 180^{o} – 55^{o}

⇒ ∠DAB + ∠ABD = 125^{o}

^{}

**13. ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ADC = 130 ^{o}, find ∠BAC.**

**Solution**

From the fig. its seem that,

∠ACB = 90^{o} **[Angle in a semi-circle is 90 ^{o}]**

Also,

∠ABC = 180^{o} – ∠ADC = 180^{o} – 130^{o} = 50^{o}

**[Pair of opposite angles in a cyclic quadrilateral are supplementary]**

By angle sum property of the right triangle ACB, we have

∠BAC = 90^{o} – ∠ABC

= 90^{o} – 50^{o}

Thus, ∠BAC = 40^{o}

^{}

**14. In the figure given alongside, AOB is a diameter of the circle and ∠AOC = 110 ^{o}, find ∠BDC.**

**Solution**

Let’s join AD first.

So, we have

∠ADC = ½ ∠AOC = ½ x 110^{o} = 55^{o}

**[Angle at the centre is double the angle at the circumference subtended by the same chord]**

Also, we know that

∠ADB = 90^{o}

**[Angle in the semi-circle is a right angle]**

∴ ∠BDC = 90^{o} – ∠ADC = 90^{o} – 55^{o}

∠BDC = 35^{o}

^{}

**15. In the following figure, O is the centre of the circle;** ∠**AOB = 60 ^{o} and **∠

**BDC = 100**

^{o}, find ∠OBC.**Solution**

Form the figure, we have

∠ACB = ½ ∠AOB = ½ x 60^{o} = 30^{o}

**[Angle at the centre is double the angle at the circumference subtended by the same chord]**

Now, by applying angle sum property in ∆BDC,

∠DBC = 180^{o }– 100^{o} – 30^{o} = 50^{o}

∴ ∠OBC = 50^{o}

^{}

**16. In ABCD is a cyclic quadrilateral in which ∠DAC = 27 ^{o}, ∠DBA = 50^{o} and ∠ADB = 33^{o}. Calculate (i) ∠DBC, (ii) ∠DCB, (iii) ∠CAB.**

**Solution**

**(i)** It’s seen that,

∠DBC = ∠DAC = 27^{o}

**[Angles subtended by the same chord on the circle are equal]**

**(ii)** It’s seen that,

∠ACB = ∠ADB = 33^{o}

And,

∠ACD = ∠ABD = 50^{o}

**[Angles subtended by the same chord on the circle are equal]**

Thus,

∠DCB = ∠ACD + ∠ACB = 50^{o} + 33^{o} = 83^{o}

**(iii)** In quad. ABCD,

∠DAB + ∠DCB = 180^{o}

27^{o} + ∠CAB + 83^{o} = 180^{o}

Thus,

∠CAB = 180^{o} – 110^{o} = 70^{o}

^{}

**17. In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80 ^{o} and ∠CDE = 40^{o}. Find the number of degrees in: (i) ∠DCE; (ii) ∠ABC.**

**Solution**

**(i)** Form the fig. its seen that,

∠DCE = 90^{o} – ∠CDE = 90^{o} – 40^{o} = 50^{o}

∴ ∠DEC = ∠OCB = 50^{o}

**(ii)** In ∆BOC, we have

∠AOC = ∠OCB + ∠OBC **[Exterior angle property of a triangle]**

∠OBC = 80^{o} – 50^{o} = 30^{o} **[Given ∠AOC = 80 ^{o}]**

∴ ∠ABC = 30^{o}

^{}

**18. In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.**

**Solution**

Firstly, join OB.

Then, ∠OBA = 90^{o} [**Angle in a semi-circle is a right angle]**

That is, OB is perpendicular to AE.

Now, we know that the perpendicular draw from the centre to a chord bisects the chord.

∴ AB = BE

**19. In the following figure,**

**(i) if ∠BAD = 96 ^{o}, find ∠BCD and ∠BFE.**

**(ii) Prove that AD is parallel to FE.**

**Solution**

**(i) **ABCD is a cyclic quadrilateral

So, ∠BAD + ∠BCD = 180^{o}

**[Pair of opposite angles in a cyclic quadrilateral are supplementary]**

∠BCD = 180^{o} – 96^{o} = 84^{o}

And, ∠BCE = 180^{o} – 84^{o} = 96^{o} **[Linear pair of angles]**

Similarly, BCEF is a cyclic quadrilateral

So, ∠BCE + ∠BFE = 180^{o}

**[Pair of opposite angles in a cyclic quadrilateral are supplementary]**

∠BFE = 180^{o} – 96^{o }= 84^{o}

**(ii)** Now, ∠BAD + ∠BFE = 96^{o} + 84^{o} = 180^{o}

But these two are interior angles on the same side of a pair of lines AD and FE.

∴ AD || FE.

**20. Prove that:**

**(i) the parallelogram, inscribed in a circle, is a rectangle.**

**(ii) the rhombus, inscribed in a circle, is a square.**

**Solution**

**(i)** Let’s assume that ABCD is a parallelogram which is inscribed in a circle.

So, we have

∠BAD = ∠BCD **[Opposite angles of a parallelogram are equal]**

And ∠BAD + ∠BCD = 180^{o}

**[Pair of opposite angles in a cyclic quadrilateral are supplementary]**

So, 2∠BAD = 180^{o}

Thus, ∠BAD = ∠BCD = 90^{o}

Similarly, the remaining two angles are 90^{o} each and pair of opposite sides are equal.

∴ ABCD is a rectangle.

Hence, Proved

**(ii)** Let’s assume that ABCD is a rhombus which is inscribed in a circle.

So, we have

∠BAD = ∠BCD **[Opposite angles of a rhombus are equal]**

And ∠BAD + ∠BCD = 180^{o}

**[Pair of opposite angles in a cyclic quadrilateral are supplementary]**

So, 2∠BAD = 180^{o}

Thus, ∠BAD = ∠BCD = 90^{o}

Similarly, the remaining two angles are 90^{o} each and all the sides are equal.

∴ ABCD is a square.

Hence, Proved

**21. In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.**

**Solution**

Give, AB = AC

So, ∠B = ∠C **… (1)**

**[Angles opposite to equal sides are equal]**

And, DECB is a cyclic quadrilateral.

So, ∠B + ∠DEC = 180^{o}

**[Pair of opposite angles in a cyclic quadrilateral are supplementary]**

∠C + ∠DEC = 180^{o} **…. (Using 1)**

But this is the sum of interior angles on one side of a transversal.

DE || BC.

But, ∠ADE = ∠B and ∠AED = ∠C **[Corresponding angles]**

Thus, ∠ADE = ∠AED

AD = AE

⇒ AB – AD = AC = AE **[As AB = AC]**

⇒ BD = CE

Hence, we have DE || BC and BD = CE

∴ DECB is an isosceles trapezium.

**22. Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.**

**Solution**

Let O and O’ be the centres of two intersecting circles, where points of the intersection are P and Q and PA and PB are their diameters respectively.

Join PQ, AQ and QB.

Thus, ∠AQP = 90^{o} and ∠BQP = 90^{o}

**[Angle in a semicircle is a right angle]**

Now, adding both these angles we get

∠AQP + ∠BQP = 180^{o}

∠AQB = 180^{o}

∴ the points A, Q and B are collinear.

**23. The figure given below, shows a circle with centre O.**

**(i) Find the relationship between a and b**

**(ii) Find the measure of angle OAB, if OABC is a parallelogram.**

**Solution**

Given: ∠AOC = a and ∠ABC = b.

**(i)** It’s seen that,

∠ABC = ½ Reflex (∠COA)

**[Angle at the centre is double the angle at the circumference subtended by the same chord]**

So, b = ½ (360^{o} - a)

a + 2b = 180^{o} **….. (1)**

**(ii)** As OABC is a parallelogram, the opposite angles are equal.

So, a = b

Now, using the above relationship in (1)

3a = 180^{o}

⇒ a = 60^{o}

Also, OC || BA

∠COA + ∠OAB = 180^{o}

⇒ 60^{o} + ∠OAB = 180^{o}

∴ ∠OAB = 120^{o}

^{}

**24. Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD as the center O is equal to twice the angle APC**

**Solution**

Required to prove: ∠AOC + ∠BOD = 2∠APC

OA, OB, OC and OD are joined.

Also, AD is joined.

Now, it’s seen that

∠AOC = 2∠ADC **…. (1)**

**[Angle at the centre is double the angle at the circumference subtended by the same chord]**

Similarly,

∠BOD = 2∠BAD **…. (2)**

Adding (1) and (2), we have

∠AOC + ∠BOD = 2∠ADC + 2∠BAD

= 2(∠ADC + ∠BAD) **….. (3)**

And in ∆PAD,

Ext. ∠APC = ∠PAD + ∠ADC

= ∠BAD + ∠ADC **…. (4)**

So, from (3) and (4) we have

∠AOC + ∠BOD = 2∠APC

**25. In the figure given RS is a diameter of the circle. NM is parallel to RS and MRS = 29 ^{o}**

**Calculate: (i) ∠RNM;**

**(ii) ∠NRM.**

**Solution**

**(i)** Join RN and MS

∠RMS = 90^{o} **[Angle in a semi-circle is a right angle]**

So, by angle sum property of ∆RMS

∠RMS = 90^{o} – 29^{o} = 61^{o}

And, ∠RNM = 180^{o} – ∠RSM = 180^{o} – 61^{o} = 119^{o}

**[Pair of opposite angles in a cyclic quadrilateral are supplementary]**

**(ii)** Now as RS || NM,

∠NMR = ∠MRS = 29^{o} **[Alternate angles]**

⇒ ∠NMS = 90^{o} + 29^{o} = 119^{o}

Also, we know that

∠NRS + ∠NMS = 180^{o}

**[Pair of opposite angles in a cyclic quadrilateral are supplementary]**

∠NRM + 29^{o} + 119^{o} = 180^{o}

⇒ ∠NRM = 180^{o} – 148^{o}

∴ ∠NRM = 32^{o}

^{}

**26. In the figure given alongside, AB || CD and O is the center of the circle. If **∠**ADC = 25 ^{o}; find the angle AEB. Give reasons in support of your answer.**

**Solution**

Join AC and BD.

So, we have

∠CAD = 90^{o} and ∠CBD = 90^{o}

**[Angle is a semicircle is a right angle]**

And, AB || CD

So, ∠BAD = ∠ADC = 25^{o} **[Alternate angles]**

∠BAC = ∠BAD + ∠CAD = 25^{o} + 90^{o} = 115^{o}

Thus,

∠ADB = 180^{o} – 25^{o} – ∠BAC = 180^{o} – 25^{o} – 115^{o} = 40^{o}

**[Pair of opposite angles in a cyclic quadrilateral are supplementary]**

Finally,

∠AEB = ∠ADB = 40^{o}

**[Angles subtended by the same chord on the circle are equal]**

**27. Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.**

**Solution**

Let’s join AC, PQ and BD.

As ACQP is a cyclic quadrilateral

∠CAP + ∠PQC = 180^{o} **… (i)**

**[Pair of opposite angles in a cyclic quadrilateral are supplementary]**

Similarly, as PQDB is a cyclic quadrilateral

∠PQD + ∠DBP = 180^{o} **…(ii)**

Again, ∠PQC + ∠PQD = 180^{o} **… (iii) [Linear pair of angles]**

Using (i), (ii) and (iii) we have

∠CAP + ∠DBP = 180^{o}

Or ∠CAB + ∠DBA = 180^{o}

We know that, if the sum of interior angles between two lines when intersected by a transversal are supplementary.

Then, AC || BD.

**28. ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.**

**Solution**

Let’s assume that ABCD be the given cyclic quadrilateral.

Also, PA = PD **[Given]**

So, ∠PAD = ∠PDA **…(1)**

**[Angles opposite to equal sides are equal]**

And,

∠BAD = 180^{o} – ∠PAD **[Linear pair of angles]**

Similarly,

∠CDA = 180^{o} – ∠PDA = 180^{o} – ∠PAD **[From (1)]**

As the opposite angles of a cyclic quadrilateral are supplementary,

∠ABC = 180^{o} – ∠CDA = 180^{o} – (180^{o} – ∠PAD) = ∠PAD

And, ∠DCB = 180^{o} – ∠BAD = 180^{o} – (180^{o} – ∠PAD) = ∠PAD

Thus,

∠ABC = ∠DCB = ∠PAD = ∠PDA

Which is only possible when AD || BC.

**Exercise 17(B) **

**1. In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.**

**Prove it.**

**Solution**

Let ABCD be the cyclic trapezium in which AB || DC, AC and BD are the diagonals.

Required to prove:

(i) AD = BC

(ii) AC = BD

Proof:

It’s seen that chord AD subtends ∠ABD and chord BC subtends ∠BDC at the circumference of the circle.

But, ∠ABD = ∠BDC **[Alternate angles, as AB || DC with BD as the transversal]**

So, Chord AD must be equal to chord BC

AD = BC

Now, in ∆ADC and ∆BCD

DC = DC **[Common]**

∠CAD = ∠CBD **[Angles in the same segment are equal]**

AD = BC **[Proved above]**

Hence, by SAS criterion of congruence

∆ADC ≅ ∆BCD

∴ by CPCT

AC = BD

**2. In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CD are equal. If ∠DEF = 110 ^{o}, calculate:**

**(i) ∠AFE, (ii) ∠FAB.**

**Solution**

Join AE, OB and OC.

**(i)** As AOD is the diameter

∠AED = 90^{o} **[Angle in a semi-circle is a right angle]**

But, given ∠DEF = 110^{o}

So, ∠AEF = ∠DEF – ∠AED = 110^{o} – 90^{o} = 20^{o}

**(ii) **Also given, Chord AB = Chord BC = Chord CD

So,

∠AOB = ∠BOC = ∠COD **[Equal chords subtends equal angles at the centre]**

But,

∠AOB + ∠BOC + ∠COD = 180^{o} **[Since, AOD is a straight line]**

Thus, ∠AOB = ∠BOC = ∠COD = 60^{o}

Now, in ∆OAB we have

OA = OB **[Radii of same circle]**

So, ∠OAB = ∠OBA **[Angles opposite to equal sides]**

But, by angle sum property of ∆OAB

∠OAB + ∠OBA = 180^{o} – ∠AOB

= 180^{o} – 60^{o}

= 120^{o}

∴ ∠OAB = ∠OBA = 60^{o}

Now, in cyclic quadrilateral ADEF

∠DEF + ∠DAF = 180^{o}

∠DAF = 180^{o} – ∠DEF

= 180^{o} – 110^{o}

= 70^{o}

Thus,

∠FAB = ∠DAF + ∠OAB

= 70^{o} + 60^{o} = 130^{o}

^{}

**3. If two sides of a cycli-quadrilateral are parallel; prove that:**

**(i) its other two sides are equal.**

**(ii) its diagonals are equal.**

**Solution**

Let ABCD is a cyclic quadrilateral in which AB || DC. AC and BD are its diagonals.

Required to prove:

(i) AD = BC

(ii) AC = BD

Proof:

(i) As AB || DC **(given)**

∠DCA = ∠CAB **[Alternate angles]**

Now, chord AD subtends ∠DCA and chord BC subtends ∠CAB at the circumference of the circle.

So,

∠DCA = ∠CAB

Hence, chord AD = chord BC or AD = BC.

(ii) Now, in ∆ABC and ∆ADB

AB = AB **[Common]**

∠ACB = ∠ADB **[Angles in the same segment are equal]**

BC = AD **[Proved above]**

Hence, by SAS criterion of congruence

∆ACB ≅ ∆ADB

∴ by CPCT

AC = BD

**4. The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°.**

**Calculate:**

**(i) ∠POS,**

**(ii) ∠QOR,**

**(iii) ∠PQR.**

**Solution**

Join OP, OQ, OR and OS.

Given, PQ = QR = RS

So, ∠POQ = ∠QOR = ∠ROS **[Equal chords subtends equal angles at the centre]**

Arc PQRS subtends ∠POS at the centre and ∠PTS at the remaining part of the circle.

Thus,

∠POS = 2×∠PTS = 2×75^{o} = 150^{o}

⇒ ∠POQ + ∠QOR + ∠ROS = 150^{o}

⇒ ∠POQ = ∠QOR = ∠ROS = 150^{o}/3 = 50^{o}

In ∆OPQ we have,

OP = OQ **[Radii of the same circle]**

So, ∠OPQ = ∠OQP **[Angles opposite to equal sides are equal]**

But, by angle sum property of ∆OPQ

∠OPQ + ∠OQP + ∠POQ = 180^{o}

⇒ ∠OPQ + ∠OQP + 50^{o} = 180^{o}

⇒ ∠OPQ + ∠OQP = 130^{o}

⇒ 2∠OPQ = 130^{o}

⇒ ∠OPQ = ∠OPQ = 130^{o}/2 = 65^{o}

Similarly, we can prove that

In ∆OQR,

∠OQR = ∠ORQ = 65^{o}

And in ∆ORS,

∠ORS = ∠OSR = 65^{o}

Hence,

(i) ∠POS = 150^{o}

(ii) ∠QOR = 50^{o} and

(iii) ∠PQR = ∠PQO + ∠OQR = 65^{o} + 65^{o} = 130^{o}

**5. In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of:**

**(i) **∠**AOB,**

**(ii) **∠**ACB,**

**(iii) **∠**ABC.**

**Solution**

**(i) **Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∠ACB = ½ ∠AOB

And as AB is the side of a regular hexagon, we have

∠AOB = 60^{o}

**(ii)** Now,

∠ACB = ½ (60^{o}) = 30^{o}

**(iii)** Since AC is the side of a regular octagon,

∠AOC = 360^{o}/ 8 = 45^{o}

Again, arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.

∠ABC = ½ ∠AOC

⇒ ∠ABC = 45^{o}/2 = 22.5^{o}

^{}

**Exercise 17(C) **

**1. In the given circle with diameter AB, find the value of x.**

**Solution**

Now,

∠ABD = ∠ACD = 30^{o}** [Angles in the same segment]**

In ∆ADB, by angle sum property we have

∠BAD + ∠ADB + ∠ABD = 180^{o}

But, we know that angle in a semi-circle is 90^{o}

∠ADB = 90^{o}

So,

x + 90^{o} + 30^{o} = 180^{o}

⇒ x = 180^{o} – 120^{o}

Hence, x = 60^{o}

^{}

**2. In the given figure, ABC is a triangle in which ∠BAC = 30 ^{o}. Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose center is O.**

**Solution**

Firstly, join OB and OC.

Proof:

∠BOC = 2∠BAC = 2×30^{o} = 60^{o}

Now, in ∆OBC

OB = OC **[Radii of same circle]**

So, ∠OBC = ∠OCB **[Angles opposite to equal sides]**

And in ∆OBC, by angle sum property we have

∠OBC + ∠OCB + ∠BOC = 180^{o}

⇒ ∠OBC + ∠OBC + 60^{o} = 180^{o}

⇒ 2∠OBC = 180^{o} – 60^{o} = 120^{o}

⇒ ∠OBC = 120^{o}/2 = 60^{o}

So, ∠OBC = ∠OCB = ∠BOC = 60^{o}

Thus, ∆OBC is an equilateral triangle.

So, BC = OB = OC

But, OB and OC are the radii of the circum-circle.

∴ BC is also the radius of the circum-circle.

**3. Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.**

**Solution**

Let’s consider ∆ABC, AB = AC and circle with AB as diameter is drawn which intersects the side BC and D.

And, join AD

Proof:

It’s seen that,

∠ADB = 90^{o} **[Angle in a semi-circle]**

And,

∠ADC + ∠ADB = 180^{o} **[Linear pair]**

Thus, ∠ADC = 90^{o}

Now, in right ∆ABD and ∆ACD

AB = AC **[Given]**

AD = AD **[Common]**

∠ADB = ∠ADC = 90^{o}

Hence, by R.H.S criterion of congruence.

∆ABD ≅ ∆ACD

Now, by CPCT

BD = DC

∴ D is the mid-point of BC.

**4. In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65 ^{o}, calculate ∠DEC.**

**Solution**

Join OE.

Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.

∠EOC = 2∠EBC = 2 x 65^{o} = 130^{o}

Now, in ∆OEC

OE = OC **[Radii of the same circle]**

So, ∠OEC = ∠OCE

But, in ∆EOC by angle sum property

∠OEC + ∠OCE + ∠EOC = 180^{o} **[Angles of a triangle]**

⇒ ∠OCE + ∠OCE + ∠EOC = 180^{o}

⇒ 2 ∠OCE + 130^{o} = 180^{o}

⇒ 2 ∠OCE = 180^{o} – 130^{o}

⇒ ∠OCE = 50^{o}/2 = 25^{o}

And, AC || ED **[Given]**

∠DEC = ∠OCE **[Alternate angles]**

Thus, ∠DEC = 25^{o}

^{}

**5. The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.**

**Solution**

Let ABCD be a cyclic quadrilateral and PQRS be the quadrilateral formed by the angle bisectors of angle ∠A, ∠B, ∠C and ∠D.

Required to prove: PQRS is a cyclic quadrilateral.

Proof:

By angle sum property of a triangle

In ∆APD,

∠PAD + ∠ADP + ∠APD = 180^{o} **…. (i)**

And, in ∆BQC

∠QBC + ∠BCQ + ∠BQC = 180^{o} **…. (ii)**

Adding (i) and (ii), we get

∠PAD + ∠ADP + ∠APD + ∠QBC + ∠BCQ + ∠BQC = 180^{o} + 180^{o} = 360^{o} **… (iii)**

But,

∠PAD + ∠ADP + ∠QBC + ∠BCQ = ½ [∠A + ∠B + ∠C + ∠D]

= ½ ×360^{o} = 180^{o}

∴ ∠APD + ∠BQC = 360^{o} – 180^{o} = 180^{o} **[From (iii)]**

But, these are the sum of opposite angles of quadrilateral PRQS.

∴ Quadrilateral PQRS is also a cyclic quadrilateral.

**6. In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:**

**(i) ∠BDC**

**(ii) ∠BEC**

**(iii) ∠BAC**

**Solution**

**(i)** Given that BD is a diameter of the circle.

And, the angle in a semicircle is a right angle.

So, ∠BCD = 90°

Also given that,

∠DBC = 58°

In ∆BDC,

∠DBC + ∠BCD + ∠BDC = 180^{o}

⇒ 58° + 90° + ∠BDC = 180^{o}

⇒ 148^{o} + ∠BDC = 180^{o}

⇒ ∠BDC = 180^{o} – 148^{o}

Thus, ∠BDC = 32^{o}

**(ii)** We know that, the opposite angles of a cyclic quadrilateral are supplementary.

So, in cyclic quadrilateral BECD

∠BEC + ∠BDC = 180^{o}

⇒ ∠BEC + 32^{o} = 180^{o}

⇒ ∠BEC = 148^{o}

**(iii)** In cyclic quadrilateral ABEC,

∠BAC + ∠BEC = 180^{o} [Opposite angles of a cyclic quadrilateral are supplementary]

⇒ ∠BAC + 148^{o} = 180^{o}

⇒ ∠BAC = 180^{o} – 148^{o}

Thus, ∠BAC = 32^{o}

^{}

**7. D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.**

**Solution**

Given,

∆ABC, AB = AC and D and E are points on AB and AC such that AD = AE.

And, DE is joined.

Required to prove: Points B, C, E and D are concyclic

Proof:

In ∆ABC,

AB = AC **[Given]**

So, ∠B = ∠C **[Angles opposite to equal sides]**

Similarly,

In ∆ADE,

AD = AE **[Given]**

So, ∠ADE = ∠AED **[Angles opposite to equal sides]**

Now, in ∆ABC we have

AD/AB = AE/AC

Hence, DE || BC **[Converse of BPT]**

So,

∠ADE = ∠B **[Corresponding angles]**

⇒ (180^{o} – ∠EDB) = ∠B

⇒ ∠B + ∠EDB = 180^{o}

But, it’s proved above that

∠B = ∠C

So, ∠C + ∠EDB = 180^{o}

Thus, opposite angles are supplementary.

Similarly,

∠B + ∠CED = 180^{o}

Hence, B, C, E and D are concyclic.

**8. In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If **∠**ADC = 92 ^{o}, **∠

**FAE = 20**∠

^{o}; determine**BCD. Given reason in support of your answer.**

**Solution**

Given,

In cyclic quad. ABCD

AF || CB and DA is produced to E such that ∠ADC = 92^{o} and ∠FAE = 20^{o}

So, ∠B + ∠D = 180^{o}

∠B + 92^{o} = 180^{o}

⇒ ∠B = 88^{o}

As AF || CB, ∠FAB = ∠B = 88^{o}

But, ∠FAD = 20^{o} **[Given]**

Ext. ∠BAE = ∠BAF + ∠FAE

= 88^{o} + 22^{o} = 108^{o}

But, Ext. ∠BAE = ∠BCD

∴ ∠BCD = 108^{o}

^{}

**9. If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66 ^{o }and ∠ABC = 80^{o}. Calculate:**

**(i) ∠DBC,**

**(ii) ∠IBC,**

**(iii) ∠BIC**

**Solution**

Join DB and DC, IB and IC.

Given, if ∠BAC = 66^{o }and ∠ABC = 80^{o}, I is the incentre of the ∆ABC.

**(i) **As it’s seen that ∠DBC and ∠DAC are in the same segment,

So, ∠DBC = ∠DAC

But, ∠DAC = ½ ∠BAC = ½ x 66^{o} = 33^{o}

Thus, ∠DBC = 33^{o}

**(ii)** And, as I is the incentre of ∆ABC, IB bisects ∠ABC.

∴ ∠IBC = ½ ∠ABC = ½ x 80^{o }= 40^{o}

(iii) In ∆ABC, by angle sum property

∠ACB = 180^{o} – (∠ABC + ∠BAC)

⇒ ∠ACB = 180^{o} – (80^{o} + 66^{o})

⇒ ∠ACB = 180^{o} – 156^{o}

⇒ ∠ACB = 34^{o}

And since, IC bisects ∠C

Thus, ∠ICB = ½ ∠C = ½ x 34^{o} = 17^{o}

Now, in ∆IBC

∠IBC + ∠ICB + ∠BIC = 180^{o}

⇒ 40^{o} + 17^{o} + ∠BIC = 180^{o}

⇒ 57^{o} + ∠BIC = 180^{o}

⇒ ∠BIC = 180^{o} – 57^{o}

∴ ∠BIC = 123^{o}

^{}

**10. In the given figure, AB = AD = DC = PB and ∠DBC = x ^{o}. Determine, in terms of x:**

**(i) ∠ABD, (ii) ∠APB.**

**Hence or otherwise, prove that AP is parallel to DB.**

**Solution**

Given, AB = AD = DC = PB and ∠DBC = x^{o}

Join AC and BD.

Proof:

∠DAC = ∠DBC = x^{o} **[Angles in the same segment]**

And, ∠DCA = ∠DAC = x^{o} **[As AD = DC]**

Also, we have

∠ABD = ∠DAC **[Angles in the same segment]**

And, in ∆ABP

Ext. ∠ABD = ∠BAP + ∠APB

But, ∠BAP = ∠APB **[Since, AB = BP]**

⇒ 2 x^{o} = ∠APB + ∠APB = 2∠APB

⇒ 2∠APB = 2x^{o}

So, ∠APB = x^{o}

Thus, ∠APB = ∠DBC = x^{o}

But these are corresponding angles,

∴ AP || DB.

**11. In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.**

**Solution**

Join EB.

Then, in cyclic quad.ABEP

∠APE + ∠ABE = 180^{o} **….. (i) [Opposite angles of a cyclic quad. are supplementary]**

Similarly, in cyclic quad.BCQE

∠CQE + ∠CBE = 180^{o} **….. (ii) [Opposite angles of a cyclic quad. are supplementary]**

Adding (i) and (ii), we have

∠APE + ∠ABE + ∠CQE + ∠CBE = 180^{o }+ 180^{o} = 360^{o}

⇒ ∠APE + ∠ABE + ∠CQE + ∠CBE = 360^{o}

But, ∠ABE + ∠CBE = 180^{o} **[Linear pair]**

⇒ ∠APE + ∠CQE + 180^{o} = 360^{o}

⇒ ∠APE + ∠CQE = 180^{o}

∴ ∠APE and ∠CQE are supplementary.

**12. In the given, AB is the diameter of the circle with centre O.**

**If ∠ADC = 32 ^{o}, find angle BOC.**

**Solution**

Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle.

Thus, ∠AOC = 2∠ADC

∠AOC = 2×32^{o} = 64^{o}

As ∠AOC and ∠BOC are linear pair, we have

∠AOC + ∠BOC = 180^{o}

⇒ 64^{o} + ∠BOC = 180^{o}

⇒ ∠BOC = 180^{o} – 64^{o}

∴ ∠BOC = 116^{o}