# 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 30o and 40o 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) – (40o + 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,

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 ×(360o – 112o) = 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 = 90o [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. O1 and O2 are the centers of two circles.

Solution

It’s seen that,

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

∠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,

(iii) ∠ABC.

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 180o]

(ii) Now,

∠ADC and ∠DAB are corresponding angles.

So,

(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]

∠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

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

⇒ 60o = ∠CBA + 20°

∴ ∠CBA = 40°

9. Calculate:

(i) ∠CDB,

(ii) ∠ABC,

(iii) ∠ACB.

Solution

Here, we have

∠CDB = ∠BAC = 49°

[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 = 75o; ∠ABD = 58o and ∠ADC = 77o.

Find:

(i) ∠BDC,

(ii) ∠BCD,

(iii) ∠BCA.

Solution

(i) By angle sum property of triangle ABD,

∠ADB = 180o – 75o – 58o = 47o

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

(ii) ∠BAD + ∠BCD = 180o

[Sum of opposite angles of a cyclic quadrilateral is 180o]

Thus, ∠BCD = 180o – 75o = 105o

(iii) ∠BCA = ∠ADB = 47o

[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:

(ii) ∠AEB

Solution

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

So, ∠ADB = ∠ACB = 60o

(ii) Now, join OA and OB.

And, we have

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

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

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

Solution

Given, ∠CAB = 75o and ∠CBA = 50o

In ∆ABC, by angle sum property we have

∠ACB = 180o – (∠CBA + ∠CAB)

= 180o – (50o + 75o) = 180o – 125o

= 55o

And,

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

Now, taking ∆ABD

∠DAB + ∠ABD + ∠ADB = 180o

⇒ ∠DAB + ∠ABD + 55o = 180o

⇒ ∠DAB + ∠ABD = 180o – 55o

⇒ ∠DAB + ∠ABD = 125o

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

Solution

From the fig. its seem that,

∠ACB = 90o [Angle in a semi-circle is 90o]

Also,

∠ABC = 180o – ∠ADC = 180o – 130o = 50o

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

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

∠BAC = 90o – ∠ABC

= 90o – 50o

Thus, ∠BAC = 40o

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

Solution

So, we have

∠ADC = ½ ∠AOC = ½ x 110o = 55o

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

Also, we know that

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

∴ ∠BDC = 90o – ∠ADC = 90o – 55o

∠BDC = 35o

15. In the following figure, O is the centre of the circle; ∠AOB = 60o and BDC = 100o, find ∠OBC.

Solution

Form the figure, we have

∠ACB = ½ ∠AOB = ½ x 60o = 30o

[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– 100o – 30o = 50o

∴  ∠OBC = 50o

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

Solution

(i) It’s seen that,

∠DBC = ∠DAC = 27o

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

(ii) It’s seen that,

And,

∠ACD = ∠ABD = 50o

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

Thus,

∠DCB = ∠ACD + ∠ACB = 50o + 33o = 83o

∠DAB + ∠DCB = 180o

27o + ∠CAB + 83o = 180o

Thus,

∠CAB = 180o – 110o = 70o

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

Solution

(i) Form the fig. its seen that,

∠DCE = 90o – ∠CDE = 90o – 40o = 50o

∴ ∠DEC = ∠OCB = 50o

(ii) In ∆BOC, we have

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

∠OBC = 80o – 50o = 30o [Given ∠AOC = 80o]

∴ ∠ABC = 30o

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 = 90o [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 = 96o, find ∠BCD and ∠BFE.

(ii) Prove that AD is parallel to FE.

Solution

(i) ABCD is a cyclic quadrilateral

So, ∠BAD + ∠BCD = 180o

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

∠BCD = 180o – 96o = 84o

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

Similarly, BCEF is a cyclic quadrilateral

So, ∠BCE + ∠BFE = 180o

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

∠BFE = 180o – 96= 84o

(ii) Now, ∠BAD + ∠BFE = 96o + 84o = 180o

But these two are interior angles on the same side of a pair of lines AD and 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 = 180o

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

Thus, ∠BAD = ∠BCD = 90o

Similarly, the remaining two angles are 90o 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 = 180o

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

Thus, ∠BAD = ∠BCD = 90o

Similarly, the remaining two angles are 90o 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 = 180o

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

∠C + ∠DEC = 180o …. (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]

⇒ 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 = 90o and ∠BQP = 90o

[Angle in a semicircle is a right angle]

Now, adding both these angles we get

∠AQP + ∠BQP = 180o

∠AQB = 180o

∴ 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 (360o - a)

a + 2b = 180o ….. (1)

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

So, a = b

Now, using the above relationship in (1)

3a = 180o

⇒ a = 60o

Also, OC || BA

∠COA + ∠OAB = 180o

⇒ 60o + ∠OAB = 180o

∴ ∠OAB = 120o

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.

Now, it’s seen that

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

Similarly,

Adding (1) and (2), we have

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 = 29o

Calculate: (i) ∠RNM;

(ii) ∠NRM.

Solution

(i) Join RN and MS

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

So, by angle sum property of ∆RMS

∠RMS = 90o – 29o = 61o

And, ∠RNM = 180o – ∠RSM = 180o – 61o = 119o

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

(ii) Now as RS || NM,

∠NMR = ∠MRS = 29o [Alternate angles]

⇒ ∠NMS = 90o + 29o = 119o

Also, we know that

∠NRS + ∠NMS = 180o

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

∠NRM + 29o + 119o = 180o

⇒ ∠NRM = 180o – 148o

∴ ∠NRM = 32o

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

Solution

Join AC and BD.

So, we have

∠CAD = 90o and ∠CBD = 90o

[Angle is a semicircle is a right angle]

And, AB || CD

Thus,

∠ADB = 180o – 25o – ∠BAC = 180o – 25o – 115o = 40o

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

Finally,

[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 = 180o … (i)

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

Similarly, as PQDB is a cyclic quadrilateral

∠PQD + ∠DBP = 180o …(ii)

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

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

∠CAP + ∠DBP = 180o

Or ∠CAB + ∠DBA = 180o

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]

[Angles opposite to equal sides are equal]

And,

Similarly,

∠CDA = 180o – ∠PDA = 180o – ∠PAD [From (1)]

As the opposite angles of a cyclic quadrilateral are supplementary,

∠ABC = 180o – ∠CDA = 180o – (180o – ∠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:

(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

DC = DC [Common]

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

Hence, by SAS criterion of congruence

∴ 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 = 110o, calculate:

(i) ∠AFE, (ii) ∠FAB.

Solution

Join AE, OB and OC.

(i) As AOD is the diameter

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

But, given ∠DEF = 110o

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

(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 = 180o [Since, AOD is a straight line]

Thus, ∠AOB = ∠BOC = ∠COD = 60o

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 = 180o – ∠AOB

= 180o – 60o

= 120o

∴ ∠OAB = ∠OBA = 60o

∠DEF + ∠DAF = 180o

∠DAF = 180o – ∠DEF

= 180o – 110o

= 70o

Thus,

∠FAB = ∠DAF + ∠OAB

= 70o + 60o = 130o

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:

(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

(ii) Now, in ∆ABC and ∆ADB

AB = AB [Common]

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

Hence, by SAS criterion of congruence

∴ 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 = 2×75o = 150o

⇒ ∠POQ + ∠QOR + ∠ROS = 150o

⇒ ∠POQ = ∠QOR = ∠ROS = 150o/3 = 50o

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 = 180o

⇒ ∠OPQ + ∠OQP + 50o = 180o

⇒ ∠OPQ + ∠OQP = 130o

⇒ 2∠OPQ = 130o

⇒ ∠OPQ = ∠OPQ = 130o/2 = 65o

Similarly, we can prove that

In ∆OQR,

∠OQR = ∠ORQ = 65o

And in ∆ORS,

∠ORS = ∠OSR = 65o

Hence,

(i) ∠POS = 150o

(ii) ∠QOR = 50o and

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

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 = 60o

(ii) Now,

∠ACB = ½ (60o) = 30o

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

∠AOC = 360o/ 8 = 45o

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

∠ABC = ½ ∠AOC

⇒ ∠ABC = 45o/2 = 22.5o

### Exercise 17(C)

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

Solution

Now,

∠ABD = ∠ACD = 30o [Angles in the same segment]

In ∆ADB, by angle sum property we have

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

So,

x + 90o + 30o = 180o

⇒ x = 180o – 120o

Hence, x = 60o

2. In the given figure, ABC is a triangle in which ∠BAC = 30o. 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 = 2×30o = 60o

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 = 180o

⇒ ∠OBC + ∠OBC + 60o = 180o

⇒ 2∠OBC = 180o – 60o = 120o

⇒ ∠OBC = 120o/2 = 60o

So, ∠OBC = ∠OCB = ∠BOC = 60o

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.

Proof:

It’s seen that,

∠ADB = 90o [Angle in a semi-circle]

And,

Now, in right ∆ABD and ∆ACD

AB = AC [Given]

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 = 65o, 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 65o = 130o

Now, in ∆OEC

OE = OC [Radii of the same circle]

So, ∠OEC = ∠OCE

But, in ∆EOC by angle sum property

∠OEC + ∠OCE + ∠EOC = 180o [Angles of a triangle]

⇒ ∠OCE + ∠OCE + ∠EOC = 180o

⇒ 2 ∠OCE + 130o = 180o

⇒ 2 ∠OCE = 180o – 130o

⇒ ∠OCE = 50o/2 = 25o

And, AC || ED [Given]

∠DEC = ∠OCE [Alternate angles]

Thus, ∠DEC = 25o

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,

And, in ∆BQC

∠QBC + ∠BCQ + ∠BQC = 180o …. (ii)

Adding (i) and (ii), we get

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

But,

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

= ½ ×360o = 180o

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

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

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 = 180o

⇒ 58° + 90° + ∠BDC = 180o

⇒ 148o + ∠BDC = 180o

⇒ ∠BDC = 180o – 148o

Thus, ∠BDC = 32o

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

∠BEC + ∠BDC = 180o

⇒ ∠BEC + 32o = 180o

⇒ ∠BEC = 148o

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

⇒ ∠BAC + 148o = 180o

⇒ ∠BAC = 180o – 148o

Thus, ∠BAC = 32o

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,

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

Now, in ∆ABC we have

Hence, DE || BC [Converse of BPT]

So,

⇒ (180o – ∠EDB) = ∠B

⇒ ∠B + ∠EDB = 180o

But, it’s proved above that

∠B = ∠C

So, ∠C + ∠EDB = 180o

Thus, opposite angles are supplementary.

Similarly,

∠B + ∠CED = 180o

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 = 92oFAE = 20o; determine BCD. Given reason in support of your answer.

Solution

Given,

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

So, ∠B + ∠D = 180o

∠B + 92o = 180o

⇒ ∠B = 88o

As AF || CB, ∠FAB = ∠B = 88o

Ext. ∠BAE = ∠BAF + ∠FAE

= 88o + 22o = 108o

But, Ext. ∠BAE = ∠BCD

∴ ∠BCD = 108o

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

(i) ∠DBC,

(ii) ∠IBC,

(iii) ∠BIC

Solution

Join DB and DC, IB and IC.

Given, if ∠BAC = 66and ∠ABC = 80o, 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 66o = 33o

Thus, ∠DBC = 33o

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

∴ ∠IBC = ½ ∠ABC = ½ x 80= 40o

(iii) In ∆ABC, by angle sum property

∠ACB = 180o – (∠ABC + ∠BAC)

⇒ ∠ACB = 180o – (80o + 66o)

⇒ ∠ACB = 180o – 156o

⇒ ∠ACB = 34o

And since, IC bisects ∠C

Thus, ∠ICB = ½ ∠C = ½ x 34o = 17o

Now, in ∆IBC

∠IBC + ∠ICB + ∠BIC = 180o

⇒ 40o + 17o + ∠BIC = 180o

⇒ 57o + ∠BIC = 180o

⇒ ∠BIC = 180o – 57o

∴ ∠BIC = 123o

10. In the given figure, AB = AD = DC = PB and ∠DBC = xo. 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 = xo

Join AC and BD.

Proof:

∠DAC = ∠DBC = xo [Angles in the same segment]

And, ∠DCA = ∠DAC = xo [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 xo = ∠APB + ∠APB = 2∠APB

⇒ 2∠APB = 2xo

So, ∠APB = xo

Thus, ∠APB = ∠DBC = xo

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.

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

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

Adding (i) and (ii), we have

∠APE + ∠ABE + ∠CQE + ∠CBE = 180+ 180o = 360o

⇒ ∠APE + ∠ABE + ∠CQE + ∠CBE = 360o

But, ∠ABE + ∠CBE = 180o [Linear pair]

⇒ ∠APE + ∠CQE + 180o = 360o

⇒ ∠APE + ∠CQE = 180o

∴ ∠APE and ∠CQE are supplementary.

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

If ∠ADC = 32o, find angle BOC.

Solution

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