## RS Aggarwal Solutions for Class 9 Maths Exercise 12C PDF

Geometrical structures like Circles are of greater importance in order to understand some of the fundamental concepts in Mathematics. The solutions are done with the aim of making the students comfortable in solving exercise wise problems. This is the third exercise of Chapter 12 which contains solutions designed by subject experts at BYJUâ€™S. The solutions are explained in simple language which match the understanding capacity of students. This exercise has problems which are solved based on cyclic quadrilaterals. RS Aggarwal Solutions for Class 9 Maths Chapter 12 Circles Exercise 12C are provided here.

## RS Aggarwal Solutions for Class 9 Chapter 12: Circles Exercise 12C Download PDF

## Access RS Aggarwal Solutions for Class 9 Chapter 12: Circles Exercise 12C

### Exercise 12(C) PAGE: 482

**1. In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that âˆ DBC = 60 ^{o} and âˆ BAC = 40^{o}. Find **

**(i) âˆ BCD,**

**(ii) âˆ CAD.**

**Solution:**

(i) We know that the angles in the same segment are equal

So we get

âˆ BDC = âˆ BAC = 40^{o}

Consider â–³ BCD

Using the angle sum property

âˆ BCD + âˆ BDC + âˆ DBC = 180^{o}

By substituting the values

âˆ BCD + 40^{o} + 60^{o} = 180^{o}

On further calculation

âˆ BCD = 180^{o} – 40^{o} – 60^{o}

By subtraction

âˆ BCD = 180^{o} – 100^{o}

So we get

âˆ BCD = 80^{o}

(ii) We know that the angles in the same segment are equal

So we get

âˆ CAD = âˆ CBD

So âˆ CAD = 60^{o}

**2. In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If âˆ PSR = 150 ^{o}, find âˆ RPQ.**

**Solution:**

We know that PQRS is a cyclic quadrilateral

It can be written as

âˆ PSR+ âˆ PQR = 180^{o}

By substituting the values

150^{o} + âˆ PQR = 180^{o}

On further calculation

âˆ PQR = 180^{o} – 150^{o}

By subtraction

âˆ PQR = 30^{o}

We know that the angle in semi-circle is a right angle

âˆ PRQ = 90^{o}

Consider â–³ PRQ

Using the angle sum property

âˆ PQR + âˆ PRQ + âˆ RPQ = 180^{o}

By substituting the values

30^{o} + 90^{o} + âˆ RPQ = 180^{o}

On further calculation

âˆ RPQ = 180^{o} – 30^{o} – 90^{o}

By subtraction

âˆ RPQ = 180^{o} – 120^{o}

So we get

âˆ RPQ = 60^{o}

Therefore, âˆ RPQ = 60^{o}.

**3. In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130 ^{o} at the centre. If AB is extended to P, find âˆ PBC.**

**Solution:**

Consider a point D on the arc CA and join DC and AD

We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment

So we get

âˆ 2 = 2 âˆ 1

By substituting the values

130^{o} = 2 âˆ 1

So we get

âˆ 1 = 65^{o}

From the figure we know that the exterior angle of a cyclic quadrilateral = interior opposite angle

âˆ PBC = âˆ 1

So we get âˆ PBC = 65^{o}

Therefore, âˆ PBC = 65^{o}.

**4. In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced to F. If âˆ ABC = 92 ^{o} and âˆ FAE = 20^{o}, find âˆ BCD.**

**Solution:**

We know that ABCD is a cyclic quadrilateral

So we get

âˆ ABC + âˆ ADC = 180^{o}

By substituting the values

92^{o} + âˆ ADC = 180^{o}

On further calculation

âˆ ADC = 180^{o} – 92^{o}

By subtraction

âˆ ADC = 88^{o}

We know that AE || CD

From the figure we know that

âˆ EAD = âˆ ADC = 88^{o}

We know that the exterior angle of a cyclic quadrilateral = interior opposite angle

So we get

âˆ BCD = âˆ DAF

We know that

âˆ BCD = âˆ EAD + âˆ EAF

It is given that âˆ FAE = 20^{o}

By substituting the values

âˆ BCD = 88^{o} + 20^{o}

By addition

âˆ BCD = 108^{o}

Therefore, âˆ BCD = 108^{o}.

**5. In the given figure, BD = DC and âˆ CBD = 30 ^{o}, find âˆ BAC.**

**Solution:**

It is given that BD = DC

From the figure we know that

âˆ BCD = âˆ CBD = 30^{o}

Consider â–³ BCD

Using the angle sum property

âˆ BCD + âˆ CBD + âˆ CDB = 180^{o}

By substituting the values

30^{o} + 30^{o} + âˆ CDB = 180^{o}

On further calculation

âˆ CDB = 180^{o} – 30^{o} – 30^{o}

By subtraction

âˆ CDB = 180^{o} – 60^{o}

So we get

âˆ CDB = 120^{o}

We know that the opposite angles of a cyclic quadrilateral are supplementary

It can be written as

âˆ CDB + âˆ BAC = 180^{o}

By substituting the values

120^{o} + âˆ BAC = 180^{o}

On further calculation

âˆ BAC = 180^{o} – 120^{o}

By subtraction

âˆ BAC = 60^{o}

Therefore, âˆ BAC = 60^{o}.

**6. In the given figure, O is the centre of the given circle and measure of arc ABC is 100 ^{o}. Determine âˆ ADC and âˆ ABC.**

**Solution:**

We know that the angle subtended by an arc is twice the angle subtended by it on the circumference

From the figure we know that

âˆ AOC = 100^{o}

So we get

âˆ AOC = 2 âˆ ADC

It can be written as

âˆ ADC = Â½ âˆ AOC

By substituting the values

âˆ ADC = Â½ (100^{o})

So we get

âˆ ADC = 50^{o}

We know that the opposite angles of a cyclic quadrilateral are supplementary

It can be written as

âˆ ADC + âˆ ABC = 180^{o}

By substituting the values

50^{o} + âˆ ABC = 180^{o}

On further calculation

âˆ ABC = 180^{o} – 50^{o}

By subtraction

âˆ ABC = 130^{o}

Therefore, âˆ ADC = 50^{o} and âˆ ABC = 130^{o}.

**7. In the given figure, â–³ ABC is equilateral. Find **

**(i) âˆ BDC,**

**(ii) âˆ BEC.**

**Solution:**

It is given that â–³ ABC is equilateral

We know that

âˆ BAC = âˆ ABC = âˆ ACB = 60^{o}

(i) We know that the angles in the same segment of a circle are equal

âˆ BDC = âˆ BAC = 60^{o}

So we get

âˆ BDC = 60^{o}

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

So we get

âˆ BAC + âˆ BEC = 180^{o}

By substituting the values

60^{o} + âˆ BEC = 180^{o}

On further calculation

âˆ BEC = 180^{o} – 60^{o}

By subtraction

âˆ BEC = 120^{o}

**8. In the adjoining figure, ABCD is a cyclic quadrilateral in which âˆ BCD = 100 ^{o} and âˆ ABD = 50^{o}. Find âˆ ADB.**

**Solution:**

It is given that ABCD is a cyclic quadrilateral

We know that the opposite angles of a cyclic quadrilateral are supplementary

So we get

âˆ A + âˆ C = 180^{o}

By substituting the values

âˆ A + 100^{o} = 180^{o}

On further calculation

âˆ A = 180^{o} – 100^{o}

By subtraction

âˆ A = 80^{o}

Consider â–³ ABD

Using the angle sum property

âˆ A + âˆ ABD + âˆ ADB = 180^{o}

By substituting the values

80^{o} + 50^{o} + âˆ ADB = 180^{o}

On further calculation

âˆ ADB = 180^{o} – 80^{o} – 50^{o}

By subtraction

âˆ ADB = 180^{o} – 130^{o}

So we get

âˆ ADB = 50^{o}

Therefore, âˆ ADB = 50^{o}.

**9. In the given figure, O is the centre of a circle and âˆ BOD =150 ^{o}. Find the values of x and y.**

**Solution:**

It is given that O is the centre of a circle and âˆ BOD =150^{o}

We know that

Reflex âˆ BOD = (360^{o} – âˆ BOD)

By substituting the values

Reflex âˆ BOD = (360^{o} â€“ 150^{o})

By subtraction

Reflex âˆ BOD = 210^{o}

Consider x = Â½ (reflex âˆ BOD)

By substituting the value

x = 210/2

So we get

x = 105^{o}

We know that

x + y = 180^{o}

By substituting the values

105^{o} + y = 180^{o}

On further calculation

y = 180^{o} – 105^{o}

By subtraction

y = 75^{o}

Therefore, the value of x is 105^{o} and y is 75^{o}.

**10. In the given figure, O is the centre of the circle and âˆ DAB = 50 ^{o}. Calculate the values of x and y.**

**Solution:**

It is given that O is the centre of the circle and âˆ DAB = 50^{o}

We know that the radii of the circle are equal

OA = OB

From the figure we know that

âˆ OBA = âˆ OAB = 50^{o}

Consider â–³ OAB

Using the angle sum property

âˆ OAB + âˆ OBA + âˆ AOB = 180^{o}

By substituting the values

50^{o} + 50^{o} + âˆ AOB = 180^{o}

On further calculation

âˆ AOB = 180^{o} – 50^{o} – 50^{o}

By subtraction

âˆ AOB = 180^{o} – 100^{o}

So we get

âˆ AOB = 80^{o}

From the figure we know that AOD is a straight line

It can be written as

x = 180^{o} – âˆ AOB

By substituting the values

x = 180^{o} – 80^{o}

By subtraction

x = 100^{o}

We know that the opposite angles of a cyclic quadrilateral are supplementary

So we get

âˆ DAB + âˆ BCD = 180^{o}

By substituting the values

50^{o} + âˆ BCD = 180^{o}

On further calculation

âˆ BCD = 180^{o} – 50^{o}

By subtraction

y = âˆ BCD = 130^{o}

Therefore, the value of x is 100^{o} and y is 130^{o}.

**11. In the given figure, sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If âˆ CBF = 130 ^{o} and âˆ CDE = x^{o}, find the value of x.**

**Solution:**

In a cyclic quadrilateral we know that the exterior angle is equal to the interior opposite angle

So we get

âˆ CBF = âˆ CDA

It can be written as

130^{o} = 180^{o} – x

On further calculation

x = 180^{o} – 130^{o}

By subtraction

x = 50^{o}

**12. In the given figure, AB is a diameter of a circle with centre O and DO || CB. If âˆ BCD = 120 ^{o}, calculate**

**(i) âˆ BAD,**

**(ii) âˆ ABD,**

**(iii) âˆ CBD,**

**(iv) âˆ ADC.**

**Also, show that â–³ AOD is an equilateral triangle.**

**Solution:**

It is given that AB is a diameter of a circle with centre O and DO || CB

(i) We know that ABCD is a cyclic quadrilateral

It can be written as

âˆ BCD + âˆ BAD = 180^{o}

By substituting the values

120^{o} + âˆ BAD = 180^{o}

On further calculation

âˆ BAD = 180^{o} – 120^{o}

By subtraction

âˆ BAD = 60^{o}

(ii) We know that the angle in a semi-circle is right angle

âˆ BDA = 90^{o}

Consider â–³ ABD

By using the angle sum property

âˆ BDA + âˆ BAD + âˆ ABD = 180^{o}

By substituting the values

90^{o} + 60^{o} + âˆ ABD = 180^{o}

On further calculation

âˆ ABD = 180^{o} – 90^{o} – 60^{o}

By subtraction

âˆ ABD = 180^{o} – 150^{o}

So we get

âˆ ABD = 30^{o}

(iii) We know that OD = OA

So we get âˆ ODA = âˆ OAD = âˆ BAD = 60^{o}

From the figure we know that

âˆ ODB + âˆ ODA = 90^{o}

By substituting the values

âˆ ODB + 60^{o} = 90^{o}

On further calculation

âˆ ODB = 90^{o} – 60^{o}

By subtraction

âˆ ODB = 30^{o}

It is given that DO || CB

We know that the alternate angles are equal

âˆ CDB = âˆ ODB = 30^{o}

(iv) From the figure we know that

âˆ ADC = âˆ ADB + âˆ CDB

By substituting the values

âˆ ADC = 90^{o} + 30^{o}

By addition

âˆ ADC = 120^{o}

Consider â–³ AOD

By using the angle sum property

âˆ ODA + âˆ OAD + âˆ AOD = 180^{o}

By substituting the values

60^{o} + 60^{o} + âˆ AOD = 180^{o}

On further calculation

âˆ AOD = 180^{o} – 60^{o} – 60^{o}

By subtraction

âˆ AOD = 180^{o} – 120^{o}

So we get

âˆ AOD = 60^{o}

We know that all the angles of the â–³ AOD is 60^{o}

Therefore, it is proved that â–³ AOD is an equilateral triangle.

**13. Two chords AB and CD of a circle intersect each other at P outside the circle. If AB = 6cm, BP = 2cm and PD = 2.5cm, find CD.**

**Solution:**

It is given that AB and CD of a circle intersect each other at P outside the circle. If AB = 6cm, BP = 2cm and PD = 2.5cm

So we get

AP Ã— BP = CP Ã— DP

From the figure we know that CP = CD + DP

By substituting the values

8 Ã— 2 = (CD + 2.5) Ã— 2.5 cm

Consider x = CD

So we get

8 Ã— 2 = (x + 2.5) Ã— 2.5

On further calculation

16 = 2.5x + 6.25

It can be written as

2.5x = 16 â€“ 6.25

By subtraction

2.5x = 9.75

By division

x = 9.75/2.5

So we get

x = 3.9cm

Therefore, CD = 3.9cm.

**14. In the given figure, O is the centre of a circle. If âˆ AOD = 140 ^{o} and âˆ CAB = 50^{o}, calculate**

**(i) âˆ EDB,**

**(ii) âˆ EBD.**

**Solution:**

(i) We know that

âˆ BOD + âˆ AOD = 180^{o}

By substituting the values

âˆ BOD + 140^{o} = 180^{o}

On further calculation

âˆ BOD = 180^{o} – 140^{o}

By subtraction

âˆ BOD = 40^{o}

We know that OB = OD

So we get âˆ OBD = âˆ ODB

Consider â–³ OBD

By using the angle sum property

âˆ BOD + âˆ OBD + âˆ ODB = 180^{o}

We know that âˆ OBD = âˆ ODB

So we get

40^{o} + 2 âˆ OBD = 180^{o}

On further calculation

2 âˆ OBD = 180^{o} – 40^{o}

By subtraction

2 âˆ OBD = 140^{o}

By division

âˆ OBD = 70^{o}

We know that ABCD is a cyclic quadrilateral

âˆ CAB + âˆ BDC = 180^{o}

âˆ CAB + âˆ ODB + âˆ ODC = 180^{o}

By substituting the values

50^{o} + 70^{o} + âˆ ODC = 180^{o}

On further calculation

âˆ ODC = 180^{o} – 50^{o} – 70^{o}

By subtraction

âˆ ODC = 180^{o} – 120^{o}

So we get

âˆ ODC = 60^{o}

Using the angle sum property

âˆ EDB + âˆ ODC + âˆ ODB = 180^{o}

By substituting the values

âˆ EDB + 60^{o} + 70^{o} = 180^{o}

On further calculation

âˆ EDB = 180^{o} – 60^{o} – 70^{o}

By subtraction

âˆ EDB = 180^{o} – 130^{o}

So we get

âˆ EDB = 50^{o}

(ii) We know that

âˆ EDB + âˆ OBD = 180^{o}

By substituting the values

âˆ EDB + 70^{o} = 180^{o}

On further calculation

âˆ EDB = 180^{o} – 70^{o}

By subtraction

âˆ EDB = 110^{o}

**15. In the given figure, â–³ ABC is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E respectively. Prove that DE || BC.**

**Solution:**

We know that â–³ ABC is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E

So AB = AC

We get

âˆ ACB = âˆ ABC

It can be written as

âˆ ADE = âˆ ACB = âˆ ABC

So we get

âˆ ADE = âˆ ABC

DE || BC

Therefore, it is proved that DE || BC.

**16. In the given figure, AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove that â–³ AEB is isosceles.**

**Solution:**

It is given that AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E

We know that in a cyclic quadrilateral exterior angle is equal to the interior opposite angle.

It can be written as

Exterior âˆ EDC = âˆ A

Exterior âˆ DCE = âˆ B

We know that AB || CD

So we get

âˆ EDC = âˆ B and âˆ DCE = âˆ A

We get

âˆ A = âˆ B

Therefore, it is proved that â–³ AEB is isosceles.

**17. In the given figure, âˆ BAD = 75 ^{o}, âˆ DCF = x^{o} and âˆ DEF = y^{o}. Find the values of x and y.**

**Solution:**

In a cyclic quadrilateral we know that the exterior angle is equal to the interior opposite angle

So we get

âˆ BAD = âˆ DCF = 75^{o}

It can be written as

âˆ DCF = x = 75^{o}

We get x = 75^{o}

We know that the opposite angles of a cyclic quadrilateral is 180^{o }

So we get

âˆ DCF + âˆ DEF = 180^{o}

By substituting the values

75^{o} + âˆ DEF = 180^{o}

On further calculation

âˆ DEF = 180^{o} – 75^{o}

By subtraction

âˆ DEF = y = 105^{o}

Therefore, the values of x and y is 75^{o} and 105^{o}.

**18. In the given figure, ABCD is a quadrilateral in which AD = BC and âˆ ADC = âˆ BCD. Show that the points A, B, C, D lie on a circle.**

**Solution:**

It is given that ABCD is a quadrilateral in which AD = BC and âˆ ADC = âˆ BCD

Construct DE âŠ¥Â AB and CF âŠ¥Â AB

Consider â–³ ADE and â–³ BCF

We know that

âˆ AED + âˆ BFC = 90^{o}

From the figure it can be written as

âˆ ADE = âˆ ADC – 90^{o} = âˆ BCD – 90^{o} = âˆ BCF

It is given that

AD = BC

By AAS congruence criterion

â–³ ADE â‰… â–³ BCF

âˆ A = âˆ B (c. p. c. t)

We know that the sum of all the angles of a quadrilateral is 360^{o}

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

By substituting the values

2 âˆ B + 2 âˆ D = 360^{o}

By taking 2 as common

2 (âˆ B + âˆ D) = 360^{o}

By division

âˆ B + âˆ D = 360/2

So we get

âˆ B + âˆ D = 180^{o}

So, ABCD is a cyclic quadrilateral.

Therefore, it is proved that the points A, B, C and D lie on a circle.

**19. Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.**

**Solution:**

Consider ABCD as a cyclic quadrilateral with centre O passing through the points A, B, C, D

We know that AB, BC, CD and DA are the chords of the circle and its right bisector passing through the centre O

So we know that the right bisectors of AB, BC, CD and DA pass through the centre O and are concurrent.

Therefore, it is proved that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

**20. Prove that the circles described with the four sides of a rhombus as diameters pass through the point of intersection of its diagonals.**

**Solution:**

Consider ABCD as a rhombus

We know that the diagonals AC and BD intersect at the point O

From the figure we know that the diagonals of the rhombus bisect at right angles

It can be written as

âˆ BOC = 90^{o}

Thus, âˆ BOC lies in the circle

We know that the circle can be drawn with BC as the diameter having the centre O

In the same way, all the circles with AB, AD and CD as diameters will pass through the centre O.

Therefore, it is proved that the circles described with the four sides of a rhombus as diameters pass through the point of intersection of its diagonals.

**21. ABCD is a rectangle. Prove that the centre of the circle through A, B, C, D is the point of intersection of its diagonals.**

**Solution:**

Consider ABCD as a rectangle

We know that O is the point of intersection of the diagonals AC and BD

The diagonals of a rectangle are equal and bisect each other

So we get

OA = OB = OC = OD

We get O as the centre of the circle through A, B, C and D

Therefore, it is proved that the centre of the circle through A, B, C, D is the point of intersection of its diagonals.

**22. Give a geometrical construction for finding the fourth point lying on a circle passing through three given points, without finding the centre of the circle. Justify the construction.**

**Solution:**

Consider A, B, C as the points

Taking B as the centre and radius equal to AC construct an arc

Taking C as the centre and radius equal to AB construct another arc which cuts the arc at D

So we get D as the required point BD and CD

Consider â–³ ABC and â–³ DCB

We know that

AB = DC and AC = DB

BC and CB are common i.e. BC = CB

By SSS congruence criterion

â–³ ABC â‰… â–³ DCB

âˆ BAC = âˆ CDB (c. p. c. t)

We know that BC subtends equal angles âˆ BAC and âˆ CDB on the same side

So we get A, B, C, D are cyclic.

Therefore, the points A, B, C, D are cyclic.

**23. In a cyclic quadrilateral ABCD, if (âˆ B – âˆ D) = 60 ^{o}, show that the smaller of the two is 60^{o}.**

**Solution:**

It is given that ABCD is a cyclic quadrilateral

(âˆ B – âˆ D) = 60^{o} â€¦â€¦.. (1)

We know that

(âˆ B + âˆ D) = 180^{o} â€¦â€¦. (2)

By adding both the equations

âˆ B – âˆ D + âˆ B + âˆ D = 60^{o} + 180^{o}

So we get

2 âˆ B = 240^{o}

By division

âˆ B = 120^{o}

By substituting equation it in equation (1)

(âˆ B – âˆ D) = 60^{o}

120^{o} – âˆ D = 60^{o}

On further calculation

âˆ D = 120^{o} – 60^{o}

By subtraction

âˆ D = 60^{o}

Therefore, the smaller of the two angles âˆ D = 60^{o}.

**24. The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.**

**Solution:**

Consider ABCD as a cyclic quadrilateral with diagonals AC and BD intersecting at right angles

We know that OL âŠ¥ AB so that âˆ O meets the line CD at the point M

From the figure we know that the angles in the same segment are equal

âˆ 1 = âˆ 2

We know that âˆ OLB = 90^{o} so we get

âˆ 2 + âˆ 3 = 90^{o} â€¦â€¦ (1)

We know that OLM is a straight line and âˆ BOC = 90^{o}

So we get

âˆ 3 + âˆ 4 = 90^{o} â€¦â€¦â€¦ (2)

It can be written as

âˆ 2 + âˆ 3 = âˆ 3 + âˆ 4

On further calculation

âˆ 2 = âˆ 4

So we get

âˆ 1 = âˆ 2 and âˆ 2 = âˆ 4

It can be written as

âˆ 1 = âˆ 4

We get

OM = CM

In the same way

OM = MD

So CM = MD

Therefore, it is proved that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

**25. On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that âˆ BAC = âˆ BDC.**

**Solution:**

We know that AB is the common hypotenuse of â–³ ACB and â–³ ADB

So we get

âˆ ACB = 90^{o}

âˆ BDC = 90^{o}

It can be written as

âˆ ACB + âˆ BDC = 180^{o}

We know that the opposite angles of a quadrilateral ABCD are supplementary

So we get ABCD as a cyclic quadrilateral which means that a circle passes through the points A, C, B and D

From the figure we know that the angles in the same segment are equal

âˆ BAC = âˆ BDC

Therefore, it is proved that âˆ BAC = âˆ BDC.

**26. ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that âˆ CBD + âˆ CDB = Â½ âˆ BAD.**

**Solution:**

Consider a point E on the circle and join BE, DE and BD

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference

So we get

âˆ BAD = 2 âˆ BED

It can be written as

âˆ BED = Â½ âˆ BAD â€¦â€¦ (1)

Consider EBCD as a cyclic quadrilateral

So we get

âˆ BED + âˆ BCD = 180^{o}

We can write it as

âˆ BCD = 180^{o} – âˆ BED

Substituting equation (1)

âˆ BCD = 180^{o} – Â½ âˆ BAD â€¦â€¦ (2)

Consider â–³ BCD

Using the angle sum property

âˆ CBD + âˆ CDB + âˆ BCD = 180^{o}

By using the equation (2)

âˆ CBD + âˆ CDB + 180^{o} – Â½ âˆ BAD = 180^{o}

So we get

âˆ CDB + âˆ CDB – Â½ âˆ BAD = 180^{o} – 180^{o}

On further calculation

âˆ CDB + âˆ CDB = Â½ âˆ BAD

Therefore, it is proved that âˆ CDB + âˆ CDB = Â½ âˆ BAD.

### Access other exercise solutions of Class 9 Maths Chapter 12: Circles

Exercise 12A Solutions 23 Questions

Exercise 12B Solutions 18 Questions

### RS Aggarwal Solutions Class 9 Maths Chapter 12 – Circles Exercise 12C

RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles Exercise 12C is the third exercise which contains problems solved using theorems and some of the results on cyclic quadrilaterals.

### Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 12: Circles Exercise 12C

- The RS Aggarwal solutions are primarily based on frequently asked questions in the board exam.
- The solutions are explained in simple language based on CBSE guidelines and exam pattern.
- The PDF of solutions can be used by the students as a vital resource to boost their exam preparation.
- Comprehensive explanation of solutions mainly help to improve conceptual knowledge among students.