# OPPOSITE ANGLES OF A CYCLIC QUADRILATERAL ARE SUPPLEMENTARY PROOF

Theorem :

Opposite angles of a cyclic quadrilateral are supplementary (or) The sum of opposite angles of a cyclic quadrilateral is 180°

Given : O is the centre of circle. ABCD is the cyclic quadrilateral.

To prove : ∠BAD + ∠BCD  =  180°, ∠ABC + ∠ADC  =  180°

Construction : Join OB and OD

Proof:

(The angle substended by an arc at the centre is double the angle on the circle.)

(ii) BCD  =  (1/2) reflex BOD.

(iii) BAD + BCD  =  (1/2)BOD + (1/2) reflex BOD.

BAD + BCD  =  (1/2)(BOD + reflex BOD)

BAD + BCD  =  (1/2) ⋅ (360°)

(Complete angle at the centre is 360°)

(iv) Similarly ∠ABC + ∠ADC  =  180°.

## Practice Problems

Problem 1 :

In the figure given below, O is the center of a circle and ∠ADC  =  120°. Find the value of x.

Solution :

ABCD is a cyclic quadrilateral. we have

ABC  =  180° - 120°

∠ABC  =  60°

Also ACB  =  90° (angle on a semi circle).

In triangle ABC we have,

BAC + ACB + ABC  =  180°

BAC + 90° + 60°  =  180°

BAC  =  180° - 150°

∠BAC  =  30°

So, the value of x is 30.

Problem 2 :

In the figure given below, ABCD is a cyclic quadrilateral in which AB || DC. If BAD  =  100° find.

(i) BCD

(iii) ABC

Solution :

100° + BCD  =  180°

BCD  =  180° - 100°

BCD  =  80°

Because AB || DC and AD is tranversal,

Subtract 180° from each side.

80° + ABC  =  180°

ABC  =  180° - 80°

ABC  =  100°

Problem 3 :

In the figure given below, ABCD is a cyclic quadrilateral in which BCD = 100° and ABD = 50° find ADB.

Solution :

DAB + DCB  =  180°

DAB  +  100°  =  180°

DAB  =  180° - 100°

DAB  =  80°

DAB + ABD + ∠ADB  =  180°

80° + 50° ∠ADB  =  180°

Subtract 130° from each side.

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