**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:**

(i) ∠BAD = (1/2)∠BOD.

(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.

Add (i) and (ii).

∠BAD + ∠BCD = (1/2)(∠BOD + reflex ∠BOD)

∠BAD + ∠BCD = (1/2) ⋅ (360°)

(Complete angle at the centre is 360°)

∠BAD + ∠BCD = 180°

(iv) Similarly ∠ABC + ∠ADC = 180°.

**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 + ∠ADC = 180°

∠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

(ii) ∠ADC

(iii) ∠ABC

**Solution :**

∠BAD + ∠BCD = 180°

100° + ∠BCD = 180°

∠BCD = 180° - 100°

∠BCD = 80°

Because AB || DC and AD is tranversal,

∠BAD + ∠ADC = 180°

100° + ∠ADC = 180°

Subtract 180° from each side.

∠ADC = 80°

∠ADC + ∠ABC = 180°

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°

In triangle ADB,

∠DAB + ∠ABD + ∠ADB = 180°

80° + 50° + ∠ADB = 180°

130° + ∠ADB = 180°

Subtract 130° from each side.

∠ADB = 180°

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