# Theorem 6 - Angles in the Same Segment of a circle are Equal

**Theorem 6 : ** Angles in the same segment of a circle are equal.

**Given : **Two angles ACB and ADB are in the same segment of a circle C(O, r).

**To Prove : ** âˆ ACB = âˆ ADB

**Construction : **Join OA and OB.

**Proof :** We know that, angle subtended by an arc of a circle at the centre is double the angle subtended by the arc in the alternate segment.

Hence, âˆ AOB = 2 âˆ ACB

âˆ AOB = 2 âˆ ADB

So, âˆ ACB = âˆ ADB

In Fig. (ii), we have,

Reflex âˆ AOB = 2âˆ ACB

and Reflex âˆ AOB = 2âˆ ADB

⇒ 2âˆ ACB = 2 âˆ ADB

⇒ âˆ ACB = âˆ ADB.

In the Fig. P is a point on the chord BC such that AB = AP. Prove that CP = CQ.

AB = AP [given]

âˆ ABP =âˆ APB [Angles opposite to equal sides of a triangle are equal]

In Î”ABP and Î”AQC, âˆ ABP = âˆ AQC [Angle in the same segment of a circle]

In Î”ABP and Î”CPQ, âˆ APB = âˆ CPQ [Vertically opposite angles are equal]

.^{.}.âˆ CPQ = âˆ AQC

.^{.}. CQ = PC [Sides opposite equal angles are equal]

or CP = CQ