Prove that, the angle in the semicircle is right angle.

In the given figure, P O Q is a straight line passing through centre O . Therefore, angle subtended by the arc P Q at O is ∠ P O Q = 180 ° . We know, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Therefore, ∠ P O Q = 2 ∠ P A Q ⇒ ∠ P O Q 2 = ∠ P A Q ⇒ 180 ° 2 = ∠ P A Q ⇒ 90 ° = ∠ P A Q Hence, proved.

Prove that, the angles in the same segment of a circle are equal to each other.

In a circle, with centre O , ∠ ABC and ∠ ADC are in same segment To prove:- ∠ ABC = ∠ ADC Construction → Join OA and OC . Proof:- ∴ Arc AC subtends ∠ AOC on centre of circle and ∠ ABC on remaining part of circle. And ∠ AOC = 2 ∠ ABC - ( i ) Similarly ∠ AOC = 2 ∠ ADC - ( ii ) Now, from equation ( i ) and ( ii ) ⇒ 2 ∠ ABC = 2 ∠ ADC ⇒ ∠ ABC = ∠ ADC

Prove that, the length of two tangents drawn from an external point to a circle are equal.

Given: Two tangents A P and A Q are drawn from a point A to a circle with centre O . To prove: A P = A Q . Construction: Join O P , O Q and O A . Proof: Since A P and A Q are the tangents to the circle, O P ⊥ A P and O Q ⊥ A Q . [Tangent is perpendicular to the radius at the point of contact]. In ∆ O P A and ∆ O Q A , ∠ O P A = ∠ O Q A = 90 ° . O A = O A . . . (common side). O P = O Q . . . (radius of the same circle). ⇒ ∆ O P A ≅ ∆ O Q A . . . . ( S A S congruence criteria). So, A P = A Q . Hence, proved.