Question

A and B are two distinct points on a circle with centre 0. AB is not a diameter of the circle. The tangents at
A and B intersect in point P. Show that angle AOB and angle APB are supplementary angles. Also show that PA = PB.​

Answer 1

Answer :

The answer is explained in the image given below.

Hope it helps!!!

Answer 2

Angle AOB and angle APB are supplementary angles. (Proved)

PA = PB (Proved)

Step-by-step explanation:

See the attached diagram.

PA and PB are two tangents to the circle with center O.

So, ∠ OAP = ∠ OBP = 90°

Now, considering quadrilateral AOBP,

∠ AOB + ∠ OBP + ∠ APB + ∠ OAP = 360°

⇒ ∠ AOB + 90° + ∠ APB + 90° = 360°

⇒ ∠ AOB + ∠ APB = 180°

So, angle AOB and angle APB are supplementary angles. (Proved)

Now, join O and P.

Take Δ OPB  and Δ OPA,

(i) ∠ OAP = ∠ OBP = 90°

(ii) OP common side and

(iii) OB = OA = Radius of the circle

Therefore, by SSA criteria, we have  Δ OPB  ≅ Δ OPA.

So, PA = PB {Corresponding sides} (Proved)