Question

If PA and PB are two tangents drawn from a point P to a circle with center O touching it at A and B respectively, prove that OP is the perpendicular bisector of AB?

Answer 1

hope this helps you.

Answer 2

Answer:

The proof is explained below.

Step-by-step explanation:

Given PA and PB are two tangents drawn from a point P to a circle with center O touching it at A and B respectively. we have to prove that OP is the perpendicular bisector of AB.

Let PO intersect AB at a point C.

Since, Tangents from an external point are equally inclined to the line joining the point to the center

⇒ ∠APO = ∠BPO

In ΔAPC and ΔBPC

AP = BP               (Tangents from an external point are equal)

PC = PC               (Common side)

∠APO = ∠BPO      (Proved above)

∴ By SAS rule, ΔAPC≅ΔBPC.

Hence, by CPCT, AC = CB and ∠ACP= ∠BCP

∠ACP+∠BCP=180° ⇒ 2∠ACP=180° ⇒ ∠ACP=90°

Hence OP is the perpendicular bisector of AB.