Pls answer question number 74

Let AB be a chord of a circle with centre O, and let AP and BP be the tangents at A and B respectively.

Suppose the tangents meet at P .

Join OP. Suppose OP meets AB at C . We have to prove that ∠PAC=∠PBC .

In triangles PCA and PCB , We have

PA = PB

∠APC=∠BPC
[The tangents are equally inclined to line joining external point and centre of circle.]

PC = PC [Common]

So, by SAS - criterion of congruence, we have

ΔPAC≅ΔPBC

⇒∠PAC=∠PBC

Alternative Solution:

In ΔPAB, PA=PB ( ∵ tangents drawn from an external point to a circle are equal.)

∵ PA = PB, ∠PAB = ∠ PBA (Equal opposite sides make equal angles)

⇒ ∠PAC = ∠PBC