Given :A circle with center O, P is a point outside the circle and PA and PB are two tangents to the circle from P

To prove: PA = PB

Proof: join OA , OB, OP

∠OAP = ∠OBP = °

Now in the two right triangles

△OAP and △OBP

OA = (radii of same circle)

OP = (common)

Therefore, by R.H.S Congruency axiom,

△OAP similarly △OBP

This gives PA =

Hence proved