Question

in figure given alongside AP and bp are two tangents from the point P to the circle with Centre O show that a p equals to b p Hind triangle aob is concurrent to triangle b o p

Answer 1

ANGLE OAP=ANGLEOBP.......TANGENT THEOREM. .....op=op....common side.... AO =OB... radius of circle.... .... AOP= BOP..... byssa test of congruency..... this is the answer of your asked question

Answer 2

From the Pb:
Given, triangle AOP is congruent to triangle BOP.
By CPCT, we get AO = BO, PA = PB.
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