Question

I've got 2 things to prove that triangle OPA and triangle OPB are congruent but I can't find the 3 point

Answer 1

Construction:OA and OB are joined
In ΔOPA and ΔOPB
$OP = OP (Common side) \\ \angle BOP = \angle AOP( OA = ON \: and\: OP = OP hence\: these \:angles\: are\: also \: equal \\ OA = ON(Radii \: of \: same \: circle) \\ By SAS, \\ \Delta OPA congruent\: to \Delta OPB \\ Hence, AP = BP by CPCT$

Answer 2

Answer:

OB and OA are joined forming a triangle

where

OP=OP

AP=PB

and angle p is at 90 degree

On the condition of SAS