Let $A$ and $B$ be two points on a circle, and let the tangents to the circle at $A$ and $B$ meet at $P$. Prove that $AP = BP$.
Hint: Consider a reflection in the line which passes through P and the centre of the circle.
Work: Let the center of the circle be $O$, Then $OA = OB$ (radius of a circle), $AP,BP$ tangent implies that $\angle OAP=\angle OBP = 90^o$
From here, normally I can just say that these two triangles are congruent and therefore $AP = BP$
But this is the beginning of a upper division geometry course and I can't use the congruent properties yet.
I know the angles and the lengths will be preserved after I take the reflection as the hint given. But I don't see clearly how I can continue the proof.