Given a circle with center O, PQ AND PR are two tangents from point P outside the circle. Prove that angle POQ = angle POR.
So the triangles share a common hypotenuse $\displaystyle PO$. We know the two radii are congruent: $\displaystyle OQ\cong OR$. Since $\displaystyle \angle PQO=90^o=\angle PRO$ (tangent lines to a circle form right angles with the radius), the two triangles are congruent by HL, and the two desired angles are congruent by CPCTC.