Two Tangent Theorem

If P is any point lying outside of a circle O, and if A and B are points such that PA and PB are tangent to O, then $PA = PB$.

Proof: As we have seen before, $m\angle OAP = m\angle OBP = 90^\circ$. Because A and B are points on the circle, $OA = OB$. By the Reflexive Property, $OP = OP$. So, by the HL-Theorem (Hypotenuse-Leg Theorem), $\triangle POA \cong \triangle POB$. CPCTC (Corresponding Parts of Congruent Triangles are Congruent) states that $PA = PB$, so the above theorem is proven. $\square$