Circles: Tangents from External Point to Circle

Let a circle with center $O$ have an external point $P$. Two tangents $PQ$ and $PR$ are drawn touching the circle at $Q$ and $R$ respectively. Join $OP$, $OQ$, and $OR$. We need to prove that $PQ = PR$.

In $\triangle OQP$ and $\triangle ORP$: 1. $\angle OQP = \angle ORP = 90^\circ$ (Tangent at any point is perpendicular to the radius through point of contact). 2. $OP = OP$ (Common hypotenuse). 3. $OQ = OR$ (Radii of the same circle). Therefore, $\triangle OQP \cong \triangle ORP$ by RHS Congruence Criterion.

Since the triangles are congruent: $PQ = PR \quad (\text{Corresponding parts of congruent triangles})$ Hence proved.