This is an exercise from Morris Kline's "Calculus: An Intuitive and Physical Approach":
Prove that if the chord (Fig. 4-23) joining the points of tangency of two tangents to a parabola goes through the focus, the tangents are perpendicular to each other. Suggestion: Use the reflection property of the parabola.
The text does not explain the solution completely, so I would just like to verify that mine is correct.
Let $R$ be the intersection of the two tangent lines below $F$.
Let $P'$ be a point on the tangent line to the parabola at $P$ to the right of $P$, and $Q'$ be a point on the tangent line to the parabola at $Q$ to the left of $Q$.
Since $QD'$ and $PD$ are parallel, $\angle D'QF + \angle DPF = 180 ^{\circ}$. Due to the reflection property of the parabola, $\angle FPR = \angle DPP'$ and $\angle Q'QD' = \angle FQR$. Then $$2 * \angle FQR + \angle D'QF = 180 ^{\circ}\text{ and } 2 * \angle FPR + \angle DPF = 180^{\circ}$$
It follows that \begin{align} 2 * \angle FQR + \angle D'QF + 2 * \angle FPR + \angle DPF &= 360^{\circ} \\ 2 * \angle FQR + 2 * \angle FPR &= 360^{\circ} - (\angle D'QF + \angle DPF) \\ 2 * \angle FQR + 2 * \angle FPR &= 360^{\circ} - 180^{\circ} \\ \angle FQR + \angle FPR &= 90 \end{align}
Since $\angle FQR$ and $\angle FPR$ are two of the angles of $\triangle QPR$, the measure of $\angle QRP$ must be $90^{\circ}$.