Question 17
In a triangle $ABC$, $\angle BAC = 90^\circ$. Point $D$ lies on the side $BC$ and satisfies $\angle BDA = 2 \angle BAD$. Prove that
$$
\frac{1}{|AD|} = \frac{1}{2} \left( \frac{1}{|BD|} + \frac{1}{|CD|} \right)
$$
$$
\frac{1}{|AD|} = \frac{1}{2} \left( \frac{1}{|BD|} + \frac{1}{|CD|} \right)
$$