In a triangle $ABC$, $\angle BAC =90^{\circ}$. Point $D$ lies on the side $BC$ and satisfies $\angle BDA=2\angle BAD$. Prove that \[\frac{2}{AD}=\frac{1}{BD}+\frac{1}{CD} \]
Source: Baltic Way 1998
Tags: geometry, circumcircle, geometry proposed
In a triangle $ABC$, $\angle BAC =90^{\circ}$. Point $D$ lies on the side $BC$ and satisfies $\angle BDA=2\angle BAD$. Prove that \[\frac{2}{AD}=\frac{1}{BD}+\frac{1}{CD} \]