BALTIC-WAY 1998 Problem 12

In a triangle $A B C, \angle B A C=90^{\circ}$. Point $D$ lies on the side $B C$ and satisfies $\angle B D A=2 \angle B A D$. Prove that

$$\frac{1}{|A D|}=\frac{1}{2}\left(\frac{1}{|B D|}+\frac{1}{|C D|}\right)$$