Question 72
Let $\triangle ABC$ be a right-angled triangle with $\angle B = 90^\circ$. Let $D$ be a point on $AC$ such that the in-radii of the triangles $ABD$ and $CBD$ are equal. If this common value is $r'$ and if $r$ is the in-radius of triangle $ABC$, prove that
$$
\frac{1}{r'} = \frac{1}{r} + \frac{1}{BD}.
$$
$$
\frac{1}{r'} = \frac{1}{r} + \frac{1}{BD}.
$$