# Inradius in Right angled triangles.

Let $AD$ be an altitude in right angled $\triangle{ABC}$ with $\angle{A}=90^{o}$ and $D$ on $BC$. Suppose that the radii of incircles of triangles $ABD$ and $ACD$ are $33$ $(r_1)$ and $56$ $(r_2)$ respectively. Let $r$ be the radius of incircle of the $\triangle{ABC}$. Find the value of $3(r+7)$.

(Figure is rough)

Both the triangles $DBA$ and $DAC$ are similar to the original triangle $ABC$.
It follows that the ratio of their inradii, $\frac{33}{56}$, equals $\frac{AB}{AC}$. By the Pythagorean theorem, since $\sqrt{33^2+56^2}=65$, we have that $\frac{BC}{AB}=\frac{65}{33}$, hence $r_{ABC}=\color{red}{65}$.
• please provide links for "the ratio of their inradii, $\frac{33}{56}$, equals $\frac{AB}{AC}$" or prove here if possible. – mnulb Sep 12 '16 at 17:19