Question

In a right triangle $ABC$, $\angle B$ is a right angle and $BD$ is an altitude drawn to the hypotenuse $AC$. Prove that $BD$ is equal to the sum of the radii of the circles inscribed in the triangles $ABC,ADB$ and $CDB$.

Answer 1

Step $1:$ Formula to be used.

The radius of the incircle of a right angles triangle is given by:

$\text{R=}\frac{Perpendicular+Base-Hypotenuse}{2}$

Step $2:$ Find the inradii of the triangles.

The radius of the incircle of the triangle $ABC$.

$R_{ABC}=\frac{AB+BC-AC}{2}$

The radius of the incircle of the triangle $ADB$.

$R_{ADB}=\frac{AD+BD-AB}{2}$

The radius of the incircle of the triangle $CDB$.

$R_{CDB}=\frac{CD+BD-BC}{2}$

Step $3:$Compute the sum of radii.

The sum of the radii of the circles inscribed in the triangles $ABC,ADB$ and $CDB$ is given by:

$$\begin{gathered} \Rightarrow R_{ABC}+R_{ADB}+R_{CDB}=\frac{AB+BC-AC}{2}+\frac{AD+BD-AB}{2}+\frac{CD+BD-BC}{2} \\ \Rightarrow R_{ABC}+R_{ADB}+R_{CDB}=\frac{AB+BC-AC+AD+BD-AB+CD+BD-BC}{2} \\ \Rightarrow R_{ABC}+R_{ADB}+R_{CDB}=\frac{-AC+AD+2BD+CD}{2} \end{gathered}$$

Since, $AD+CD=AC$.

$$\begin{gathered} \Rightarrow R_{ABC}+R_{ADB}+R_{CDB}=\frac{-AC+2BD+AC}{2} \\ \Rightarrow R_{ABC}+R_{ADB}+R_{CDB}=\frac{2BD}{2} \\ \Rightarrow R_{ABC}+R_{ADB}+R_{CDB}=BD \end{gathered}$$

Hence prove, $BD$ is equal to the sum of the radii of the circles inscribed in the triangles $ABC,ADB$ and $CDB$.