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

Hint: We recall that the length in-radius of a right angled triangle is half of the sum of the length of perpendicular sides minus the lengths of hypotenuse. It means if $p$ and $b$ are lengths of the perpendicular sides and $h$ is the hypotenuse then length of in-radius is $r=\dfrac{1}{2}\left( p+b-h \right)$. We use this formula in different triangles given in the question to express BD in terms of in-radii of the triangles.

Complete step-by-step solution:

We are given the question that ABC is a right angled triangle ABC where $\angle B$ is a right angle and BD is an altitude drawn to the hypotenuse AC. Then the triangles ADB and CDB are the right angles $\angle ADB,\angle CDB$.
We see that in the right angled triangle ABC the hypotenuse opposite to the right angle $\angle B$ that is AC and the perpendicular sides are AB and BC. We use the formula of in-radius and have the in-radius of the right angled triangle ABC as
\[{{r}_{ABC}}=\dfrac{AB+BC-AC}{2}\]
We see that in the right angled triangle ABD the hypotenuse opposite to the right angle $\angle ABD$ that is AB and the perpendicular sides are AD and BD. We use the formula of in-radius and have the in-radius of the right angled triangle ABD as
\[{{r}_{ABD}}=\dfrac{AD+BD-AB}{2}\]
We see that in the right angled triangle CBD the hypotenuse opposite to the right angle $\angle B$ that is BC and the perpendicular sides are CD and BD. We use the formula of in-radius and have the in-radius of the triangles right angled triangle CBD as
\[{{r}_{CBD}}=\dfrac{CD+BD-BC}{2}\]
 Let us find the sum of in-radii of the triangles. We have
\[\begin{align}
  & {{r}_{ABC}}+{{r}_{ABD}}+{{r}_{CBD}} \\
 & \Rightarrow \dfrac{AB+BC-AC}{2}+\dfrac{AD+BD-AB}{2}+\dfrac{CD+BD-BC}{2} \\
 & \Rightarrow \dfrac{AB+BC-AC+AD+BD-AB+CD+BD-BC}{2} \\
 & \Rightarrow \dfrac{2BD+AD+CD-AC}{2} \\
\end{align}\]
We put $AD+CD=AC$ in the above step to have
\[\begin{align}
  & \Rightarrow \dfrac{2BD+AC-AC}{2} \\
 & \Rightarrow \dfrac{2BD}{2}=BD \\
\end{align}\]
Hence it is proved that BD is equal to the sum of the radii of the circles inscribed in the triangles ABC, ADB, and CDB that is
\[BD={{r}_{ABC}}+{{r}_{ABD}}+{{r}_{CBD}}\]

Note: We note that in-radius is the radius of the in-circle drawn at the in-centre inscribed in a triangle. The in-centre is the point of intersection of angle bisectors of internal angles of a triangle. The in-radius of any triangle is given by $r=\dfrac{\Delta }{s}$ where $\Delta $ is the area of the triangle and $s$ is the semi-perimeter. So we can also find the in-radius of right triangle with perpendicular sides $p,b$ as $r=\dfrac{pb}{p+b+\sqrt{{{p}^{2}}+{{b}^{2}}}}$.