Properties of Triangles 3 Question 17
17. Let $A B C$ be a triangle with incentre $I$ and inradius $r$. Let $D, E, F$ be the feet of the perpendiculars from $I$ to the sides $B C, C A$ and $A B$, respectively. If $r _1, r _2$ and $r _3$ are the radii of circles inscribed in the quadrilaterals $A F I E, B D I F$ and CEID respectively, then prove that
$\frac{r _1}{r-r _1}+\frac{r _2}{r-r _2}+\frac{r _3}{r-r _3}=\frac{r _1 r _2 r _3}{\left(r-r _1\right)\left(r-r _2\right)\left(r-r _3\right)}$.
(2000 3M)
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Solution:
- The quadrilateral $H E K J$ is a square, because all four angles are right angles and $J K=J H$.
Therefore, $\quad H E=J K=r _1 \quad$ and $\quad I E=r \quad$ [given]
$$ \Rightarrow \quad I H=r-r _1 $$
Now, in right angled $\triangle I H J$,
$$ \angle J I H=\pi / 2-A / 2 $$
$\left[\because \angle I E A=90^{\circ}, \angle I A E=A / 2\right.$ and $\left.\angle J I H=\angle A I E\right]$
In $\triangle J I H$,
$$ \tan \frac{\pi}{2}-\frac{A}{2}=\frac{r _1}{r-r _1} \Rightarrow \cot \frac{A}{2}=\frac{r _1}{r-r _1} $$
Similarly, $\quad \cot \frac{B}{2}=\frac{r _2}{r-r _2}$ and $\cot \frac{C}{2}=\frac{r _3}{r-r _3}$
On adding above results, we get $\cot A / 2+\cot B / 2+\cot C / 2$
$=\cot A / 2 \cot B / 2 \cot C / 2$
$\Rightarrow \frac{r _1}{r-r _1}+\frac{r _2}{r-r _2}+\frac{r _3}{r-r _3}=\frac{r _1 r _2 r _3}{\left(r-r _1\right)\left(r-r _2\right)\left(r-r _3\right)}$
