Question

Let ABC be a triangle with incentre I and inradius r.
Let D, E ,F be the feet of the perpendiculars from I to
the sides BC, CA and AS, respectively. If $r_1, r_2 \, and \, r_3$
are the radii of circles inscribed in the quadrilaterals
AFIE, BDIF and CEID respectively, then prove th at
$ \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 }{ (r - r_1) \, (r - r_2) \, (r - r_3)} $ .

Answer 1

The quadrilateral HEKJ is a square, because all four
angles are right angles and J K = J H .
Therefore, HE = JK = $ r_1 \, and \, IE = r $ $$ [ given ]
$\Rightarrow $ $$ $ IH = r - r_1 $
Now, in right angled $\triangle$ IHJ,
$$ $ \angle JIH = \pi / 2 - A/ 2 $
$ [ \because \angle JEA = 90^\circ, \, \angle IAE = A/ 2 \, and \, \angle JIH = \angle AIE ] $
In $ \triangle $ JIH,
tan $ \bigg( \frac{ \pi}{ 2} - \frac{A}{ 2} \bigg) = \frac{ r_1}{ r - r_1} $
$\Rightarrow cot \frac{ A}{2} = \frac{ r_1}{ r - r_1} $
Similarly, 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 }{ (r - r_1) \, (r - r_2) \, (r - r_3)} $