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 r1, r2 and r3 are the radii of circles inscribed in the quadrilaterals AFIE, BDIF and CEID respectively, then prove th at ( r1/ r - r1 ) + ( r2/ r - r2 ) + ( r3/ r - r3 ) = ( r1 r2 r3 / (r - r1) (r - r2) (r - r3)) .

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 r1, r2 and r3 are the radii of circles inscribed in the quadrilaterals AFIE, BDIF and CEID respectively, then prove th at ( r1/ r - r1 ) + ( r2/ r - r2 ) + ( r3/ r - r3 ) = ( r1 r2 r3 / (r - r1) (r - r2) (r - r3)) . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The quadrilateral HEKJ is a square, because all four
angles are right angles and J K = J H .
Therefore, HE = JK =

r_{1} an d I E = r

[ given ]

\Rightarrow

I H = r - r_{1}

Now, in right angled

△

IHJ,

∠ J I H = π /2 - A /2

[∵ ∠ J E A = 9 0^{\circ}, ∠ I A E = A /2 an d ∠ J I H = ∠ A I E]

In

△

JIH,
tan

(\frac{π}{2} - \frac{A}{2}) = \frac{r _{1}}{r - r _{1}}

\Rightarrow co t \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} )}