In the figure, $P$, $Q$ and $I$ are the incenters of the
triangles $\triangle AHB$, $\triangle BHC$ and $\triangle ABC$ respectively. Calculate the area of the shaded region if $MN = a$.(Answer: $\frac{a^2}{2}$)
My progress:
$S_{BPQI} = S_{\triangle BPQ}-S_{\triangle BQI}.$
$P$ is incenter, therefore $BP$ is angle bisector of $\angle ABH$.
Let $\angle ABP = \angle PBI = \alpha$.
But $JM \parallel JB \implies \angle BPM = \alpha$.
Therefore $ \triangle MPB$ is isosceles.
Similarly $\triangle BNQ$ is isosceles.
….??
Best Answer
Say, $x$ is inradius of $\triangle ABH$, $y$ is inradius of $\triangle CBH$ and $z$ is inradius of $\triangle ABC$. As $\triangle AHB, \triangle BHC$ and $\triangle ABC$ are similar, their hypotenuse are in ratio,
$AB:BC:AC = x:y:z~$ and $~z^2 = x^2 + y^2$
Now we extend $JM$ and $KN$ to $RS \parallel AC$.
As $\triangle BRM \sim \triangle ABC, RM = y$. Also, $SN = x$. So we have $BM = BN = z$.
Now as $BM = BN = z$ which is inradius of right triangle $\triangle ABC$, $IM$ and $IN$ must be perp to $AB$ and $BC$ respectively. That leads to $BI = a$ and $BI \perp MN$.
$ \displaystyle [BPIQ] = \frac 12 \cdot MN \cdot BI = \frac{a^2}{2}$