Prove that the circumcenter of $\triangle PIQ$ is on the hypotenuse $AC$. In right angled $\triangle ABC$ with $\angle B=90 ^{\circ}$, $BD$ is an altitude on $AC$. $P,Q,I$ are the incenters of $\triangle ABD,\triangle CBD$ and $\triangle ABC$ respectively. Prove that the circumcenter of $\triangle PIQ$ lies on the hypotenuse $AC$.
This question was asked today in the Indian National Mathematics Olympiad (INMO) 2015 in which I appeared.I totally had no idea and infact have no idea on how to proceed. Any help would be appreciated.
 A: It is not a very elegant technique, but we can just prove that the perpendicular bisector of $PI$ intersects $AC$ in the same point where the perpendicular bisector of $QI$ intersects $AC$. 

Let $AB=c,AC=b,BC=a$ and $r,r_P,r_Q$ be the inradii of $ABC,ABD,CBD$. We have:
$$BD = \frac{ac}{b},\quad r=\frac{ac}{a+b+c},\quad r_P = r\frac{BD}{BC}=r\frac{c}{b},\quad r_Q=r\frac{BC}{AC}=r\frac{a}{b}$$
hence the distance between the midpoint of $PI$ and the $AC$-line is:
$$\frac{r_P+r}{2}=\frac{ac}{2b}\frac{b+c}{a+b+c} $$
while the distance between the midpoint of $QI$ and the $AC$-line is:
$$\frac{r_Q+r}{2}=\frac{ac}{2b}\frac{a+b}{a+b+c} $$
so the perpendicular bisector of $PI$ cuts $AC$ in a point $P'$ for which:
$$ P'I = P'P = \frac{1}{\cos(\widehat{A}/2)}\frac{ac}{2b}\frac{b+c}{a+b+c}$$
and the perpendicular bisector of $QI$ cuts $AC$ in a point $Q'$ for which:
$$ Q'I = Q'Q = \frac{1}{\cos(\widehat{C}/2)}\frac{ac}{2b}\frac{a+b}{a+b+c}$$
so we just need to check that:
$$ \frac{a+b}{\cos(\widehat{C}/2)}=\frac{b+c}{\cos(\widehat{A}/2)} $$
that is trivial.

Edit: Using similarities, it is way easier to prove that, given that $J$ is the projection of $I$ on $AC$,
$$ JP=JI=JQ=r $$
holds.
A: We note that $\triangle ABC \sim \triangle ADB \sim \triangle BDC$, with corresponding segments within these triangles having lengths in the proportion $1 : \sin C : \sin A$. In particular, if $r$ is the radius of incircle $\bigcirc I$ of $\triangle ABC$, then $r\sin C$ and $r \sin A$ are radii of incircles $\bigcirc P$ and $\bigcirc Q$, respectively.

With $J$ and $K$ (and $L$) the points of tangency of $\bigcirc I$ with the sides of $\triangle ABC$ as shown, we see that $\square BJIK$ is a square of side-length $r$. The perpendiculars from $J$ and $K$ to $\overline{BD}$ have length $r\sin C$ and $r\sin A$, respectively; these match the lengths of perpendiculars from $P$ and $Q$ to $\overline{BD}$, so that $\overline{JP}$ and $\overline{KQ}$ are parallel to $\overline{BD}$. Moreover, $\overline{JP}$ is the hypotenuse of a right triangle with a leg of length $r \sin C$ opposite an angle congruent to $\angle C$; that hypotenuse must have length $r$. Likewise for $\overline{KQ}$.
The above shows that points $P$ and $Q$ are translates of $J$ and $K$ by distance $r$ in direction $\overrightarrow{BD}$. But point $L$ is a translate of $I$ in the same way. Consequently,
$$|\overline{LP}| = |\overline{IJ}| = r = |\overline{IK}| = |\overline{LQ}| \qquad\text{and}\qquad |\overline{LI}| = r$$
so that the circumcenter of $\triangle IPQ$ is $L$. $\square$
A: Let $X,Y$ be points of tangency of incircle of $ABC$ with sides $AB, AC$ respectively. Let $Z$ be reflection of $I$ with respect to $XY$. Then we have $ZX=ZY=IX=IY=BX$ and $ZX \parallel IY \parallel BD$. 
Angle chasing gives us $\angle ZBX = \angle XZB = \angle DBZ$, so $BZ$ is bisector of angle $DBA$. Obviously $Z$ lies on bisector of angle $BAC$, so we see that $Z$ coincides with $P$. 
So we proved that $YP=YI$. Analogously we show that $YQ=YI$. Thus $Y$ is the circumcenter of triangle $PIQ$, which lies on $AC$.
