Given a triangle $ABC$ and its incenter I. If $P,Q$ are points on $BI$ and $CI$ so that $2\angle PAQ=\angle BAC$. If $D$ is the point where the inscribed circle of $ABC$ intersects $BC$, prove that $\angle PDQ=90$.
If $X\in BI$ we state $f(X)$ is the only point on $CI$ which is collinear with $A$ and is the points which becomes of X if we rotate $X$ around $A$ by angle $\frac{1}{2}\angle ABC$. Hence $f(P)=Q$. We also have that $f(B)=I$ and $f(I)=C$. Also if $I_1$ and $I_2$ the incenter of $\triangle ABD, \triangle ACD$ respectively then $f(I_1)=f(I_2)$
This is as far as I got. I believe it should be able to finish like this, but I've gotten stuck. Could you please help me finish it off?
To continue your idea: the map $f$ is a projective map from the set of points on $BI$ to the set of points on $CI$. We can consider another map $g$ that sends $X\in BI$ to the point $Y\in CI$ such that $\angle XDY=90^{\circ}$. This map $g$ is also projective, so to show that $f\equiv g$, it suffices to check that they are equal at three points.
But this is easy: $f(B)=I=g(B)$, $f(I)=C=g(I)$ and $f(I_1)=I_2=g(I_1)$.
An alternative approach: $\angle DIQ+\angle AIB=180^{\circ}$, so $I$ has an isogonal conjugate wrt quadrilateral $ABDQ$. But it's easy to see that the isogonal conjugate must in fact be $P$.
So $\angle PDQ=\angle IDB=90^{\circ}$.