What is the value of $BC^2$ in the following diagram?

Question

Let $ABC$ be a triangle with $AB=20$ and $AC=22$. Suppose its incircle touches $BC,CA$ and $AB$ at $D, E$ and $F$ respectively, and $P$ is the foot of the perpendicular from $D$ to $EF$. If $\angle BPC=90$ degrees, then compute $BC^2$.

I couldn't figure out where to start. Can I get a hint?

Answer 1

Hint (and more):

Let us consider an arbitrary triangle $\Delta ABC$, let $I$ be its incenter, $D,E,F$ the projections of $I$ on the sides, $A',B',C'$ the mid points of the sides of $\Delta DEF$, $P$ such that $DP$ is the height from $D$ in $\Delta DEF$, and draw the Euler circle $PA'B'C'$ for it.

Let $R,S$ be the intersection of the angle bisector $AA'I$ with the lines $DE$, and respectively $DF$. Then $EA'PF\|BR\|CS$.

Let $Q,T,U$ be the intersections of $DP$ with $BR$, $CS$, and respectively $AB$.
Denote by a star the reflection w.r.t. the line $DP$. So $Q=Q^*$, $T=T^*$, $U=U$, and also consider the reflected points $F^*$, $B^*$, $C^*$. Then $B^*$ is on $PC$, and $C^*$ is on $PB$.

Then finally: $PD$ is the angle bisector of $\widehat{BPC}$.