Inspired by and dedicated to Ayush Rai

Consider a triangle $ABC$ such that $P$ and $Q$ are the feet of the perpendiculars drawn from point $A$ onto the internal bisectors of angles $B$ and $C$ , respectively.

Let $a,b,c$ denote thee sides opposite to the vertices $A,B,C$ , respectively.

Denote $X$ as the length $PQ$ .

Compute $56 \left( \dfrac{X+ \frac{a+b+c}2}{2(b+c)}\right)$ .