Internal angle bisector of

$\mathrm{\angle}A$ of triangle

$\mathrm{\Delta}ABC$, meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at P and the side AB at Q. If a, b, c represent the sides of

$\mathrm{\Delta}ABC$ then.

(a)

$AD=\frac{2bc}{b+c}\mathrm{cos}\frac{A}{2}$(b)

$PQ=\frac{4bc}{b+c}\mathrm{sin}\frac{A}{2}$(c) The triangle

$\mathrm{\Delta}APQ$ is isosceles

(d) AP is HM of b and c

My approach is as follow

This is the rough image that I have drawn

$\frac{\mathrm{cos}\frac{A}{2}}{AD}=\frac{\mathrm{sin}{90}^{o}}{{b}_{1}}$$\frac{\mathrm{sin}X}{{b}_{2}}=\frac{\mathrm{sin}C}{PD}$$PD=AD\mathrm{tan}\frac{A}{2}$$\frac{\mathrm{sin}X}{{b}_{2}}=\frac{\mathrm{sin}C}{PD}\Rightarrow \frac{\mathrm{sin}X}{{b}_{2}}=\frac{\mathrm{sin}C}{AD\mathrm{tan}\frac{A}{2}}$The official answer is a,b,c and d.

I am not able to approach from here