Question related to internal angle bisector Internal angle bisector of $\angle A$ of triangle $\Delta ABC$, 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 ∆ABC then.
(a) $AD = \frac{{2bc}}{{b + c}}\cos \frac{A}{2}$
(b) $PQ = \frac{{4bc}}{{b + c}}\sin \frac{A}{2}$
(c) The triangle ∆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
$\frac{{\cos \frac{A}{2}}}{{AD}} = \frac{{\sin {{90}^o}}}{{{b_1}}}$
$\frac{{\sin X}}{{{b_2}}} = \frac{{\sin C}}{{PD}}$
$PD = AD\tan \frac{A}{2}$
$\frac{{\sin X}}{{{b_2}}} = \frac{{\sin C}}{{PD}} \Rightarrow \frac{{\sin X}}{{{b_2}}} = \frac{{\sin C}}{{AD\tan \frac{A}{2}}}$
The official answer is a,b,c and d.
I am not able to approach from here
 A: You know that $\displaystyle \small \frac{BD}{DC} = \frac{c}{b}$. Add $1$ to both sides and you get,
$\displaystyle \small \frac{a}{DC} = \frac{b+c}{b}$
Now we know that $\displaystyle \small \frac{a}{\sin A} = \frac{c}{\sin C} \implies \sin C = \frac{c}{a} \sin A$
In $\triangle ADC, \displaystyle \small \frac{\sin C}{AD} = \frac{\sin (A/2)} {DC}$
$AD = \displaystyle \small \frac{2 \cdot DC \cdot c}{a} \cos {\frac{A}{2}} = \frac{2 bc}{b+c} \cos \frac{A}{2}$
Next, in $\triangle APQ$, $AD$ is angle bisector of $\angle A$ and is also  perpendicular to base $PQ$, therefore $\triangle APQ$ must be isosceles and $QD = DP = \frac{PQ}{2}$.
Can you take the rest of it from here?
A: You can use sine-rule in $\triangle ADC$ to find $AD$ as in $(a)$,
$$\frac{AD}{\sin C}=\frac{CD}{\sin A/2}$$
where $CD$ is found using $BD:CD=c:b$. See another straightforward method using areas here.
Next $\triangle ADP \cong \triangle ADQ$ by ASA congruence. Hence $PD=QD$ and $AP=AQ \Rightarrow (c)$.
Now in right $\triangle ADP$,
$$AP = AD \sec A/2 \Rightarrow (d)$$
Also $$PQ=2PD =2AD \tan A/2 \Rightarrow (b)$$
