<s>[INST] 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
 [/INST] 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?
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