A problem on Angle bisectors of a triangle and perpendiculars. 
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*From $A$ perpendiculars $AX, AY$ are drawn to the bisectors of the exterior angles of $B$ and $C$ of $\Delta ABC$. Prove that $XY || BC$.


*Prove that the feet of the four perpendiculars dropped from a vertex of a triangle upon the four bisectors of the other two angle are collinear.


*In triangle $\Delta ABC, X $and $Y$ be the feet of perpendiculars from vertex $A$ to the internal angle bisector of $∠B$ and $∠C$ respectively. Line $XY$ meets $AB$ at $P$ and $AC$ at $Q$. If $AB = 7$ cm, $BC = 8$ cm and $CA = 5$ cm then find $PQ$ and $XY$.
These were 3 questions with almost similar configurations; as I wasn't able to solve the first one, I couldn't solve the next two either (I don't know if it's just psychological or something but anyway).
So, could someone please provide the SOLUTION for the first one and just hints for the next two
A solution using just elementary geometry would be ideal but any solution would be really helpful,
Thanks
 A: Here is a nice solution using trigonometry; I will name the points according to your diagram (i.e., using $AD$ and $AE$ instead of $AX$ and $AY$). So let $F$ and $G$ be the feet of the perpendiculars from $D$ and $E$ onto $BC$, respectively. It suffices to show that $d(BC, D)=|DF|=|EG|=d(BC, E)$.
Notice that $\Delta DGB$ and $\Delta ADB$ are similar and hence
$$\frac{|DF|}{|AD|}=\frac{|DB|}{|AB|}\Leftrightarrow |DF|=\frac{|AD||DB|}{|AB|}$$
However, using $\alpha, \beta, \gamma$ to denote the interior angles of the triangle, as usual, we have $|AD|=|AB|\cos(\beta/2)$ and $|DB|=|AB|\sin(\beta/2)$, so
$$|DF|=|AB|\sin(\beta/2)\cos(\beta/2)=\frac{1}{2}|AB|\sin(\beta)$$
Finally, the sine rule in $\Delta ABC$ then gives
$$|DF|=\frac{1}{2}|BC|\frac{\sin(\beta)\sin(\gamma)}{\sin(\alpha)}$$
and by symmetry, the same argument for $E$ instead of $D$ gives
$$|EG|=\frac{1}{2}|BC|\frac{\sin(\beta)\sin(\gamma)}{\sin(\alpha)}$$
and we are done.
For problem 2, call the feet onto the internal bisectors $H$ and $I$. Then a similar computation should give exactly the same values for $d(H, BC)$ and $d(I, BC)$ as above and you are done.
For problem 3, use that the length of the altitude from $A$ in $\Delta ABC$ is given by $|BC|\frac{\sin(\beta)\sin(\gamma)}{\sin(\alpha)}$.
A: 
The dashed lines are angle bisectors. Using angle chasing, congruency and mid-point theorem, bunch of isosceles triangles and various lengths can be found. I'll leave the work to you.
$RS$ contains the midpoints of $AB, AC$. The final results should be :
$$RS = \dfrac{1}{2}A_3A_4 = s$$
$$PQ = \dfrac{1}{2}A_1A_2 = \dfrac{1}{2}(CA_2+BA_1 - BC) = s-a$$
