Triangle and the altitude

Question

Let $ABC$ be a triangle, altitude from $A$ intersects $BC$ at $D.$ $P$ be an arbitrary point on $AD. $ $BP$ and $CP$ intersects $AC$ and $AB$ at $X$ and $Y$ respectively. Prove that $\angle ADX=\angle ADY.$

I almost tried everything to solve this. But it is too hard for me.

Any help will be appreciated.

Answer 1

Set coordinate system as $A(0,1)$, $B(b,0)$ and $C(0,c)$. Then $D(0,0)$ and $P(0,p)$ for some $p$.

Then $X({(p-1)bc\over pc-b},{p(b-c)\over pc-b})$ and so the slope from $D$ through $X$ is $k = {p(c-b)\over (p-1)bc} $.

Also $Y({(p-1)bc\over pb-c},{p(c-b)\over pb-c})$ and so the slope from $D$ through $Y$ is $k' = {p(b-c)\over (p-1)bc} = -k$.

So $\angle ADX = \angle ADY$.