Geometry / Iran Second Round 2015

Problem :  (Iran Second Round 2015)

In the quadrilateral $ABCD$, $AC$ is the bisector of $\angle A$ and $\angle ADC=\angle ACB$. $X,Y$ are feet of perpendicular from $A$ to $BC,CD$ respectively.  Prove that the orthocenter of triangle $AXY$ is on $BD$.

Solution :

Easy to see that the circle $(AXY)$ through $C$.

Let $E,F$ be the intersection of $(AXY)$ and $AB,AC$ respectively.

Let $H$ be the intersection of $EY$ and $FX$.

We apply Pascal theorem for six points $H,A,G,X,C,Y$, we get that $B,H,D$ are conlinear.

On the other hand,

$$\angle AFX=\angle ACX=\angle ADC$$.

So we have $XF$ is parrallel to $C$, implies $XF\perp AY$.

Similary, $YE\perp AX$.

Consequenlty, $H$ be the orthocenter of triangle $AXY$.

We are done.