In a quadrilateral $ABCD$ diagonal $AC$ is a bisector of $\angle BAD$ and $\angle ADC = \angle ACB$. The points $X$ and $Y$ are the feet of the perpendiculars from $A$ to $BC$ and $CD$ respectively. Prove that the orthocenter of $\triangle AXY$ lies on the line $BD$.
Problem
Source: IX International Festival of Young Mathematicians Sozopol, Theme for 10-12 grade
Tags: geometry
sunken rock
23.09.2018 22:01
My solution @ https://www.facebook.com/photo.php?fbid=1916125101801720&set=a.784793394934902&type=3&theater Best regards, sunken rock
Skibizi123
05.07.2019 09:30
sunken rock wrote: My solution @ https://www.facebook.com/photo.php?fbid=1916125101801720&set=a.784793394934902&type=3&theater Best regards, sunken rock I can't use this link
Pinko
26.08.2019 21:02
Hey Skibizi123! Sorry for the late reply! Here is my solution to the problem: It is easily seen that $\Delta ADC \sim \Delta ACB$ (By angles). Let $k$ be their similarity coefficient so that $\frac{AD}{AC}=k$. Since $AY$ and $AX$ are altitudes in $\Delta ADC$ and $\Delta ACB$ respectively, then $\angle AYC=\angle AXC=90^\circ$ which leads to $AXCY$ being cyclic. Hence $\angle CAX=\angle XYC=\alpha$ and $\angle YAC=\angle YXC=\beta$. Let $H$ be the orthocenter of $\Delta AXY$. We see that $YH\parallel CB$, since both are perpendicular to $AX$. In a similar way $XH\parallel CD$, since both are perpendicular to $AY$. So $HXCY$ is parallelogram and $\angle HYX=\angle YXC=\beta$, $\angle HXY=\angle XYC=\alpha$ . Now $\angle DYH=\angle HXB=180^\circ -\alpha -\beta$. Also $\frac{DY}{YH}=\frac{DY}{CX}=k=\frac{YC}{XB}=\frac{XH}{XB}$. Hence $\Delta DYH \sim \Delta HXB$ from where it follows that $\angle DHB=\angle DHY+\angle BHX +\angle YHX=\angle HBX+\angle BHX+\angle HXB=180^\circ$. So $H\in BD$ as desired.