MRF2017 wrote: In quadrilateral $ABCD$ , $AC$ is bisector of $\hat{A}$ and $\widehat{ADC}=\widehat{ACB}$. $X$ and $Y$ are feet of perpendicular from $A$ to $BC$ and $CD$,respectively.Prove that orthocenter of triangle $AXY$ is on $B \color{red}{ D }\normalcolor $. Typo corrected My solution : Let $ Z $ be the point on $ BD $ such that $ XZ \parallel CD $ . Easy to see $ C $ is the antipode of $ A $ in $ \odot (AXY) $ . ... $ (\star) $ From $ \angle BAC=\angle CAD, \angle ACB=\angle ADC \Longrightarrow \triangle ABC \sim \triangle ACD $ , so $ BZ:ZD=BX:XC=CY:YD \Longrightarrow YZ \parallel BC \Longrightarrow XZYC$ is a parallelogram . Since $ Z $ is the reflection of $ C $ in the midpoint of $ XY $ , so combine $ (\star) $ we get $ Z $ is the orthocenter of $ \triangle AXY $ . Q.E.D

MRF2017 wrote: In quadrilateral $ABCD$ , $AC$ is bisector of $\hat{A}$ and $\widehat{ADC}=\widehat{ACB}$. $X$ and $Y$ are feet of perpendicular from $A$ to $BC$ and $CD$,respectively.Prove that orthocenter of triangle $AXY$ is on $BD$. My 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.

MRF2017 wrote: In quadrilateral $ABCD$ , $AC$ is bisector of $\hat{A}$ and $\widehat{ADC}=\widehat{ACB}$. $X$ and $Y$ are feet of perpendicular from $A$ to $BC$ and $CD$,respectively.Prove that orthocenter of triangle $AXY$ is on $BD$. My solution: Let $H$ is orthocenter of triangle $AXY$. Easy to see that $XHYC$ is parallelogram. Thus, \[\widehat{BXH}=\widehat{HYD}\] Now we have to prove that $\dfrac{BX}{HX}=\dfrac{HY}{YD}$. It's equivalent to $\dfrac{BX}{CY}=\dfrac{CX}{YD}$. Easy to see that triangle $BAC$ is similar to triangle $CAD$. So that triangle $BAX$ is simliar to triangle $CAY$, triangle $CAX$ is similar to triangle $DAY$, thus $\dfrac{BX}{CY}=\dfrac{AX}{AY}$ and $\dfrac{CX}{YD}=\dfrac{AX}{AY}$, therefore $\dfrac{BX}{CY}=\dfrac{CX}{YD}$. Thus, triangle $BXH$ is similar to triangle $HYD$. So, \[\widehat{BHX}+\widehat{XHY}+\widehat{YHD}=\widehat{BHX}+\widehat{BXH}+\widehat{XBH}=180^o\] Hence, $B, H, D$ are collinear. We are done.

Let $H$ be the intersection point of $BD$ with the perpendicular from $X$ to $AY$. Then $HX \parallel DC $, so $\frac{BH}{HD}=\frac{BX}{XC} \displaystyle {(1)}$. Triangles $ADY$ and $ACX$ are similar because they are right and $\angle ADY= \angle ACX$. Hence $\frac{DY}{XC}=\frac{AY}{AX} \displaystyle {(2)}$. Also triangles $AYC$ and $AXB$ are similar because they are right and $ \angle YAC= \angle DAC - \angle DAY = \angle CAB - \angle CAX= \angle BAX$. Therefore $\frac{YC}{BX}=\frac{AY}{AX} \displaystyle {(3)}$. By $\displaystyle {(2)}$ and $\displaystyle {(3)}$ it is $\frac{DX}{XC}=\frac{YC}{BX} \Rightarrow \frac{BX}{XC}=\frac{CY}{YD}$. Then by $\displaystyle {(1)}$ it is $\frac{DH}{HB}=\frac{DY}{YC}$. Thus, we obtain that $HY \parallel BC$. But $AX\perp BC$ yields that $HY\perp AX$. So $H$ is the orthocenter of triangle $AXY$ and it lies on $BD$.

Let perpendicular from X to AY meet BD at S. we have to show S is the orthocenter so let's prove YS || BC instead. ∠ADC = ∠ACB ---> BC tangent to circle ADC ---> ∠DAC = ∠DCX ∠AXC = ∠AYC = 90 ---> ∠ACXY is cyclic. step1 : AYD and AXC are similar. (1) ∠YAD = ∠YAX - ∠DAX = ∠YCX - ∠DAX = ∠DAC - ∠DAX = ∠XAC ∠AYD = 90 = ∠AXC step2 : ADC and ACB are similar. (2) ∠ADC = ∠ACB ∠DAC = ∠CAB step3 : DSY and DBC are similar. (3) from (1) we have : YD/XC = AD/AC and from (2) we have : AD/AC = DC/CB so YD/XC = DC/CB ---> DY/DC = XC/BC and from Thales theorem we have XC/BC = SD/BD. we have DY/DC = SD/BD and ∠YDS = ∠CDB. now from (3) we have SY || BC. we're Done.

Let perpendicular to $AY$ through $X$ meets $BD$ at $H.$ Clearly $ABXC\sim ACYD,$ so $$\frac{|XH|}{|CD|}=\frac{|BX|}{|BC|}=\frac{|CY|}{|CD|}.$$Thus $CXHY$ is parallelogram, and since $C$ is antipode of $A$ on $\odot (AXY),$ $H$ is orthocenter of $AXY.$ Done.