vicentev wrote: Let \( AD \) and \( BE \) be altitudes of triangle \( \triangle ABC \) that meet at the orthocenter \( H \). The midpoints of segments \( AB \) and \( CH \) are \( X \) and \( Y \), respectively. Prove that the line \( XY \) is perpendicular to line \( DE \). We have $XA=XB=XD=XE$ and $YC=YH=YD=YE$, so $XY$ is the perpendicular bisector of $DE$. This is also Argentina Cono Sur TST 2015 P4.

Trivial problem! This is well-known orthocenter config By simple angle chasing, you can prove that $XDYE$ is kite. If you want to learn more about orthocenter config see i3435’s handout

WLOGQED1729 wrote: Trivial problem! This is well-known orthocenter config By simple angle chasing, you can prove that $XDYE$ is kite. If you want to learn more about orthocenter config see i3435’s handout yeyeyye i saw the configuration in the olympiad jjsjssjsj (if you have time look the problem 6 of this year)

We will use the lemma that radical axis is perpendicular to the line connecting the center of both circles. Firstly we can see that $\angle{ADB}=\angle{AEB}$ so $ABDE$ are concyclic. Then $\angle{HDC}+\angle{HEC}=180$ , from this we get $HDCE$ concyclic. SO the radical axis is $DE$ , we can see it clearly. Now the only thing we want to show is that $X$ is the center of one circle while $Y$ is the center of the other circle. This is also clear because we know that $AB$ and $HC$ are both diameters and $X,Y$ are their midpoints respectively. so we are done here. $Q.E.D$

Let $C$ be the antipode of $C$ in $(ABC)$. Take a homothety about $H$ with scale factor 2, then $XY$ goes to $C'C$. It is well-known that $DE \perp CC'$, so the result follows.