Let $M$ be the midpoint of $BC$ so $XDMY$ is cyclic $\Longrightarrow D$ lies on Nine point circle of $\triangle ABC$ So $AD \perp BC$ Hence done.

BBAI wrote: Let $M$ be the midpoint of $BC$ so $XDMY$ is cyclic $\Longrightarrow D$ lies on Nine point circle of $\triangle ABC$ So $AD \perp BC$ Hence done. Or you could argue that since $XY\parallel DM$(in circle $XYZD$), we have $XD=YM=AX=BX$ which means that $\angle ADB=90$.

I assume this is "2011 Philippine Math Olympiad National Stage Problem 2"; if so, you better put it in the source field, like for the other four problems you just posted.

thugzmath10 wrote: In triangle $ABC$, let $X$ and $Y$ be the midpoints of $AB$ and $AC$, respectively. On segment $BC$, there is a point $D$, different from its midpoint, such that $\angle{XDY}=\angle{BAC}$. Prove that $AD\perp BC$. Okay I'm probably overlooking something, but could someone tell me if my following approach is valid. Also, before I begin, sorry for the bump.

Thanks in advance.

This problem can use Miquel's Theorem to solve

HYP135peppers wrote: Okay I'm probably overlooking something, but could someone tell me if my following approach is valid. Also, before I begin, sorry for the bump. No it is correct. Just drop the "if" part

D belong to nine point circle by angle condition so if it s not the midpoint of BC it must be the foot of perpendicular from A

NKzk wrote: This problem can use Miquel's Theorem to solve Can someone further explain?