Given an acute triangle $ABC$, let $X$ be point on the side $AB$ such that $|AX|=|BX|$. Furthermore, drop heights from points $A$ (to point $D\in BC$) and $B$ (to point $E\in AC$). Then, let $P$ be a point such that $PX\perp AD$, and, if $G$ is the intersection between $PX$ and $AD$, that $|PG|=|GX|$. Similarly, let $Q$ be a point such that $QX\perp BE$, and, if $H$ is the intersection between $QX$ and $BE$, that $|QH|=|HX|$. Now: is it always true, that if $F$ is the center of $ED$, then the points $P$, $Q$ and $F$ are colinear?
I'm looking mainly for hints, but:
- I've noticed that if the statement is true, then the points $P$, $E$, $Q$ and $D$ form a parallelogram, thus the thesis is equivalent to the fact that $PE||DQ$ and $EQ||PD$.
- It's pretty straightforward to prove that $AE||XQ$ and $XP||BD$, but I found it much more difficult to exhibit that $AP||XD$ (which, coupled with $AE||XQ$ would prove that $PE||DQ$).
$XQ$ is parallel to $AC$ and $XP$ is parallel to $BC$.
$X$ is the midpoint of $AB$. Then $H$ is the midpoint of $BE$ and $G$ is the midpoint of $AD$.
Since $H$ is by assumption the midpoint of $QX$ and $G$ is the midpoint of $PX$ the quads $BXPD$ and $AXQE$ are parallelograms. Hence both $PD$ and $EQ$ are parallel to $AB$ and $PD = AX = BX = EQ$.
Since $PD$ parallel to $EQ$ and $PD = EQ$ then $EQDP$ is a parallelogram and its diagonals intersect at a point which is a midpoint for both diagonals. Hence if $F$ is the midpoint of $ED$ then $F$ lies on $PQ$ and it is its midpoint.