Triangle with a circle having an altitude as its diameter: proof of a property without using crossratios

Question

I´ve been thinking about this problem for 3 days:

Let $ABC$ be an acute triangle, $D, E,$ and $F$ the feet of the altitudes of $A, B$ and $C$, respectively. Let;

1. $O$ be the midpoint of segment $AD$

2. $c$ be the circle with center $O$ passing through $A$ and $D$

3. $X$ and $Y$ the intersections of $c$ with $AB$ and $AC$, respectively

4. $P$ the intersection of $XY$ with $AD$, and $Q$ the intersection of $AD$ and $EF$.

Prove that $P$ is the midpoint of the segment $QD$.

I have proved that $XY$ and $EF$ are parallel lines by using inscribed angles. Someone helped me solve this problem via cross ratios. The following is his solution;

It's easy to show that $BCXY$ is cyclic, then $EF\parallel XY$ so if $XY,EF$ intersect $BC$ at $G$ and $S$ respectively and if $T$ is the reflection of $D$ in $G$ then; $$GX\cdot GY=GB\cdot GC=GD^2$$ $$\implies (B,C;D,T)=-1.$$ Thus $S$ and $T$ coincide. Since $G$ is midpoint of $DS,$ we know that $P$ is the midpoint of $QD.$

I don´t understand how he applies cross ratio, and I would be grateful if someone could explain to me how he solves the problem or if someone could solve the problem without using projective geometry.

This is my visual proof that EF and XY are parallels

Answer 1

See this link from AoPS blog (https://artofproblemsolving.com/community/c1642h1016535_harmonic_divisions__a_powerful_amp_rarely_used_tool). Proof of Theorem $4$ uses a construction that can be used to prove that $P$ is the midpoint of $QD$. But instead of repeating that work, I will use theorem $4$ to show that points $T,C, D$ and $B$ are in harmonic range i.e. $~\displaystyle \frac{CT}{CD} = \frac{BT}{BD}$. Then use theorem $1$ to conclude that $S$ is the midpoint of $TD$.

As $CDHE$ is cyclic, $\angle HED = \angle HCD = 90^\circ - \angle B$ and hence $\angle CED = \angle B$. Also as $BCEF$ is cyclic, $\angle BEF = \angle BCF = 90^\circ - \angle B$ and $\angle CET = \angle B$

So $(i) ~EC$ bisects $ \angle DET $ and $(ii) ~\angle BEC = 90^\circ$. Those two statements mean that pencil $~E~ (TD; CB)$ forms a harmonic range as shown in Theorem $4$.

Now using $BCEF$ as cyclic, note that $\angle CFE = \angle CBE = 90^\circ - \angle C$.
So, $\angle AYX = \angle AFE = \angle C ~~(\text {given } EF \parallel XY$, which you have already shown).

That leads to $BCXY$ being cyclic. Hence,
$SC \cdot SB = SX \cdot SY = SD^2$

Using converse of theorem $1$, $S$ must be the midpoint of $TD$ and using $EF \parallel XY$, that proves $P$ is the midpoint of $QD$.