Basically all we need to show is that $\angle ACF = \angle ACH$, which can be done using harmonic pencils and La Hire.

I used coordinates, which is not that hard.

Easy to show $PQ\perp FH \Leftrightarrow \angle ACF=\angle ACH$ we can prove this by Pascal theorem and using Appolonius circle

Can someone post a full solution, please? I can't seem to find a solution. Edit: Nvm, I just angle chased my way through and got a fairly simple solution.

Let $CH$ and $DF$ cuts in $X$ and $CA$ and $DF$ cuts in $Y$. Let $AD$ and $CF$ cuts in $R$ .It is easy to see that $DF$ and $FA$ are tangents to $(BDGA)$. We have $AD$ the polar of point $F$ , since $H$ is on $AD$ , $H$ is on polar of point $F$.So $F$ is on polar of point $H$.Let $DG$ cuts $AB$ in $T$ . By Brocard theorem $TC$ is polar of point $H$.So $T,C,F$ are collinear .We have $(T, S ,B,A)=-1$ , where $S$ is the intersection between $CH$ and $AB$. $(T,S,B,A)$=$(R,H,D,A)$=$(F,X,D,Y)=-1$.Then we have $CA$ the bisector of angle $HCF$.

lminsl wrote: Triangle $ABC$ satisfies $\angle C=90^{\circ}$. A circle passing $A,B$ meets segment $AC$ at $G(\neq A,C)$ and it meets segment $BC$ at point $D(\neq B)$. Segment $AD$ cuts segment $BG$ at $H$, and let $l$, the perpendicular bisector of segment $AD$, cuts the perpendicular bisector of segment $AB$ at point $E$. A line passing $D$ is perpendicular to $DE$ and cuts $l$ at point $F$. If the circumcircle of triangle $CFH$ cuts $AC$, $BC$ at $P(\neq C),Q(\neq C)$ respectively, then prove that $PQ$ is perpendicular to $FH$. Observe that $F=\overline{AA} \cap \overline{DD}$ for circle $\odot(AGDB)$. Define $F'=\overline{GG} \cap \overline{BB}$. Claim: Points $C,F,F'$ are collinear and lines $\overline{CF}, \overline{GD}, \overline{AB}$ concur. (Proof) Follows from a homography making $AGDB$ a rectangle with the same circumcircle. $\blacksquare$ Hence $(CF, CH, CA, CB)=-1 \implies (CF, CH, CP, CQ)=-1 \implies PFQH$ is harmonic. Hence $\overline{PQ} \perp \overline{FH}$.

We can use Analytical geometry to prove this problem too.

just use pascal and blanchet’s thm

We first of all remove unnecessary points (like $E$), and restate the problem a bit by using $\angle PCQ=90^{\circ}$ as follows:- Restated Problem wrote: Let $\triangle ABC$ be a right-angled triangle with $\angle ACB=90^{\circ}$. $D,E$ are points on $BC,AC$ respectively such that $A,B,D,E$ lie on circle $\omega$. Suppose the tangents to $\omega$ at $A$ and $D$ meet at $F$, and let $AD \cap BE=H$. Also let $\odot (CFH) \cap AC,BC=P,Q$. Then show that $PQ$ is the perpendicular bisector of $FH$. Note that by Pascal on $AAGDDB$ and $ABBDGG$, we get that the tangents to $\omega$ at $B$ and $G$ also meet on $CF$, thus giving that $CF$ is the polar of $H$ wrt $\omega$. Let $CF \cap BG=T$. Then $$-1=(B,G;H,T) \overset{C}{=} (Q,P;H,F) \Rightarrow FPHQ \text{ is a harmonic quadrilateral} \Rightarrow PQ \text{ is the perpendicular bisector of } FH$$

By angle chasing, it suffices to prove that $\angle ACH = \angle ACF$ As $\angle GCD = 90$, we only need to prove that $(F, H; CA \cap HF, CD \cap HF) = -1$. Let $X$ lie on $\odot(AGD)$ such that $XDGA$ is a harmonic quadrilateral. Taking perspectivity at $A$ projecting from $\odot(AGD)$ onto line $HF$, we only need to prove that $AX, CD, HF$ concur. Pascal's in $DDBBAX$ shows that this is equivalent to proving $BA, XD, HF$ concur. Pascal's in $GXDAAB$ proves this, and we're done.

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