Let $M$ be the midpoint of $BC$, $AE$ be diameter in the circle $ABC$. $EM=MH$ so $M$ is equidistant from the projections $A$ and $P$ of the points $E$ and $H$ on $\ell_a$ and if $N$ is the projection of $M$ on $\ell_a$ then $\angle AMN=\angle PMN$. Thus $$\angle(HP, PQ)=\angle(HA, AQ)= \angle(MA, AE) = \angle(AM, MN) = \angle(NM, MP) = \angle(HP, PM)$$which means that $P, Q, M$ are collinear Q.E.D.

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parmenides51 wrote: Given a scalene triangle $\vartriangle ABC$ has orthocenter $H$ and circumcircle $\Omega$. The tangent lines passing through $A,B,C$ are $\ell_a,\ell_b,\ell_c$. Suppose that the intersection of $\ell_b$ and $\ell_c$ is $D$. The foots of $H$ on $\ell_a,AD$ are $P,Q$ respectively. Prove that $PQ$ bisects segment $BC$. By a $\sqrt{AH\cdot AD}$ Inversion where $D$ is the Orthogonal Projection of $A$ onto $BC$ it boils down to prove the following property Property wrote: In a $\triangle ABC$ the interstion point of the tangent at $A$ and $BC$, the intersection of $A$-Symmedian and $BC$, then $A-$ HM point and $A$ are concyclic. Now applying $\sqrt{AB\cdot AC}$ Inversion to this property boils down to prove the following property which is well known Property wrote: In a $\triangle ABC$ , the line $\|$ to $BC$ though $A(\ell)\cap \odot(ABC)$, the $A-$ Median $\cap\odot(ABC)$ and the Pole of $BC$ are collinear. Let $\ell\cap\odot(ABC)=K$ , $M$ be the midpoint of $BC$ and $AM\cap\odot(ABC)=N$ and $T$ be the Pole of $BC$. Then $-1=(BC;M\infty{BC})\overset{A}{=}(BC;NK)$. So, $\overline{K-N-T}$. Inverting back twice we solve the problem. $\blacksquare$

$\definecolor{A}{RGB}{0,148,79}\color{A}\fbox{Coordinates}$ Let $a,b,c$ be complex numbers lying on unit circle centered at $0$ such that they are vertices of the triangle $ABC$. $\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Solution}$ Let $M$ be the midpoint of segment $BC$.$$m=\frac{b+c}{2},\ h=a+b+c,\ d=\frac{2ab}{a+b}$$$$p=a+b+c-\frac{(b+c)(a^2+bc)}{2bc}$$$$q=\frac{\overline{a}d-a\overline{d}+\overline{h}(a-d)+h\overline{(a-d)}}{2\overline{(a-d)}}=\frac{a^2(b-c)^2+bc(b+c)(b+c-2a)}{2bc(b+c-2a)}$$Therefore $$q-m=\frac{a^2(b-c)^2}{2bc(b+c-2a)},\ p-m=\frac{2bc-ab-ac}{2bc}\cdot a$$and $$\frac{p-m}{q-m}=\frac{(b+c-2a)(2bc-ab-ac)}{a(b-c)^2}=\overline{\left(\frac{(b+c-2a)(2bc-ab-ac)}{a(b-c)^2}\right)}=\overline{\left(\frac{p-m}{q-m}\right)}\implies \definecolor{A}{RGB}{255,0,255}\color{A}M\in PQ.\blacksquare$$#1762

Let $E$, $F$ be the foot of altitudes from $B$, $C$ respectively and let $N=AD\cap EF$. It is sufficient to prove $EFPQ$ is harmonic because $MF$, $ME$ are tangent to $(AH)$. Notice that $EF$ is antiparallel to $BC$ and $AD$ is the $A-$symmedian therefore $AD$ is the median of $\triangle AFE$. Notice that \[ -1=(FE;N\infty)\stackrel{A}{=}(EF;PQ) \]So done.

Let $K$ be $A-$humpty point and $l_c\cap BC=T,AD\cap BC=L$. Let $M$ be the midpoint of $BC$. Let $P$ be the foot of the altitude from $A$ to $BC$. Let $AD\cap (ABC)=N$. $TN$ is tangent to $(ABC)$ since $AN$ is the polar of $T$ according to $(ABC)$. By inversion centered at $A$ with radius $\sqrt{AH.AP}, \ Q\leftrightarrow L,K\leftrightarrow M,P\leftrightarrow T$. Hence we want to prove that $A,T,L,K$ are cyclic. By inversion centered at $T$ with radius $TA=TK=TN, \ K\leftrightarrow K,N\leftrightarrow N,A\leftrightarrow A,B\leftrightarrow C,L\leftrightarrow M$. Since $A,K,M$ are collinear, $A,K,L,T$ are cyclic as desired.$\blacksquare$

Let $\Delta DEF$ be the orthic triangle. It suffices to show that $(P,Q;E,F)=-1$ but it is easy by projecting at $A$.