Isn't it reaaaally easy or am I blind this evening? $\angle {APQ}=\angle {ABH}+\angle {PHB}=\angle {ACH}+\angle {CHQ}=\angle AQP$ So triangle $APQ$ is isosceles. Looking at triangles $PAM$ and $QAM$, we have $AP=AQ $, $AM$ common and $\angle {PAM}=\angle {QAM}$, so they are congruent . $\implies MP=MQ$.

You probably are blind for not calling $APMQ$ a kite as $A, M$ lie on perpendicular bisector of $PQ$ (just kidding )

Okay So here's my solution. My Solution. Let $\angle PHB=\angle CHQ=\theta.$ Then $\angle HPA=\angle QHA=90-A-\theta,$ as $\angle HBA=\angle HCA=90-A.$ As $P,H,Q$ are collinear, we can say that $\triangle APQ$ is isosceles. Now, by fact 5, we can say that $AM$ bisects $\angle A.$ Now, let us denote $AM\cap PQ=X.$ As $\triangle APQ$ is isosceles, $AX\perp PQ,$ and $PM=MQ.$ Which means $AM$ is the perpendicular bisector of $PQ.$ Now, we know that the set of all points equidistant from $2$ given points is the line joining the $2$ vertices. As $M$ lies on the perpendicular bisector of $PQ,$ $M$ must be equidistant from $P,Q.$ Which means $MP=MQ.~\Square$

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can someone post a solution with complex numbers? it is interesting if there is a solution with complex baash

How is the fact that M is the midpoint of arc BC l used?

First, let $E$ be the foot of the perpendicular from $B$ to $AC$ and $F$ be the foot of the perpendicular from $C$ to $AB$. Let $N$ be the intersection of $PQ$ and $AM.$ Note that $\angle FHB=\angle EHC$, so $PQ$ is just the angle bisector of $\angle FHB$ and $\angle EHC$. So, $\triangle FHP \sim \triangle EHQ \implies \angle FPH=\angle EQH$. Combined with $\angle AQN=\angle PAN$ and $AN=AN$, it follows by AAS that $\triangle PAN \cong \triangle QAN$. In particular, $AM \perp PQ$ and $PN=QN$. Combined with $NM=NM$, we get that $\triangle PNM \cong \triangle QNM$, so $MP=MQ$ as desired

Simple one... $\angle APQ = \angle PHB + \angle PBH = \angle QHC + \angle HCQ = \angle AQP$ and so $AP = AQ$. So $APQ$ is isosceles and so its angle bisector is its perpendicular bisector. But since $AM$ is the angle bisector, $M$ must lie on the perpendicular bisector of $PQ$ and so $MP = MQ$

Its easy becuase its P1 Claim 1: The triangle $\triangle APQ$ is isosceles. Proof: Its known that $\angle PBH=\angle QCH$, if we add the fact that $\angle PHB=\angle QHC$ then: $\angle PBH+\angle PHB=\angle QCH+\angle QHC \implies \angle APQ=\angle AQP \implies \triangle APQ \; \text{isosceles}$ By Claim 1 we know that $AM$ is perpendicular bisector of $PQ$ thus $APQM$ is a kite and then $MP=MQ$ and we are done

The line $CH$ meets $AB$ at $D$ and $BH$ meets $AC$ at $E$. Since $\angle BDC = \angle BEC = 90^{\circ}$ we get $BDEC$ is cyclic quadrilateral. Hence, $\angle PBH = \angle DBE = \angle DCE = \angle HCQ$ so that $$\angle APQ = \angle PBH + \angle PHB = \angle HCQ + \angle CHQ = \angle PQA$$ Therefore, $AP = AQ$. Finally, since $M$ lie on the internal bisector of $\angle PAQ$, it turns out that $AM$ is perpendicular bisector of $PQ$, thus $MP = MQ \blacksquare$

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