Come on, this seems known :/ (same point in ESL G4) Let $D, E, F$ be feet of altitudes, let $AH$ hit $(ABC)$ again at $H'$, and define $X = EF \cap BC \cap AP$ so that we have $-1=(X, D; B, C) \stackrel{A}{=} (P, H'; B, C)$. Then the tangents to $(ABC)$ at $P, H'$ meet on $BC$, i.e. $QH'$ is tangent to $(ABC)$. So $QP=QH'=QH$, done.

rocketscience wrote: Come on, this seems known :/ (same point in ESL G4) Let $D, E, F$ be feet of altitudes, let $AH$ hit $(ABC)$ again at $H'$, and define $X = EF \cap BC \cap AP$ so that we have $-1=(X, D; B, C) \stackrel{A}{=} (P, H'; B, C)$. Then the tangents to $(ABC)$ at $P, H'$ meet on $BC$, i.e. $QH'$ is tangent to $(ABC)$. So $QP=QH'=QH$, done. Pretty known, indeed. :/

Similar approach, Let $AH$ meet $BC,(ABC)$ at $D,L$. $K$ the antipode of $A$ in $ABC$, $M$ the midpoint of $BC$ and $T$ the intersection of the tangents to $(ABC)$ at $B,C$. 1- It is well known that $K,M,H,P$ are collinear. 2- $\overarc KC=\overarc BL$ then $PL$ is the $P$-symmedian in $\triangle BPC$, then P,L,T are collinear. 3- the pole of $BC$ is in $PL$, then the pole of $PL$ is in $BC$, then the tangents at $P,L$ meet at $Q$ in $BC$. 4- the conclusion follows after the well known fact $HD=DL$.

Good proof 3th point is also called La Hire's theorem

Nice problem . Let $BH$ hit $AC$ and $PC$ at $E$, $X$ and let $CH$ hit $AB$ and $PB$ at $Z$ ,$Y$. Finally, let $M$ be the midpoint of $BC$ and $A'$ the antipode of $A$ in $(ABC)$. It is well known that $H,P,M,A'$ are on a straight line and that $BHCA'$ is a parallelogram. Claim1 $PXHY$ is cyclic. Indeed. $\angle{YPX}=\angle{BPC}=\angle{BAC}=180-\angle{ZHE}=180-\angle{YHX}$ $\Box$ Claim2 $XY \parallel BC$ Indeed. $PXHY$ is cyclic as shown above, hence $\angle{PXY}=\angle{PHY}=\angle{AHC}=\angle{HA'B}=\angle{PA'B}=\angle{PCB}$ . $\Box$ From Claim2 follows the fact that there is a homothety centered at $P$ sending $XY$ to $CB$ as well as a homothety centered at $H$ sending $XY$ to $BC$, that is $(PXHY)$ is tangent to $(ABC)$ and $(HBC)$ at $P$ and $H$ respectively. Hence, since $BC$ is their common cord applying the radical axis theorem on $(PXHY), (ABC), (HBC)$ yields the result.

Let $M$ - midpoint of $BC$, altitudes $BB_H, CC_H$ cut $(ABC)$ at $B_0, C_0$ and let $PH$ cut $(ABC)$ at $T$. It's well-known that $M \in PH$. $MB_H = BC/2 = MC_H$. With homothety $(H, k=2)$ we have $TB_0 = TC_0$. Now it's just a converse of IMO 2010 P4 on $\triangle PBC$.

Let D be the antipode of A ,E the foot of altitude from A, M the midpoint of (PH) and X the reflection of H across BC. The points P,H,D are collinear and the points A,B,C,X are cocyclic,so,from PoP,we get AH.HX=DH.PH or AH.HE=DH.HM,which means that ADEM is cyclic. If O is the circumcenter of ABC and Y is the point where the perpendicular bisector of (PH) cuts BC,since HEYM is cyclic,we get YHM = YEM or YPD = APO,so PY is tangent to the circumcircle of ABC Hence Q = Y,so QH = PQ

Let $O$ be the circumcenter of $ABC$, $N$ the midpoint of $AH$ and $M$ the midpoint of $BC$. It is well known that $P$, $H$ and $M$ are collinear, and $OM$ is parallel to $AH$, so, $\angle OMP=\angle AHP$. Also, $\angle OPQ = \angle OMQ = 90º \implies OMQP$ ciclic $\implies \angle OMP = \angle PQO$. $\angle PQO = \angle AHP$, and $\angle QPO = \angle HPA = 90º \implies OPQ $ is similar to $APH$, so, by spiral similarity, $PQH$ is similar to $POA$. But, $PO=PA$ (circumcircle of ABC), so, $PQ=PH$.

Too easy if you're familiar with projective (and I'm not a huge fan of it): The statement is equivalent to $(C,B;H',P)=-1$ where $H'$ is the reflection of $H$ in $BC$. Ratio $PB/PC = BH/CH$ because $P$ is the center of spiral similarity taking $PEF$ to $PBC$. Since $BH/CH= BH'/CH'$, then $PB/PC = BH'/CH'$ this quadrilateral is indeed harmonic.

Problem similar to Brazil MO 2011/5 Let T be the intersection of tangents to (ABC) by B and C, by the la hire's theorem we have that TP is the polar of Q, therefore, as we want to prove that QP=QH, it is enough to prove that the quadrilateral PH'BC is harmonic (with H' the symmetric of the orthocenter by BC). Let M is the midpoint of BC then ME and MF are tangent to the circumference of diameter AH (with BE, CF heights of ABC), as M,H,P are collinear we have that $-1=(P,H;F,E )\stackrel{A}{=}(P,H'; B,C)$

Jafet98 wrote: Let $ABC$ be an acute-angled triangle with $AB< AC$, and let $H$ be its orthocenter. The circumference with diameter $AH$ meets the circumscribed circumference of $ABC$ at $P\neq A$. The tangent to the circumscribed circumference of $ABC$ through $P$ intersects line $BC$ at $Q$. Show that $QP=QH$. Posting for storage since i'm noting that many came up with this sol. Claim 1: Let $AH \cap (ABC)=I \ne A$ it follows that $QI$ is tangent to $(ABC)$. Proof: Let $D,E,F$ the altitudes from $A,B,C$ respectivily. Let $EF \cap BC=G$. Since it is known that $A-P-G$ by projections we have: $$-1=(G, B; D, C) \overset{A}{=} (P, B; I, C) \implies PBIC \; \text{harmonic quadrilateral}$$By the propeties of an harmonic quadrilateral we are done. Using Claim 1 we have $QP=QI$ but since its known that $I$ is the reflection of $H$ over $BC$ we have that $QP=QI=QH$ thus we are done

Let $D,E,F$ be the foot of altitudes from $A,B,C$ respectively. Let $G$ be the another point of tangency from $Q$ to $(ABC) \implies (P,G;B,C)=-1$. Let $I = EF \cap BC$. Famous result of Miquel points is that Miquel point $(AFE) \cap (BIF) \cap (CEI)$ lies on $(ABC) \iff E,F,I$ are collinear, since $(AFE)$ intersects $(ABC)$ at $P$ thus $P$ is our required Miquel point. By angle chase $\overline{I-P-A}$ are collinear. $$\implies (I,D;B,C) = -1 \stackrel{A}{=} (P, AD \cap (ABC);B, C)$$So $G = AD \cap (ABC)$ and since $Q$ lies on the perpendicular bisector of $HG \implies QG = QH = QP$.

Another way to finish a projective solution. Let $H'$ be the symmetric of $H$ with respect to $BC$, $X=PH'\cap BC$, $D$ be the foot of the $A$-altitude of $\Delta ABC$ and $M$ be the midpoint of $\overline{BC}$. As noted by others, the problem is equivalent to proving that $(Q,X;B,C)=-1$. It is known that $P$,$H$ and $M$ are collinear. Note that quadrilateral $PAMD$ is cyclic. By angle chasing, $\angle PMX=\angle PMD=\angle PAD=\angle PAH'=\angle QPH'=\angle QPX$. This means that $QP$ is tangent to $(PMX)$. Then, by power of point, $QB\cdot QC=QP^2=QX\cdot QM$. But this is sufficient from corollary 2 of https://alexanderrem.weebly.com/uploads/7/2/5/6/72566533/projectivegeometry.pdf .