j___d wrote: An acute triangle $ABC$ in which $AB<AC$ is given. Points $E$ and $F$ are feet of its heights from $B$ and $C$, respectively. The line tangent in point $A$ to the circle escribed on $ABC$ crosses $BC$ at $P$. The line parallel to $BC$ that goes through point $A$ crosses $EF$ at $Q$. Prove $PQ$ is perpendicular to the median from $A$ of triangle $ABC$. Let $H_A$ be the A-HM point. As the polar of $M$-the midpoint of $BC$ passes through $Q$, by La Hire’s theorem we obtain that $QA=QH_A$. It’s well known that $PH_A$ is tangent to $\odot(BH_AC) $ * . So $PH_A^2=PB\cdot PC=PA^2$. So $PQ$ is the perpendicular bisector of $AH_A$. * look at Example 2 from https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2017-02/article_1_a_special_point_on_the_median.pdf

Nice problem! Let $T$ be the reflection of $H$ w.r.t. $A$, $M$ be the midpoint of $BC$ and $A'$ be the antipode of $A$ w.r.t. $\odot(ABC)$. Since $\triangle ABC\cup P\cup A'\sim\triangle AEF\cup Q\cup H$, we see that $\frac{AP}{AQ} = \frac{AA'}{AH}= \frac{AT}{AA'}$. Combining with $\angle TAQ=\angle PAA'=90^{\circ}$ gives $\triangle ATA'\sim\triangle AQP$. So $\angle(TA', PQ) = \angle(AT, AQ)=90^{\circ}$. Clearly $BHCA'$ is parallelogram so $M$ is the midpoint of $HA'$. So $TA'\parallel AM$. Combining with the previous paragraph gives $PQ\perp AM$ as desired.

Let $H_A$ be the A-Humpty point, clearly $(AH_AEF)=-1$ and because $QA$ is tangent to $\odot(AEF)$ and $Q$ lies on $EF$ we have that $QA=QH_A$. Because $P$ is the center of the A-Apollonius circle, which passes through both $A$ and $H_A$ we have that $PQ$ is perpendicular to $\overline{AH_A}$, which is precisely the A-median of $\triangle{ABC}$.

Is this correct? Take the circles $C(A,0)$ and $C(BCEF)$. We have $PA^2=PB\times PC$ and $QA^2=QE\times QF$ hence $PQ$ is the radical axis the 2 circles with centers $A$ and $M$, hence we are done.

Here is a bashy solution, in which I will prove that $\overrightarrow{AM}\bullet\overrightarrow{PQ}=0$, $M$ being the midpoint of $BC$. Note that this is equivalent to $(\overrightarrow{AB}+\overrightarrow{AC})\bullet(\overrightarrow{PA}+\overrightarrow{AQ})=0$. Now, observe these four equalities: $\overrightarrow{AB}\bullet\overrightarrow{PA}=|AB|\cdot|PA|\cdot\cos{(180-\hat{C})}$ $\overrightarrow{AB}\bullet\overrightarrow{AQ}=|AB|\cdot|AQ|\cdot\cos{(180-\hat{B})}$ $\overrightarrow{AC}\bullet\overrightarrow{PA}=|AC|\cdot|PA|\cdot\cos{\hat{B}}$ $\overrightarrow{AC}\bullet\overrightarrow{AQ}=|AC|\cdot|AQ|\cdot\cos{\hat{C}}$ Adding these with using $|PA|=|AQ|\cdot\cos{\hat{A}}$ leaves to prove that $\frac{|AC|}{|AB|}=\frac{\cos{\hat{B}}+\cos{\hat{A}}\cdot\cos{\hat{C}}}{\cos{\hat{C}}+\cos{\hat{A}}\cdot\cos{\hat{B}}}$, which is equal to $\frac{\cos{\hat{B}}+\cos{(\hat{A}-\hat{C})}}{\cos{\hat{C}}+\cos{(\hat{A}-\hat{B})}}=\frac{\sin{\hat{B}}\cdot\sin{\hat{A}}}{\sin\hat{C}\cdot\sin\hat{A}}$, the conclusion follows.

j___d wrote: An acute triangle $ABC$ in which $AB<AC$ is given. Points $E$ and $F$ are feet of its heights from $B$ and $C$, respectively. The line tangent in point $A$ to the circle escribed on $ABC$ crosses $BC$ at $P$. The line parallel to $BC$ that goes through point $A$ crosses $EF$ at $Q$. Prove $PQ$ is perpendicular to the median from $A$ of triangle $ABC$. Note that $Q$ is the radical center of $\odot(A), \odot(AEF), \odot(BC)$ hence $Q$ lies on the radical axis of $\odot(A), \odot(BC)$. Also $P$ is the radical center of $\odot(A), \odot(ABC), \odot(BC)$ hence $P$ also lies on the radical axis of $\odot(A), \odot(BC)$. Now the $A$-median passes through the center of $\odot(BC)$, the result follows.

Let $Z=EF\cap BC,K=EF\cap AH, T=AH\cap BC$. It is well known fact that $ZH\perp AM$ $\implies$ it is enough to show that $ZH\parallel PQ$. Let $F=ZH\cap AQ$. Let's show that $AQ=QF$ $\iff$ $\frac{AQ}{AF}=\frac{1}{2}$. Using parellelism we have that \[AQ=\frac{|AK||ZT|}{KT} \ ,\ AF=\frac{|AH||ZT|}{HT}\]$\implies$ \[\boxed{\frac{AQ}{AF}=\frac{|AK||KT|}{|HT||AH|}}\]It is easy to show that last relatioon is equal to $\frac{1}{2}$ (Sine's Law). Finally we have that $AQ=QF=PZ$ ($APZQ$ is alrealy a parallelogram since $AP\parallel EF$) $\implies$ $PQFZ$ is a parallelogram $\implies$ $ZH\parallel PQ$ as desired $\blacksquare$

This is a very nice algebra problem.(Codyj,USAMO 2018,P5) We can find easily with barycentric coordinate $P=(0,\frac{-2b^2}{2(c^2-b^2)},\frac{2c^2}{2(c^2-b^2)}),$ and $Q=(\frac{2(c^2-b^2)}{2(c^2-b^2)},\frac{-a^2+b^2+c^2}{2(c^2-b^2)},\frac{-(-a^2+b^2+c^2)}{2(c^2-b^2)}).$ The rest is very easy.

Let $T$ be the midpoint of $AH$, where $H$ is the orthocenter of $\triangle ABC$. Also, Let $M_A$ be the midpoint of $BC$. Note that $EF$ is the radical axis of $\odot (AEHF)$ and $\odot (BFEC) \Rightarrow TM_A$ is the perpendicular bisector of $EF$. Also, $\angle PAB = \angle ACB \Rightarrow \angle PAF = \angle AFE \Rightarrow PA \parallel EF$ But, $TM_A$ is perpendicular to $EF \Rightarrow TM_A \perp PA \Rightarrow T$ is the orthocenter of $\triangle AM_AP \Rightarrow PT \perp AM_A$ Now, Let $H_A$ be the A-Humpty point and $PT \cap AM_A = X \Rightarrow TX \perp AH_A \Rightarrow X$ is the midpoint of $AH_A$. Also, $M_AE$ and $M_AF$ are tangents to $\odot (AEHH_AF) \Rightarrow AEH_AF$ is a harmonic quadrilateral. $\Rightarrow QH_A$ is also tangent to $\odot (AEH_AF) \Rightarrow Q, X, T, P$ are collinear $\Rightarrow PQ \perp AM_A \text{ } \blacksquare$

Nice problem from Poland..Nice solutions..

Sorry for double posting, but here's a bit different ending: Let $H_A$ be the $A$-Humpty point, and let $EF \cap BC=X$. From the various properties proved here, we have that $X$ lies on $HH_A$, and that $HH_A$ is perpendicular to the $A$-median. Thus it suffices to show that $PQ$ is parallel to $XH$. Let $XH \cap AQ=T$. Now, as both $EF$ and $AP$ are antiparallel to $BC$, we get that $EF$ is parallel to $PQ$. But, as $PX$ is parallel to $AQ$, we have that $APXQ$ is a parallelogram, and so $AQ=PX$. Also, as $AEH_AF$ is a harmonic quadrilateral, and cause $AQ$ is tangent to $\odot (AEHH_AF)$ at $A$, we must have that $QH_A$ is tangent to $\odot (AEHH_AF)$ at $H_A$, which gives $QH_A=QA$. But, $TH_A$ is perpendicular to $AH_A$, which means that $Q$ is the circumcenter of $\triangle AH_AT$, or equivalently $QT=QA=PX$. As $PX \parallel QT$, quadrilateral $PXTQ$ is a parallelogram $\Rightarrow PQ \parallel TX$. Hence, done. $\blacksquare$

Let $M$ be midpoint of $BC$ then we have: $PA^2$ $-$ $QA^2$ = $\overline{PB}$ . $\overline{PC}$ $-$ $\overline{QE}$ . $\overline{QF}$ = $P_{P / (BCEF)}$ $-$ $P_{Q / (BCEF)}$ = $PM^2$ $-$ $QM^2$ or $PQ$ $\perp$ $AM$

Let T denote midpoint of AH where H is orthocenter of ABC. It easy to prove that T is orthocenter of APM since AT is perpendicular to BC and MT is perpendicular to AP. Therefore, PT is perpendicular to AM. Now, since Q lies on EF which is polar of M, then AM is polar of Q so QT is perpendicular to AM and we are done.

let $T=EF \cap BC$ ,$N=EF \cap AD$,$X=AQ \cap TH$ from brocard on $EFBC$ we have that $AM$ is perpendicular to $TH$ so it suffices to show that $QP \parallel TX$ we have $AP \parallel TQ \implies AQTP $is parallelogram so it suffices to show that $Q$ is the midpoint of $AX$ but we have $(A,H;N,D) =_T (A,X:Q,P_\infty)=-1$ and we done

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%205.pdf p. 48-50. Sincerely Jean-Louis

Let $H$ be the orthocenter of $\triangle ABC$, and let $AH$ intersect $BC$ at $D$. Let $X_A$ be the $A$-humpty point of $\triangle ABC$. Let $L$ be the midpoint of $AH$, and let $\omega$ be the circle with diameter $AH$, let $K$ be the foot of the perpendicular from $H$ to $AP$. Let $J$ be the midpoint of $EF$, which lies on line $LM$ by inversion around $\omega$. Claim: $QA = QX_A$. Proof: Note that $-1 =(AX_A;EF)$ by Three Tangents Lemma, and since $QA$ is a tangent to $\omega$, $QX_A$ must also be, as desired. Claim: $PA = PX_A$. Proof: First, remark that $PKHD$ is cyclic, so by radical axes on $(PKHD)$ and $(DHX_AM)$ we find that $PKX_AM$ is cyclic. We claim that $KLX_AM$ is also cyclic, which would imply the $L$ is the arc midpoint of $\triangle PKX_A$, and thus $\angle APL = \angle LPX_A$, which would finish the claim. To this end, we invert around $\omega$. We wish to show that $K,J,X_A$ are collinear. Redefine $J$ to be the intersection of $KX_A$ and $EF$. Remark that $AK \parallel EF$, so $-1 = (EF;AX_A) \stackrel{K}{=}(EF;\infty_{EF}J)$, so $J$ is the midpoint of $EF$, as desired. Finally, consider the circles centered around $P$ and $Q$ passing through $A$ and $X_A$. Their radical axis is $AM$, thus $PQ \perp AM$, as desired. Remarks: After reading the solutions I realized how much I overcomplicated this problem. However, I think most steps in this proof are really natural if you've seen HM point before; most points that showed up in my diagram are standard pieces of the orthocenter puzzle; the only strange one was $P$ and $K$, which was dealt with by only considering the property that $AP \parallel EF$ instead of trying to use the tangent.

Let the midpoint of $BC$ be $M$. Invert about the circle centered at $A$ with radius $\sqrt{AE \cdot AC}=\sqrt{AF \cdot AB}$. Then $E$ and $F$ go to $B$ and $C$, so $(AEF)$ gets sent to line $BC$ and $(ABC)$ gets sent to line $EF$. $P$ goes to the other intersect of line $AP$ and $(AEF)$ and $Q$ goes to the second intersection of $(ABC)$ and line $QA$. Call these two points $P'$ and $Q'$, respectively. We want to show that lines $\overline{AM} \perp \overline{PQ}$. It suffices to show that their images under the inversion are orthagonal. Since $\overline{AM}$ goes through the center of inversion ($A$), it gets sent to itself. $\overline{PQ}$ gets sent to $(P'AQ')$. For a circle and a line to be orthogonal, the line must go through the circle's center. So to finish, it suffices to show that $MP'=MA=MQ'$. The angle between $\overline{AH}$ and $\overline{BC}$ maps to the angle between $\overline{AH}$ and $\overline{Q'A}$ so $\angle Q'AH=90^\circ$ and $QA \parallel BC$. Hence $ABCQ'$ is a trapezoid. $ABCQ'$ is cyclic and consequently an isocoleces trapezoid so, combined with the fact that $M$ is a midpoint, $\overline{MA}=\overline{MQ'}$. Similarly, $\overline{PA} \parallel \overline{EF}$ so $P'AEF$ is a trapezoid. $P'AEF$ is also cyclic and consequently an isocoleces trapezoid. Now, we prove the following Claim. $M$ lies on the perpendicular bisector of $EF$. Proof. Notice that $\measuredangle BFC = \measuredangle BEC$ so $BFEC$ is cyclic. It is well known that the circumcenter of a right triangle is the midpoint of the hypotenuse. The circumcenter of both $\triangle BFC$ and $\triangle BEC$ is $M$. Hence, it must also be the circumcenter of $(BFEC)$. The circumcenter of a cyclic quadrilateral is the intersection of the perpendicular bisectors of its sides. So the perpendicular bisector of $EF$ meets $BC$ at $M$. $\square$ Hence $M$ is on the perpendicular bisector of $AP'$ and $MA=MP'$. Thus, $MP'=MA=MQ'$ and the problem is solved. $\blacksquare$

j___d wrote: An acute triangle $ABC$ in which $AB<AC$ is given. Points $E$ and $F$ are feet of its heights from $B$ and $C$, respectively. The line tangent in point $A$ to the circle escribed on $ABC$ crosses $BC$ at $P$. The line parallel to $BC$ that goes through point $A$ crosses $EF$ at $Q$. Prove $PQ$ is perpendicular to the median from $A$ of triangle $ABC$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.215056421232438cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ real xmin = -3.5003702503830105, xmax = 1.6071579602332087, ymin = -1.4512527129674466, ymax = 1.1395672197453763; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen qqffff = rgb(0.,1.,1.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pair A = (-0.47604097886087826,0.8794230986534166), B = (-0.8563048722764766,-0.5164706823388596), C = (0.8645105550903593,-0.502614663671245), F = (-0.733910614509161,-0.06717905564256275), Q = (1.0373237038345644,0.891608722376249), M = (0.004102841406941393,-0.5095426730050523), G = (-1.574602498145136,-0.5222544201846837), H = (-0.46783529604699514,-0.13966224735668792), D = (-0.4648266131396366,-0.5133184964737749), T = (-0.16048653209914282,-0.03341656968220569), P = (-3.0879671808405784,-0.5344400439075159); filldraw(A--B--C--cycle, qqffff + opacity(0.10000000149011612), linewidth(0.8) + qqffff); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(0.8)); draw(A--B, linewidth(0.8) + qqffff); draw(B--C, linewidth(0.8) + qqffff); draw(C--A, linewidth(0.8) + qqffff); draw(C--F, linewidth(0.8)); draw(B--(0.037242961582917296,0.3502544911644847), linewidth(0.8)); draw(A--Q, linewidth(0.8)); draw(Q--F, linewidth(0.8) + red); draw((0.1374494037929564,0.4044972929636135)--(0.14178399271569253,0.43054544305651604), linewidth(0.8) + red); draw((0.1374494037929564,0.4044972929636135)--(0.16162909660971167,0.3938842236771702), linewidth(0.8) + red); draw(A--M, linewidth(0.8)); draw(G--F, linewidth(0.8)); draw(A--D, linewidth(0.8)); draw(G--H, linewidth(0.8)); draw(H--T, linewidth(0.8)); draw(B--G, linewidth(0.8)); draw(circle((0.004102841406941443,-0.5095426730050523), 0.8604356054227742), linewidth(0.8)); draw(circle((-0.47193813745393676,0.36988042564836426), 0.5095591907921436), linewidth(0.8)); draw(A--P, linewidth(0.8) + red); draw((-1.7962612207204742,0.16477398696972068)--(-1.791926631797738,0.19082213706262324), linewidth(0.8) + red); draw((-1.7962612207204742,0.16477398696972068)--(-1.772081527903719,0.1541609176832774), linewidth(0.8) + red); draw(G--P, linewidth(0.8)); draw(P--Q, linewidth(0.8)); draw((0.037242961582917296,0.3502544911644847)--T, linewidth(0.8)); draw(Q--T, linewidth(0.8)); /* dots and labels */ dot(A,xdxdff); label("$A$", (-0.4772323312818748,0.960464053955932), NE * labelscalefactor,xdxdff); dot(B,xdxdff); label("$B$", (-0.9249902457554853,-0.5742647977225817), NE * labelscalefactor,xdxdff); dot(C,xdxdff); label("$C$", (0.9185371676289663,-0.5433849415519878), NE * labelscalefactor,xdxdff); dot((0.037242961582917296,0.3502544911644847),linewidth(4.pt) + uuuuuu); label("$E$", (0.06934112293763597,0.22243549147873934), NE * labelscalefactor,uuuuuu); dot(F,linewidth(4.pt) + uuuuuu); label("$F$", (-0.7798549217536943,0.030980383221057558), NE * labelscalefactor,uuuuuu); dot(Q,linewidth(4.pt) + uuuuuu); label("$Q$", (1.0729364484819355,0.9326721834023977), NE * labelscalefactor,uuuuuu); dot(M,linewidth(4.pt) + uuuuuu); label("$M$", (0.016845367447626464,-0.6236725675955318), NE * labelscalefactor,uuuuuu); dot(G,linewidth(4.pt) + uuuuuu); label("$G$", (-1.5487633404014807,-0.592792711424938), NE * labelscalefactor,uuuuuu); dot(H,linewidth(4.pt) + uuuuuu); label("$H$", (-0.43400053264304345,-0.24693832231428697), NE * labelscalefactor,uuuuuu); dot(D,linewidth(4.pt) + uuuuuu); label("$D$", (-0.4741443456648154,-0.63293652444671), NE * labelscalefactor,uuuuuu); dot(T,linewidth(4.pt) + uuuuuu); label("$T$", (-0.10049808600063008,-0.06165918529072396), NE * labelscalefactor,uuuuuu); dot(P,linewidth(4.pt) + uuuuuu); label("$P$", (-3.1545158612723596,-0.6082326395102349), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] First of all note that $(GD,BC)=-1$ since $BEFC$ is a complete quadrilateral. also $\angle PAF=\angle ACB=\angle AFE$ so $AP||QG$, now from $(GD,BC)=-1$ we understand that $(AE,Ft)=-1$ which gives that $QA=QT$ on the other hand $T$ is A-humpty point so $PA=PT$ so $PQ$ is the perpendicular bisecot of $AT$ and we are done.(You can read more about A-Humpty point here)

An excellent problem! Let $H$ denote the orthocentre of $ABC$. Let $\omega_1$ denote the circumcircle of $ABC$ First, note that $AEHf$ and $EFCB$ are cyclic quadrilaterals. Denote by $\omega_2$ the circle through $AEFH$ (diameter $AH$) and $\omega_3$ the circle through $EFCB$ (diameter $BC$ and centre $M_{BC}$). Now observe that the parallel through $A$ to $BC$ is simply the tangent at $A$ to $\omega_2$ We now define a circle of radius zero at point $A$ as $\omega_4$ and consider the radical centre of $\omega_1$, $\omega_3$, $\omega_4$ Note that the radical axes are $BC$, tangent at $A$ and the radical axis of the zero radius circle at $A$ and the circle through $EFCB$ Since $BC \cap$ tangent at $A = P$, $P$ lies on the 3rd radical axis too. We now consider the radical centre of $\omega_2$, $\omega_3$ and $\omega_4$ Note that the radical axes are $EF$, parallel at $A$ and the same radical axis of the zero radius circle at $A$ and the circle through $EFCB$ Since $EF \cap $ parallel at $A = Q$, $Q$ lies on the same radical axis too Thus line $PQ$ is just the radical axis of the radius $0$ circle at $A$ and the circle through $EFCB$. Since the line joining the centres is perpendicular to the radical axis, $AM \perp PQ$

Let $\omega_a$ denote the point circle $A$ and $M$ be the midpoint of $BC$. It's clear that $$Pow_{\omega_a}(P) = PA^2 = Pow_{(ABC)}(P) = PB \cdot PC = Pow_{(BC)}(P)$$and the Three Tangents Lemma gives $$Pow_{\omega_a}(Q) = QA^2 = Pow_{(AEF)}(Q) = QE \cdot QF = Pow_{(BC)}(Q)$$so $PQ$ is the Radical Axis of $\omega_a$ and $(BC)$. But $(BC)$ is centered at $M$, so we're done. $\blacksquare$ Remark: Utilizing the $A$-Humpty point is probably more motivated than this solution.

In order to prove PQ ⊥ AM we need to prove |PA^2 - PM^2| = |QA^2 - QM^2| or |PA^2 - QA^2| = |PM^2 - QM^2| |PA^2 - QA^2| = PB.PC - QE.QF = power of P w.r.t BFEC - power of Q w.r.t BFEC Let M be midpoint of BC. M is center of BFEC so : power of P w.r.t BFEC - power of Q w.r.t BFEC = |PM^2 - QM^2| so at the end we have |PA^2 - QA^2| = P w.r.t BFEC - power of Q w.r.t BFEC = |PM^2 - QM^2| and this will prove our problem so we're Done.

Let $M$ be the midpoint of $\overline{BC},$ $\omega_1=(M,\overline{MB}),$ and $\omega_2=(A,0)$ with $(O,r)$ denoting the circle with center $O$ and radius $r.$ Notice $\angle PAB=\angle ACB$ so $\triangle ACP\sim\triangle BAP.$ Hence, $$\text{pow}_{\omega_2}{P}=AP^2=PB\cdot PC=\text{pow}_{\omega_1}{P}.$$Similarly, $Q$ lies on the radical axis of $\omega_1$ and $\omega_2$ by the three tangents lemma. Radical Axis finishes. $\square$ Remarks: (Motivation) My solution was motivated by the three tangents lemma, which had occurred to me after seeing $A,E,F,$ and $\overline{AQ}\parallel\overline{BC}.$ Since $M$ was the center of $(BCEF),$ radical axis was natural and the solution followed.

Special Thanks to guptaamitu1 for referring and motivating me to solve this problem .

From $\angle BAP=\angle BCA=\angle EFA =\angle EAQ$ we deduce $$|QA|^2=|QE|\cdot |QF|,|PA|^2=|PB|\cdot |PC|,$$hence $PQ$ is the radical axis of $A,\odot (BCEF),$ the conclusion follows.

Sol:- Note that $P$ is the radical center of $(A, 0), (BC), (ABC)$ and $Q$ is the radical center of $(A,0), (BC), (AED)$.Hence the radax of $(A, 0)$ and $(BC)$ is $A$- median.

We claim $P$ and $Q$ lie on the radax of $(A)$ and $(BFEC)$ (which is sufficient). By PoP on $(ABC)$ we have $PA^2 = PB \cdot PC$. It is well known that $AQ$ is tangent to $A$. By PoP $AQ^2 = QE \cdot QF$. Proving our result.

Lol I complicated the solution so much although iÍ still think it is really nice. Let $X$ be the intersection of $AM$ with the nine-point circle $\gamma$. Also, let $Y$ be the intersection of $AH$ with $\gamma$. It is well-known that $MY$ is a diameter of $\gamma$, so $\angle YXM=90^\circ$. It now suffices to show that $P,Q\in XY$. It is well-known that $(AEHF)$ is tangent to the line through $A$ to $BC$. To see that, simply note $\angle QAE=\angle ACB=180^\circ-\angle BFE=\angle EFA$. It is also easy to see that $\triangle AXY$ and $\triangle AMD$ are homothetic at $A$, so $(AXY)$ is tangent to $QA$ as well. Applying the radical axis theorem to $\gamma$, $(AEHF)$ and $(AXY)$ shows that $Q\in XY$. An angle chase shows that \[\angle PAB=\angle ACB=\angle EFA=\angle AEH\]meaning that $AP\parallel EF$. As $MY\perp EF$, we also have $MY\perp AP$. Therefore, letting $Z=MY\cap AP$, we see that $Z$ lies on $(AYX)$. Therefore, $AZDM$ is cyclic, and apllying the radical axis theorem on $(AZDM)$, $\gamma$ and $(AYX)$ we see that $P\in YX$, so we are done.

Let $M$ be the midpoint of $BC$. We claim that $PQ$ is the radical axis of the circle with radius $0$ centered at $A$, and $(BFEC)$. Since by Power of a Point on $(ABC)$, we have $PA^2 = PB \cdot PC$, $P$ has equal power to both $(A)$ and $(BFEC)$. Then we have $\angle QAE = \angle C = \angle AFE$, and $\angle Q = \angle Q$, $\triangle QAF \sim \triangle QEA$. From this, we have $\frac{QA}{QE} = \frac{QF}{QA} \implies QA^2 = QE \cdot QF$ so $Q$ also lies on the radical axis of $(A)$ and $(BFEC)$. We finish by noting that $A$ and $M$ are the respective centers of $(A)$ and $(BFEC)$, so $AM \perp PQ$.

oops radius 0 is so smart i didn't think of that here is a probably isomorphic solution We first eliminate $P$. The conclusion is equivalent to $$PM^2-PA^2=QM^2-QA^2.$$However, note that $$PM^2-PA^2=PM^2-PB\cdot PC=PM^2-(PM^2-MB^2)=MB^2=\frac{a^2}{4}.$$Thus, it remains to show that $$QM^2-QA^2=\frac{a^2}{4}.$$ Note that $$\angle QAE=\angle QFA=\gamma$$since $AQ\parallel BC$ and $BFEC$ is cyclic. Thus, $$QA^2=QE\cdot QF.$$Hence, $$QE\cdot QF=Pow_{(BFEC)}(Q)=QM^2-\frac{a^2}{4},$$as desired.

Decent problem indeed. $EF \cap BC$ at $X_{A}$ $X_{A}H \cap AQ$ at $R$ $EF \cap AH$ at $L$ We have $X_{A}H$ is prependicular to $AM$ that motivates us to prove $PQ$ is parallel to $X_{A}H$ We also have $AQ$ is parallel to $BC$ and $PX_{A}$ and $AP$ is parallel to $X_{A}Q$ so that $APX_{A}Q$ is parallelogram It remains to prove $Q$ is midpoint of $AR$ From easy harmonic bundle $(A,H;L,AH \cap BC) \equiv (A,R;Q,V_\infty)=-1$. which leads conclusion.