Let $D'$, $X$ be the reflections of $D$ across $M$, $I$ respectively. It is well known that $D'$ is the point at which the excircle touches $BC$, and also that $A,X,D'$ are collinear. If $Q'$ is the second intersection of this line and the incircle, then $\angle AQ'D=\angle XQ'D=90^\circ$. So $Q'=Q$, and $Q$ lies on $AXD'$. Let $\omega$ be the incircle and $\omega_1$ be the circle with diameter $DD'$. Note that $Q$ lies on $\omega_1$, which is also the circle centered at $M$ with radius $MD$. Thus $P$ is one of the two intersections of $AI$ and $\omega_1$. Note that $Q$ is the spiral similarity center mapping $DD'$ to $XD$. The same spiral similarity then maps $\omega_1$ to $\omega$. Thus it suffices to show that this spiral similarity maps $AI$ to $EF$. This is because if this is true, $P$ maps to either $E$ or $F$, so if it WLOG maps to E, $\angle PQE=\angle D'QD=90^{\circ}$. So note that the spiral similarity has angle $90^\circ$, and that $AI\perp EF$. Thus it suffices to show that there is a point on $AI$ which maps to $EF$. Specifically, we show that $I$ maps to a point $I'$ on $EF$. Note that since the similarity maps $M$ to $I$, it then maps $I$ to the point $I'$ on $MQ$ such that $\angle I'IM=90^{\circ}$ (since $\angle IQM=90^{\circ}$). But $MQ$ is tangent to $\omega$ since $MQ=MD$, so the polar of $I'$ is the line through $Q$ parallel to $IM$. But by homothety centered at $D$ $IM\parallel AXQD'$, so said line is $A$. And thus the polar of $A$, $EF$, passes through $I'$.

I wonder whether there is any solution that does not use spiral similarity.

WLOG assume that $AC >AB.$ A-excircle $(I_a)$ touches $BC$ at $D'$ and $U$ is the antipode of $D$ WRT $(I).$ As $A$ is the exsimilicenter of $(I) \sim (I_a),$ it follows that $A,U,D'$ are collinear. Thus if $AUD'$ cuts $(I)$ again at $Q',$ we have $\angle UQ'D \equiv \angle AQ'D=90^{\circ}$ $\Longrightarrow$ $Q \equiv Q'.$ Since $M$ is also midpoint of $DD',$ then $DPQD'$ is cyclic with circumcenter $M.$ Note that $P$ is intersection of $AI$ with the B-midline of $\triangle ABC,$ thus it is on $DE$ (well-known). Hence $\angle PQE=\angle UQE+\angle UQP=\angle UDE+\angle EDC=90^{\circ},$ as desired.

Assume WLOG $AB<AC$. We will show $\angle PQE=90^\circ$. Lemma I: $AI$, $DE$, and the line through $M$ parallel to $BC$ are concurrent. Furthermore their intersection is $P$. Proof: The lines are concurrent by the Midline Concurrency Lemma (see e.g. here). Say this intersection is $P'$. Then $\angle DP'M=\angle DEC=\angle EDC=\angle P'DM$. So, $P'M=DM$ and $P'$ lies on $AI$. Hence $P'=P$. Since $\angle AQD=90^\circ$, $AQ$ passes through $D'$, the point diametrically across $D$ on the incircle. By a well-known fact of $D'$, $AQ$ passes through the tangency point of the $A$-excircle on $BC$, which we call $X$. Since $BD=CX$, $M$ is the midpoint of $DX$. Since $AQ\perp QX$, $Q$ lies on the circle with diameter $AX$. It follows that $A,P,Q,X$ are all on the circle with center $M$ and radius $DM$. Since we have $\angle DQD'=90^\circ$, it suffices to show $\angle DQP=\angle D'QE$. Since $DD'$ is tangent to the circle with center $M$ and radius $DM$, we have $\angle DQP=\angle D'DP$. We also have $\angle D'QE=\angle D'DE$. By collinearity of $D,P,E$ (Lemma I), $\angle D'DE=\angle D'DP$ and we are done.

Let $\omega$ be the incircle of $\triangle ABC$, and WLOG let $AB>AC$. We will show that $\angle PQF=90^\circ$ in this case; if this holds true then trivially $\angle PQE=90^\circ$ if $AC>AB$, completing the proof. First, we show that $M$ is the circumcenter of $\triangle DPQ$. Let $N=AQ\cap BC$ and $D'=AQ\cap\omega$. Then since $AQ\perp QD$ we have $D'$ and $D$ antipodal, so a homothety centered at $A$ sending $\omega$ to the $A$-excircle of $\triangle ABC$ sends $D'$ to the tangency point of the $A$-excircle of $\triangle ABC$ with $BC$, which is therefore $N$. It is well-known that $BN=CD$, and since $BM=MC$ trivially we have $NM=MD$. From the previously-mentioned perpendicularity $M$ is the midpoint of the hypotenuse $\overline{ND}$ of right triangle $NQD$, so $MQ=MP=MD$. Next, we show that $F$, $P$, and $D$ are collinear. Let $FD$ intersect $AI$ at a point $P'$. Angle chasing gives \[\angle AP'F=\angle BFD-\angle FAP'=\dfrac{\pi-\angle B}2-\dfrac{\angle A}2=\dfrac{\angle C}2.\] Therefore $\angle IP'F\equiv\angle AP'F=\angle IP'D$, so $ICDP'$ is a cyclic quadrilateral. Hence $\angle IP'C=\angle IDC=90^\circ$, so $P'$ is the foot of the perpendicular from $C$ to the bisector from $\angle A$. From here it is well-known that $P'M\parallel AB$, so more angle chasing gives \[\angle MP'D=\angle PAB+\angle ICD=\dfrac12\angle A+\dfrac12\angle C=\dfrac{\pi-\angle B}2=\angle FDM\equiv\angle P'DM.\] Hence $P'$ lies on the circle centered at $M$ with radius $MD\implies P\equiv P'$. To finish, note that $D'D\perp BC$ from the fact that $D'$ and $D$ are antipodes. This makes $\angle PDN$ and $\angle D'DF$ complementary, so $\angle PND=\angle D'DF=\angle D'QF.$ Therefore \[\angle FQP=\angle FQD'+\angle D'QP=\angle D'QP+\angle PQD=90^\circ,\] which is what we wanted.

lfetahu wrote: I wonder whether there is any solution that does not use spiral similarity. The solution I had in mind is the same as Luis's -- I was not actually aware of the spiral similarity solution until after the fact. This problem was created by just messing around in GeoGebra (I had a lot of time in April). I was just curious what would come up once I constructed the circle with center $M$ and radius $D$. I noticed the collinearity of $D$, $P$, $E$ (in $AB < AC$ case) after about 20 minutes, and I remember I was initially very confused about how this was not symmetric. The observation about $\angle PQE$ came about during my half-asleep attempts to prove the collinearity.

My solution: Let $ X $ be the midpoint of $ AD $ and $ Y=MQ \cap AC $ . Since $ Q \equiv (I) \cap (AD) $ , so we get $ XI $ is the perpendicular bisector of $ DQ $ . Since $ X \in IM $ (well known) , so we get $ MQ $ is tangent to $ (I) $ at $ Q $ . Since $ MP \parallel AC $ (well known) , so $ \angle QMP=\angle MYC =\angle QIE $ , hence we get $ \triangle QMP \sim \triangle QIE $ and $ \angle EQP=\angle IQM=90^{\circ} $ . Q.E.D

Some consequence from this nice configuration The circle $(M,MQ)$ intersect $AI$ at $P,R$ ($P$ is inside triangle) and intersect $BC$ again at $S$. - Note that Spiral similarity center $Q$ transform triangle $DEF$ to triangle $SPR$. So triangle $SPR$ and $DEF$ are similar. - Tangents of $(M)$ from $P,R$ intersect at $T$ then $AT\perp BC$. - Tangents of $(M)$ from $P,S$ intersect at $U$ and tangents of $(M)$ from $R,S$ intersect at $V$ then $BU\parallel CV$.

We will bary you!

General problem Let $ABC$ be a triangle and $DEF$ is pedal triangle of $P$. $(DEF)$ cut $BC$ again at $G$. The line passes through $P$ and perpendicular to $EF$ cuts $DE,DF$ at $M,N$. Circle $(DMN)$ cuts $(DEF)$ again at $Q$. $T$ is a point such that $TM\perp AC,TN\perp AB$. $AT$ cuts $BC$ at $S$. Prove that four points $A,Q,G,S$ are concyclic.

buratinogigle wrote: General problem Let $ABC$ be a triangle and $DEF$ is pedal triangle of $P$. $(DEF)$ cut $BC$ again at $G$. The line passes through $P$ and perpendicular to $EF$ cuts $DE,DF$ at $M,N$. Circle $(DMN)$ cuts $(DEF)$ again at $Q$. $T$ is a point such that $TM\perp AC,TN\perp AB$. $AT$ cuts $BC$ at $S$. Prove that four points $A,Q,G,S$ are concyclic. My solution: Let $ Y=(DMN) \cap BC $ . Since $ MN \perp EF, TM \perp AE, TN \perp AF $ , so we get $ \triangle TMN \sim \triangle AEF $ . Since $ \angle MQN=\angle EDF=\angle EQF $ , so $ Q $ is the spiral center of $ \triangle TMN \mapsto \triangle AEF $ . ... $ (\star) $ Since $ \angle YNQ=\angle GFQ $ , so $ \triangle TMN \cap Y \sim \triangle AEF \cap G $ , hence combine with $ (\star) $ we get $ \angle QGS=\angle QGY=\angle QAT=\angle QAS $ . ie. $ A, Q, G, S $ are concyclic Q.E.D

buratinogigle wrote: General problem Let $ABC$ be a triangle and $DEF$ is pedal triangle of $P.$ $(DEF)$ cut $BC$ again at $G.$ The line passes through $P$ and perpendicular to $EF$ cuts $DE,DF$ at $M,N.$ Circle $(DMN)$ cuts $(DEF)$ again at $Q.$ $T$ is a point such that $TM\perp AC,TN\perp AB.$ $AT$ cuts $BC$ at $S.$ Prove that four points $A,Q,G,S$ are concyclic. Clearly $Q$ is center of spiral similarity that carries $\triangle QMN$ into $\triangle QEF$ and it also carries $T$ into $A$ as $\triangle TMN$ and $\triangle AEF$ are directly similar. Since $MN,EF$ form right angle, it follows that the rotational angle is right $\Longrightarrow$ $\angle MQE=\angle TQA=90^{\circ}.$ Hence if $U \equiv TM \cap AC,$ then $MQUE$ and $TQUA$ are cyclic $\Longrightarrow$ $\angle TAQ=\angle MUQ=\angle DEQ=\angle CGQ$ $\Longrightarrow$ $AQGS$ is cyclic.

WLOG, $AB > AC$. Let $AQ \cap (I) = T$. Since, $\angle{TQD} = 90^{\circ}$, so $T$ is the antipode of $D$ w.r.t $(I)$. Therefore, $AQ$ intersects $BC$ at the $A$-extouch point $D'$. Since $\angle{DQD'} = 90^{\circ}$ and $M$ is the midpoint of the hypotenuse of $\triangle{DQD'}$, $M$ is the circumcentre of $\triangle{DD'Q}$. This also means that $DD'PQ$ is cyclic with centre $M$. Let $DF \cap AI = P'$. Then, it is well known that $MP'$ is the $C$-midline and that $I,P',C,D,E$ are concyclic. Let $MP' \cap (IPD) = T$. Then \[ \angle{TCD} = \angle{MP'D} = \angle{BFD} = \angle{BFD} = \angle{BDF} \] \[ \angle{MDF} = \angle{MDP'} \implies MD = MP' \]Therefore, $P = P'$, giving $MP AB$. Some more angle chasing gives \[ \angle{DQF} = 180^{\circ} - \frac{\angle{FID}}{2} = 90^{\circ}+B/2 = 90^{\circ}+ \frac{\angle{DMP}}{2} \] \[ = 90^{\circ}+ \angle{DQP} \implies \angle{PQF} = 90^{\circ} \]

buratinogigle wrote: General problem Let $ABC$ be a triangle and $DEF$ is pedal triangle of $P$. $(DEF)$ cut $BC$ again at $G$. The line passes through $P$ and perpendicular to $EF$ cuts $DE,DF$ at $M,N$. Circle $(DMN)$ cuts $(DEF)$ again at $Q$. $T$ is a point such that $TM\perp AC,TN\perp AB$. $AT$ cuts $BC$ at $S$. Prove that four points $A,Q,G,S$ are concyclic. $MN$ intersects $EF$ at $I$. We have \[\begin{aligned} (QF,QN) & \equiv (QF,QD)+(QD,QN) \pmod{ \pi} \\ & \equiv (EF,ED)+(MD,MN) \pmod{ \pi} \\ & \equiv (EI,EM)+(ME,MI) \equiv 90^{\circ} \pmod{ \pi}. \end{aligned}\] This follows $QE \perp QM$. Similarly, we obtain $QF \perp QN$. Since $(FD,FQ) \equiv (ED,EQ) \pmod{ \pi}$ so $\triangle EQM \sim \triangle FQN \; ( \text{A.A})$. From here, we can easily get $\triangle QMN \sim \triangle QEF \; ( \text{S.A.S})$. On the other hand, we can easily prove that $\triangle TMN \sim \triangle AEF \; (\text{A.A})$. Therefore, from these two couple of similar triangle, we obtain $\triangle QEA \sim \triangle QMT \; ( \text{S.A.S})$. This follows $\angle AQT= \angle MQE=90^{\circ}$ and $\tfrac{AQ}{TQ}= \tfrac{EQ}{QM}$. Hence, $\triangle AQT \sim \triangle EQM \; ( \text{S.A.S})$. We get $(AQ,AS) \equiv (EQ,ED) \equiv (GQ,GS) \pmod{ \pi}$ or $Q,A,G,S$ are concyclic. $\blacksquare$

Dear Mathlinkers, for the first problem 1. D, P and E are collinear (well known see : http://jl.ayme.pages perso-orange.fr/ vol. 4, Un unlikely concurrence…p. 4-5) 2. AQ goes thrhrough the antipole of D wrt (I) 3. the circle centered at M and passing through D is orthogonal to (I) ; it passes also by Q (according to a developpment with the Reim’s theorem) 4. then QPE is Q-rectangular… Sincerely Jean-Louis

I also have a solution. Without loss of generality let $AB \le AC$ then $Q$ lies on the minor arc $DE$ of the incircle.Let $X=AQ \cap BC$ and $Y=AQ \cap (I)$.It is well known that the homothety with center $A$ sending $Y$ to $X$ sends $I$ to $I_a \implies X$ is the point of contact of the excircle with $BC \implies BD=CX$.Thus $M$ is the midpoint of $DX$,and as a consequence $DPQX$ is cyclic with center $M$.Let $P'=DE \cap AJ$ where $AJ$ is the angle bisector of $A$.A very quick bash shows that $\frac{JP'}{JI}=\frac{JX}{JC}=\frac{(b-c)(b+c-a)}{2ab}$ so $XP' \parallel CI$.This means that $XP' \perp P'D$ or in other words $MP'=MD \implies P' \equiv P$.Thus we have $\angle{PQE}=\angle{PDI}=\angle{EQY}$ or $\angle{PQE}=90^{\circ}$ as desired. A good geometry by v_Enhance.Thank you.

I reduced the problem to showing that Q is on the circle with radius MD, but I can't seem to do that. Any help?

^ Well, that follows trivially -- (as in the other solutions...) if $AQ$ intersects $BC$ at $T$, then $T$ is the point where the $A$ excircle touches $BC$, so the circle with radius $MD$ is actually the circle with diameter $DT$. Now the result follows by noting that $DQ \perp QT$.

Solved with squareman, asdf334, jkmmm3, v4913, CyclicISLscelesTrapezoid, awesomehuman, CT18, tigerzhang, crazyeyemoody907, DottedCaculator, DottedCalculator100, asdf1434, squareman, teasaffrontaffy, cocohearts Let $D'$ be the $D-$ antipode in the incircle, and let $AI$ intersect $DE$ and $DF$ at $P_1$ and $P_2$. Let $(DP_1P_2)$ intersect the incircle again at $Q'$. We claim $Q = Q'$. Let $DD'$ and $DQ$ intersect $EF$ at $X$ and $Y$. Since $AE$ and $AF$ are tangents and $A-D'-Q$, we have $$-1 = (EF;D'Q) \overset{D}{=} (EF;XY).$$ Now, it is well known that $P_1$ and $P_2$ are inverses about $(DEF)$, so $(DP_1P_2)$ is orthogonal to $(DEF)$ and hence $ID$ and $IQ'$ are tangent to $(DP_1P_2)$. Then if $Y' = DQ'\cap EF$, we have $$-1 = (P_1P_2;DQ') \overset{D}{=} (EF;XY')$$ so $Y' = Y$ and hence $Q' = Q$, as desired. In particular, $Q$ is the center of the spiral similarity mapping $P_1P_2$ to $EF$, so if $N$ is the midpoint of $EF$, $Q$ lies on $(NEP_1)$ and $(NFP_2)$. But $(DP_1QP_2)$ is just the circle centered at $M$ through $D$, so $P$ is $P_1$ or $P_2$ and we're done.

One day I will solve a problem with complex numbers and not drop a negative The key claim is that $\overline{AI}$, $\overline{DE}$, and the line through $Q$ perpendicular to $\overline{EQ}$ concur. I will prove this by complex numbers. Set the incircle as the unit circle. Let $E=x$ and $F=\tfrac{1}{x}$, and $D=y$, so $A=\tfrac{2x}{x^2+1}$. Let $D',E'$ be the $D$ and $E$-antipodes, so $D'=-y$ and $E'=-x$. The line $\overline{E'Q}$ is the aforementioned perpendicular to $\overline{EQ}$, and it is well-known that $A,D',Q$ are collinear. Then $$q-y=a-qy\overline{a}=a-qya \implies q=\frac{a+y}{ay+1}.$$Since $\overline{AI}$ is the line joining $1$ and $-1$, by the complex intersection formula we have $$\overline{DE} \cap \overline{AI}=\frac{x+y}{xy+1},$$and $$\overline{E'Q} \cap \overline{AI}=\frac{q-x}{-xq+1}==\frac{x-\frac{a+y}{ay+1}}{\frac{ax+xy}{ay+1}-1}=\frac{axy+x-y-a}{ax-ay+xy-1}=\frac{\frac{2x^2y}{x^2+1}+x-y-\frac{2x}{x^2+1}}{\frac{2x^2}{x^2+1}-\frac{2xy}{x^2+1}+xy-1}=\frac{2x^2y+x^3+x-x^2y-y-2x}{2x^2-2xy+x^3y+xy-x^2-1}=\frac{x^3+x^2y-x-y}{x^3y+x^2-xy-1}=\frac{x+y}{xy+1},$$where we divide by $x^2-1\neq 0$ to reach the last expression. This is the same as $\overline{DE} \cap \overline{AI}$, hence the desired concurrency exists. WLOG let $AB<AC$. It suffices to show if $P'$ is the concurrency point above, then $MD=MP'$. Let $T$ be the $A$-extouch point, which lies on $\overline{AD'Q}$ and is the reflection of $D$ over $M$. Then, $$\measuredangle TDP'=\measuredangle TDE=\measuredangle DD'E=\measuredangle D'QE'=\measuredangle TQP',$$hence $TDP'Q$ is cyclic. Furthermore, since $\measuredangle DQT=\measuredangle DAQ=90^\circ$, $\overline{DT}$ is a diameter, hence the center of $(TDP'Q)$ is $M$, so $MD=MP'$ as desired. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.0331462483410454, xmax = 2.0278180221901008, ymin = -1.3654572812855146, ymax = 2.7288391158074075; /* image dimensions */ pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen ccqqqq = rgb(0.8,0,0); pen qqwwzz = rgb(0,0.4,0.6); pen cczzqq = rgb(0.8,0.6,0); /* draw figures */ draw((-1.263072240125755,2.4684982553699357)--(-1.5323170617762356,-0.911274648628153), linewidth(0.7) + blue); draw((-1.5323170617762356,-0.911274648628153)--(1.6863454116149246,-0.911274648628153), linewidth(0.7) + blue); draw((1.6863454116149246,-0.911274648628153)--(-1.263072240125755,2.4684982553699357), linewidth(0.7) + blue); draw(circle((0.07701417491934448,0.6611312133767673), 2.249979383077001), linewidth(0.7) + dotted + ubqqys); draw(circle((-0.47061928134326314,0.06920807284523972), 0.9804827214733927), linewidth(0.7) + ccqqqq); draw((-1.263072240125755,2.4684982553699357)--(-0.47061928134326314,-0.9112746486281529), linewidth(0.7) + qqwwzz); draw(circle((0.07701417491934448,-0.911274648628153), 0.5476334562626076), linewidth(0.7) + cczzqq); draw((-0.47061928134326314,-0.9112746486281529)--(-0.47061928134326314,1.0496907943186322), linewidth(0.7) + qqwwzz); draw((0.26812248670831196,0.7138839349991624)--(-0.47061928134326314,-0.9112746486281529), linewidth(0.7) + qqwwzz); draw((-1.263072240125755,2.4684982553699357)--(-0.2830595549164033,-0.49866186222872216), linewidth(0.7) + qqwwzz); draw((-0.47061928134326314,1.0496907943186322)--(0.26812248670831196,0.7138839349991624), linewidth(0.7) + qqwwzz); draw((0.26812248670831196,0.7138839349991624)--(0.36421274609671594,-0.4449921343390092), linewidth(0.7) + qqwwzz); draw((-0.2830595549164033,-0.49866186222872216)--(0.36421274609671594,-0.4449921343390092), linewidth(0.7) + qqwwzz); draw((-0.47061928134326314,-0.9112746486281529)--(0.36421274609671594,-0.4449921343390092), linewidth(0.7) + qqwwzz); draw((-0.47061928134326314,1.0496907943186322)--(0.36421274609671594,-0.4449921343390092), linewidth(0.7) + qqwwzz); draw((0.36421274609671594,-0.4449921343390092)--(0.624647631181952,-0.9112746486281542), linewidth(0.7)); draw((-0.47061928134326314,1.0496907943186322)--(-1.263072240125755,2.4684982553699357), linewidth(0.7) + qqwwzz); /* dots and labels */ dot((-1.263072240125755,2.4684982553699357),dotstyle); label("$A$", (-1.3665073026046594,2.551067766137159), NE * labelscalefactor); dot((-1.5323170617762356,-0.911274648628153),dotstyle); label("$B$", (-1.7120500020766064,-1.0654699366222117), NE * labelscalefactor); dot((1.6863454116149246,-0.911274648628153),dotstyle); label("$C$", (1.7333842241898945,-0.9932485179134306), NE * labelscalefactor); dot((-0.47061928134326314,0.06920807284523972),linewidth(4pt) + dotstyle); label("$I$", (-0.4109826820002866,0.05671351607647744), NE * labelscalefactor); dot((0.07701417491934448,-0.911274648628153),linewidth(4pt) + dotstyle); label("$M$", (0.10010994849063738,-0.8654753603379391), NE * labelscalefactor); dot((-0.47061928134326314,-0.9112746486281529),linewidth(4pt) + dotstyle); label("$D$", (-0.6420879676985127,-1.0765806459766023), NE * labelscalefactor); dot((0.26812248670831196,0.7138839349991624),linewidth(4pt) + dotstyle); label("$E$", (0.2889920075851093,0.7566882054030828), NE * labelscalefactor); dot((-1.4480055212530345,0.14707016338299173),linewidth(4pt) + dotstyle); label("$F$", (-1.6431625193218621,0.1289331268800161), NE * labelscalefactor); dot((-0.2830595549164033,-0.49866186222872216),linewidth(4pt) + dotstyle); label("$P$", (-0.47653803667953576,-0.4877112422886601), NE * labelscalefactor); dot((-0.47061928134326314,1.0496907943186322),linewidth(4pt) + dotstyle); label("$X$", (-0.4498701647550308,1.095564840711995), NE * labelscalefactor); dot((0.624647631181952,-0.9112746486281542),linewidth(4pt) + dotstyle); label("$Y$", (0.6445347070570564,-0.8654753603379391), NE * labelscalefactor); dot((0.36421274609671594,-0.4449921343390092),linewidth(4pt) + dotstyle); label("$Q$", (0.44454193860408614,-0.4488237595482931), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] W problem mainly because I got lucky and had a $45$-ish minute solve. Without loss of generality assume the following configuration with $AC > AB$. Let $X$ be the $D$-antipode. Then $X$, $Q$ and $A$ are collinear. Claim: $\overline{MQ}$ is tangent to the incircle. Proof. Let the second tangent from $M$ meet the incircle at $Q'$. Note that if $Y$ is the $A$-excircle touch point we have $A$, $X$, $Q$ and $Y$ collinear. However noting that $MX = MY$ we find $(DQY)$ has center $M$ and hence $\angle DQY = \angle DQA = 90$. $\square$ Claim: $D$, $E$ and $P$ are collinear. Proof. Let $\overline{DE}$ meet $\overline{AI}$ at a point $P'$. We will now use complex numbers. Orient such that $\overline{AI}$ is the real axis. Let $E$ have coordinate $e$, $F$ have coordinate $\frac{1}{e}$, $D$ have coordinate $d$. We compute $A = \frac{2e}{e^2 + 1}$. Then by general intersection we find $P = \overline{AI} \cap \overline{DE}$ as, \begin{align*} p' &= \frac{-a\left(\frac{e}{d} - \frac{d}{e} \right)}{\overline{a}(d - e) - a\left(\frac{1}{d} - \frac{1}{e} \right)}\\ &= \frac{-a(e^2 - d^2)}{\overline{a}de(d-e) - a(e - d)}\\ &= \frac{a(d-e)(e+d)}{(d-e)(\overline{a}de + a)}\\ &= \frac{a(d+e)}{\overline{a}de + a}\\ &= \frac{\frac{2e(d+e)}{e^2+1}}{\overline{\left(\frac{2e}{e^2+1}\right)}de + \frac{2e}{e^2+1}}\\ &= \frac{\frac{2e(d+e)}{e^2+1}}{\frac{\frac{2}{e}}{\frac{1}{e^2} + 1}de + \frac{2e}{e^2+1}}\\ &= \frac{\frac{2e(d+e)}{e^2+1}}{\frac{2e}{e^2 + 1}de + \frac{2e}{e^2+1}}\\ &= \frac{d+e}{de + 1} \end{align*}Cool. Now we can compute $b = \frac{2\frac{d}{e}}{d + \frac{1}{e}} = \frac{2d}{ed + 1}$ and $c = \frac{2de}{e+d}$. Then we can easily find that $m = \frac{d}{ed+1} + \frac{de}{e+d} = \frac{d^2 + ed + e^2d^2 + ed}{(e+d)(ed+1)} = \frac{d^2(e^2 + 1) + 2ed}{(e+d)(ed+1)}$. Now note that, \begin{align*} m - d &= \frac{d^2(e^2 + 1) + 2ed - d(e+d)(ed+1)}{(e+d)(ed+1)}\\ &= \frac{-ed(d^2-1)}{(e+d)(ed+1)} \end{align*}Also we have, \begin{align*} m - p &= \frac{d^2(e^2 + 1) + 2ed - (e+d)^2}{(e+d)(ed+1)}\\ &= \frac{e^2(d^2-1)}{(e+d)(ed+1)} \end{align*}Now it suffices to check $|m - d| = |m - p|$. Note that then, \begin{align*} |m - d| &= |m - p|\\ \frac{-ed(d^2-1)}{(e+d)(ed+1)} \cdot \overline{\left( \frac{-ed(d^2-1)}{(e+d)(ed+1)} \right)} &= \frac{e^2(d^2-1)}{(e+d)(ed+1)} \cdot \overline{\left( \frac{e^2(d^2-1)}{(e+d)(ed+1)} \right)}\\ -ed(d^2 - 1) \cdot \overline{-ed(d^2-1)} &= e^2(d^2-1) \cdot \overline{e^2(d^2-1)}\\ -ed \cdot -\overline{d} &= e^2 \cdot \frac{1}{e}\\ e &= e \end{align*}and hence our claim is proven. $\square$ Now magically we find that $Q$ is the center of the spiral similarity mapping $\overline{EX} \mapsto \overline{DP}$. Then $\angle PQD = \angle EQX$ and hence \[ \angle EQP = \angle EQX + \angle XQP = \angle XQP + \angle PQD = \angle XQD = 90 \]so we are done. $\blacksquare$

Because there are too little complex bashes... Reformulation: Quote: Let $ABC$ be a non-isosceles triangle with incenter $I$ whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$, respectively. Denote by $M$ the midpoint of $\overline{BC}$. Let $Q$ be a point on the incircle such that $\angle AQD = 90^{\circ}$. Let $X = IE \cap (DEF)$, $P=XQ \cap AI$. Prove that $MP=MD$. Let $(DEF)$ be a unit circle with $d=1$. Then $a=\frac{2ef}{e+f}, b=\frac{2f}{f+1}, c=\frac{2e}{e+1}$, so $m=\frac{f}{f+1}+\frac{e}{e+1}$. It is known that line $MQ$ is tangent to $(DEF)$, so $m+q^2\overline{m}=2q$. From this and $m+\overline{m}=2$ we can get that $q=\frac{m}{\overline{m}} = \frac{2ef+e+f}{e+f+2}$. Also we have $x=-e$, so $p-qe\overline{p} = q-e$ and $\frac{p}{\overline{p}} = \frac{a}{\overline{a}} = ef$. From these conditions we get $p=\frac{f(q-e)}{f-q}=\frac{f(f-e)(e+1)}{(f-e)(f+1)}=\frac{f(e+1)}{f+1}$. We want to check that $MD=MP$ or $(m-p)(\overline{m-p}) = (m-1)(\overline{m-1})$. But easy to see that both of sides are equals to $\frac{-(ef-1)^2}{(e+1)^2(f+1)^2}$ so we are done!

Very easy if you have seen similar configurations : WLOG assume $AB>AC$ and we will show $\angle PQF=90^{\circ}$. Let $J$ be the reflection of $D$ about $M$ and let $H$ be the reflection of $D$ about $I$ Claim 1: P lies on FD We claim that in fact P is intersection of the $A$-bisector, $B$-intouch line, and $C$-midline which are known to intersect by the Iranian Lemma. Simple angle chasing shows that this point is indeed the desired point. Claim 2: The points $J$, $P$, $Q$, and $D$ are concyclic We claim in fact that $Q$ is the foot of $D$ onto $AHJ$ which satisfies the requirments and the claim follows. Now we can finish off with easy angle-chasing.

We will assume that $AB < AC$ and show that $\angle PQE = 90^\circ$. Claim. $Q$ lies on $(M)$. Proof. Let $D'$ be the excentral touchpoint; then $Q = \overline{AD'} \cap (I)$ (lower intersection) and follows $\angle DQD' = 90^\circ$, implying $Q$ lies on the circle. $\blacksquare$ Claim. $P = \overline{DE} \cap \overline{AI} \cap \overline{MN}$, where $\overline{MN}$ is the $B$-midline. Proof. Let $P'$ be the aforementioned point (which exists by incenter chord lemma). Then $\angle MP'D = \angle CED = \angle CDE$, so $P'=P$. $\blacksquare$ Let $X = \overline{AD'} \cap (I)$ (higher intersection) be the $D$-antipode. Hence $$\angle PQE = \angle PQA + \angle XQE = \angle XQE + \angle EDC = 90^\circ.$$.

Let $T$ be the $A-$ extouch point and $S$ be the anitpode of $D$ in the incircle. Fro homothety, we know that $A,S,T$ are collinear. Moreover, $MD=MT$. Consider the circle with diameter $\overline{DT}$. Note that $P$ lies on this circle. Since $\angle PQA=90^\circ$, $P$ also lies on this circle. Hence, $D,P,Q,T$ are cyclic. Assume $AB<AC$. Let $N$ be the mid-point of $AB$. From Iran Lemma, we have that $MN,DE,AI$ concur, at point $P'$. We claim that $P'=P$. Note that $\angle P'EC=\angle CDE$ and $MN\parallel BC$, which means $\angle P'EC=\angle DP'M=\angle MDP'$, so $MD=MP'$. Moreover, since $P'=P$. Hence, $PM\parallel BC$, which means $PT\parallel CI$. Hence, \[\angle PTD=\angle PQD=C/2=\angle IDE=\angle SDE=\angle SQE\]It follows that $\angle PQE=\angle DQS=90^\circ$, as required. $\blacksquare$ If the other intersection of $AI$ with circle with diameter $DT$ is taken, call this point $P_1$, then $\angle P_1QF=90^\circ$ holds!

Let $D'$ be the extouch point. It's well known that $\overline{AD'}$ passes through $G$, the antipode of $D$ WRT $(DEF)$, so it follows that $Q$ lies on the circle $\omega$ with diameter $DD'$. Furthermore, since $D'$ is the reflection of $D$ across $M$, it follows that $P$ also lies on $\omega$. Let $R$ be the second intersection of line $AI$ with $\omega$. Consider the $90^{\circ}$ spiral similarity about $Q$ which sends $\omega$ to the incircle. Clearly, this sends $D'$ to $D$ and $D$ to $G$. It will send $\overline{PR}$ to some line parallel to $\overline{EF}$; in fact, it must precisely send $\overline{PR}$ to $\overline{EF}$ since both $QPDR$ and $QEGF$ are harmonic. (In particular, $QPDR$ is harmonic since lines $D'D'$, $PR$ and $DD$ concur at $I$; $QEGF$ is harmonic since lines $EE$, $GQ$ and $FF$ concur at $A$.) So, $\triangle QPR \sim \triangle QEF$ and the angle condition follows.

WLOG diagram as shown. It is well known that $A, D', X$ are collinear where $D'$ is the antipode of $D$ with respect to the incircle and $X$ is the $A$-extouch point. Hence $M$, $I$, and the center of $(AD)$ are collinear by a homothety centered at $D$ with ratio $\frac{1}{2}$. This implies that $M$ is the circumcenter of the cyclic quadrilateral $\square{DPQX}$ because it is well known that $M$ is the midpoint of $XD$. It also implies that $A, D', Q, X$ are collinear. We can show by a phantom point argument that $MP \parallel AB$ implying that $D, P, F$ are collinear by Iran Lemma. Hence: $\angle{PQF} = \angle{PQD'} + \angle{D'QF} = \angle{PDX} + \angle{D'EF} = \angle{PDX} + 90^{\circ} - \angle{DEF} = 90^{\circ}$ by Alternate Segment Theorem, and we are done.

$WLOG$ Let $AB<AC$. We prove $\angle PQE=90°$. We define $D'$ as $D$ antipode in the incircle and $X$ as the $A$ extouch point. It is well known that $\overline{A-D'-Q-X}$ are collinear. $\underline{Claim-1}$: $(DXQP)$ is cyclic with center $M$. $\underline{Proof}$: Since $\Delta DQX$ is right angled at $Q$, $M$ is it's circumcenter as it is the midpoint of hypotenuse and since $MP=MD$, $P$ lies on this circle. $\underline{Claim-2}:$ $P$ lies on $DE$. $\underline{Proof}:$ $\angle EDX=$ $\angle PDX$ $=90 -\frac{C}{2}$ so $\angle DMP= C$ so $ MP || AC$ implying $P$ lies on the midline and hence on $DE$ due to Iran Lemma. Now we prove $\angle PQE$ is right. Note that by alternate segment theorem, $\angle XDQ=\angle DEQ$ and $\angle EPQ = \angle QXD$ so $ \Delta PQE \sim \Delta XQD$ and the required follows.

Let $\omega$ denote the incircle and $\gamma$ the circle with center $M$ and radius $MD$. Denote by $D'$ , $E'$ , $F'$ the $D$ , $E$ and $F-$antipodal points in $\omega$. It is now well known that $Q$ lies on line $\overline{AD'}$ and $\gamma$. Note that in particular, $\overline{IM}$ is the internal $\angle QID-$bisector, which we will use extensively in the proceeding angle chase. Let $P_1=\overline{DE} \cap \overline{QE'}$ and $P_2=\overline{DF} \cap \overline{QF'}$. We start by locating these points with the following claim. Claim : Points $P_1$ and $P_2$ lie on $\gamma$. Proof : This is a direct angle chase. Note that, \begin{align*} \measuredangle DP_2Q &= \measuredangle FP_2F' \\ &= \measuredangle P_2FF' + \measuredangle FF'P_2\\ &= \measuredangle DFF' + \measuredangle FEQ\\ &= \frac{\pi}{2} + \measuredangle DF'F + \measuredangle EFQ + \measuredangle FQE\\ &= \frac{\pi}{2} + \measuredangle DQF + \measuredangle EDQ + \measuredangle FQE\\ &= \frac{\pi}{2} + \measuredangle DEQ \\ &= \frac{\pi}{2} + \measuredangle MIQ\\ &= \measuredangle IMQ \end{align*}which implies $P_2$ lies on $\gamma$. Similarly, one can show that $P_1$ lies on $\gamma$ finishing the proof of the claim. Now, note that since $D'EQF$ is a harmonic quadrilateral, \[(P_1P_2;DQ)\overset{D}{=}(EF;D'Q)=-1\]Thus, $P_1DP_2Q$ is also a harmonic quadrilateral which implies that $I$ , $P_1$ and $P_2$ are collinear. Finally, \begin{align*} \measuredangle(\overline{F'Q};\overline{AI}) &= \measuredangle AQF' + \measuredangle IAQ\\ &= \measuredangle AQF' + \measuredangle IAC + \measuredangle CAQ \\ &= \measuredangle AQF' + \measuredangle EAQ + \measuredangle IAC \\ &= \measuredangle AEF' + \measuredangle EF'Q + \measuredangle IAC\\ &= \measuredangle EFF' + \measuredangle EFQ + \measuredangle F'FE\\ &= \measuredangle EFQ \\ &= \measuredangle P_1DQ\\ &= \measuredangle P_1P_2Q \end{align*}But, this implies that $P_1P_2\parallel AI$. Further, we showed above that these lines clearly have the common point $I$, implying that they are in fact the same line. Thus, $P_1$ and $P_2$ are the intersections of $\overline{AI}$ with $\gamma$. Note that since $P_1$ lies on $\overline{E'Q}$ and $P_2$ lies on $\overline{F'Q}$, \[\measuredangle EQP_1 = \measuredangle EQE' = \frac{\pi}{2} \text{ and } \measuredangle FQP_2 = \measuredangle F'QF = \frac{\pi}{2}\]Now, since $P$ must be one of $P_1$ and $P_2$ (an intersection of $\overline{AI}$ with $\gamma$) one of $\angle PQE = \frac{\pi}{2}$ and $\angle PQF = \frac{\pi}{2}$ must be satisfied as desired.

v_Enhance wrote: Let $ABC$ be a non-isosceles triangle with incenter $I$ whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$, respectively. Denote by $M$ the midpoint of $\overline{BC}$. Let $Q$ be a point on the incircle such that $\angle AQD = 90^{\circ}$. Let $P$ be the point inside the triangle on line $AI$ for which $MD = MP$. Prove that either $\angle PQE = 90^{\circ}$ or $\angle PQF = 90^{\circ}$. Proposed by Evan Chen [asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(9); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.016122787689947, xmax = 5.363206682564721, ymin = -5.305953403874886, ymax = 5.281408081267477; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); draw((-3,4)--(-5,-2)--(4,-2)--cycle, linewidth(0.5)); /* draw figures */ draw((-3,4)--(-5,-2), linewidth(0.5)); draw((-5,-2)--(4,-2), linewidth(0.5)); draw((4,-2)--(-3,4), linewidth(0.5)); draw(circle((-1.9474945684780638,0.20012143404084015), 2.2001214340408404), linewidth(0.5)); draw((0.07259021629667628,-0.6715389280541193)--(-1.9474946214599704,-2), linewidth(0.5) + blue); draw(circle((-0.5,-2), 1.4474946214599704), linewidth(0.5)); draw((-1.5990198250473244,-1.0579829601814568)--(0.07259021629667628,-0.6715389280541193), linewidth(0.5) + ccqqqq); draw((0.07259021629667628,-0.6715389280541193)--(-0.515674565215948,1.8705781987565269), linewidth(0.5) + ccqqqq); draw((0.07259021629667628,-0.6715389280541193)--(-4.034713026635554,0.895860920093337), linewidth(0.5)); draw((-0.5,-2)--(-3.9999999612597397,1.0000001162207814), linewidth(0.5)); draw((-1.9474945154961572,2.4002428680816803)--(-1.9474946214599704,-2), linewidth(0.5) + blue); draw((0.9474946214599704,-2)--(0.07259021629667628,-0.6715389280541193), linewidth(0.5) + linetype("4 4")); draw((-0.515674565215948,1.8705781987565269)--(-1.5990198250473244,-1.0579829601814568), linewidth(0.5) + ccqqqq); draw((-1.5990198250473244,-1.0579829601814568)--(-1.9474946214599704,-2), linewidth(0.5)); draw((-3,4)--(-1.9474945154961572,2.4002428680816803), linewidth(0.5)); draw((-1.9474945154961572,2.4002428680816803)--(0.07259021629667628,-0.6715389280541193), linewidth(0.5) + blue); /* dots and labels */ dot((-3,4),linewidth(3pt) + dotstyle); label("$A$", (-2.945510073494243,4.08650901601519), NE * labelscalefactor); dot((-5,-2),linewidth(3pt) + dotstyle); label("$B$", (-4.946271299033525,-1.9157746606009538), NE * labelscalefactor); dot((4,-2),linewidth(3pt) + dotstyle); label("$C$", (4.057154215893245,-1.9157746606009538), NE * labelscalefactor); dot((-1.9474945684780638,0.20012143404084015),linewidth(3pt) + dotstyle); label("$I$", (-2.1674362635623,0.27950501742069117), NE * labelscalefactor); dot((-1.9474946214599704,-2),linewidth(3pt) + dotstyle); label("$D$", (-2.181330438739656,-2.318705740744167), NE * labelscalefactor); dot((-0.515674565215948,1.8705781987565269),linewidth(3pt) + dotstyle); label("$E$", (-0.4584527167474963,1.9607002138803054), NE * labelscalefactor); dot((-4.034713026635554,0.895860920093337),linewidth(3pt) + dotstyle); label("$F$", (-4.307139240875144,0.7519069734506654), NE * labelscalefactor); dot((-0.5,-2),linewidth(3pt) + dotstyle); label("$M$", (-0.44455854157014024,-1.9157746606009538), NE * labelscalefactor); dot((0.07259021629667628,-0.6715389280541193),linewidth(3pt) + dotstyle); label("$Q$", (0.18067934141088554,-0.7347697705260182), NE * labelscalefactor); dot((-1.5990198250473244,-1.0579829601814568),linewidth(3pt) + dotstyle); label("$P$", (-1.8895527600151776,-1.2349600769106968), NE * labelscalefactor); dot((0.9474946214599704,-2),linewidth(3pt) + dotstyle); label("$X_A$", (1.000435676874897,-1.9157746606009538), NE * labelscalefactor); dot((-3.9999999612597397,1.0000001162207814),linewidth(3pt) + dotstyle); label("$N$", (-4.1681974891015825,1.1548380535938787), NE * labelscalefactor); dot((-1.9474945154961572,2.4002428680816803),linewidth(3pt) + dotstyle); label("$D'$", (-1.8895527600151776,2.4886788706196885), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] If $AC>AB$ then $\angle PQE=90^{\circ}$. Let $N$ be the midpoint of $AB$ so we have $MN, DE, AI$ concur by Iran Lemma. As $\angle AQD=90^{\circ}$ we have $AQ\cap \odot(I)=D'$, where $D'$ is the $D-$antipode, this gives us $AQ\cap BC=X_A$, where $X_A$ is the $A-$Expoint, so $X_A$ lies on the circle $\odot{(M, MD)}.$ Let $P'$ be $AI \cap ED$, by the Iran Lemma we have $MP'$ is parallel to $AC$ so, $\angle MDP'=\angle CED=\angle DP'M \implies MP'=MD\implies P\equiv P' $. Note that $DD'$ is tangent to $\odot(PQX_AD)$ so $\angle EQD'=\angle EDD'=\angle PQD$ and because $DQ\perp AX_A$ we get $\angle PQE=90^{\circ}$. The case for $AB>AC$ is similar to this.

Why is everyone adding the $A$ excircle touch point. You can solve the problem without adding any point here is my solution. WLOG $AC>AB$ Let $N$ be the midpoint of $AB$ Iran Lemma By this lemma we know that lines $AI$, $MN$ and $DE$ concur. Call that point $P'$ Claim: Points $D-P-E$ are collinear Proof: Since $M$and $N$ are the midpoints of the sides $BC$ and $AB$ respectively we know that line $MN$ is the midsegment of the triangle $\triangle ABC \iff MP' \equiv MN \parallel AB \equiv CE \implies MP' \parallel CE.$ So $\angle MDP' \equiv \angle CDE \stackrel{CD=CE}{=} \angle CED \equiv \angle DEC \stackrel{CE \parallel MP'}{=} \angle DP'M \implies \angle MDP'= \angle DP'M \implies$ Triangle $\triangle MP'D$ is isosceles $\iff MP'=MD$. Since both $P$ and $P'$ lie on the line $AI$ $(\because \angle PAB=\angle PAC)$ and $MP'=MD, MP=MD$ we get that $P=P'$ hence Points $D-P-E$ are collinear $\square$ Claim: Points $A$,$F$,$P$ and $Q$ are concyclic Proof: Note that from the incircle we have $\angle DFE+\angle DQE=180 \implies \angle AQE=\frac{\angle C}{2},$ also $\angle FDE=\angle FQE \implies \angle FQA=\frac{\angle B}{2}$. Note that since $P$ lies on the perpendicular bisector of $EF (P \in AI)$ we get that $PE= PF \iff \angle FPA \equiv \angle FPI =\angle APE$. Now from triangle $\triangle APE$ we get that $\angle APE=\frac{\angle B}{2} \implies \angle FPA=\angle APE=\frac{\angle B}{2}=\angle FQA \implies \angle FPA=\angle FQA \implies$ Points $A$,$F$,$P$ and $Q$ are concyclic $\square$ Claim: $\angle PQE=90$ Proof: By previous claim we have: $\angle FQP=\angle FAP \equiv \angle BAI \equiv \frac{\angle A}{2} \implies \angle FQP=\frac{\angle A}{2}$. So $\angle PQE=\angle PQF+\angle FQA+\angle AQE=\frac{\angle A}{2}+\frac{\angle B}{2}+\frac{\angle C}{2}=90 \implies \angle PQE=90$ $\blacksquare$

Let $AC>AB$ and assume $P$ to be the point inside the triangle $ABC$. Let $N$ denote the midpoint of $AB$. Claim:$P=MN \cap DE$. Due to Iran Lemma $MN, DE, AI$ are concurrent at point $P'$. Note that: $\triangle DP'M \sim \triangle CEC$ which implies $MP'=MD$. Thus: $P=P'$ Let $S$ denote the antipode of $D$ in the incircle of $\triangle ABC$ with $T$ being tangency point of $A$-excircle touch with $BC$. Claim: $S, Q, T$ are collinear. Let $ST$ and incircle intersect at $Q'$. Note that: $A, S, T$ are collinear and $\angle A Q'T = \angle SQ'T = 90^\circ$. This implies $Q'=Q$ and we are done. Now, we finish the problem by showing $\angle DQP = \angle SQE$ (as $\angle DQS = 90^\circ$): Notice that: $$\angle PQD = \frac{\angle PMD}{2} = \frac{\angle BCA}{2} = 90^\circ - \angle DFE = 90^\circ - \angle DSE = \angle SQE$$and thus, we are done.