Let $M$ be the midpoint of $AB$, so $CQPDM$ is cyclic with diameter $CD$. Note $MP=MQ$, since $CM$ bisects $\angle PCQ$. Furthermore, the problem condition says $PQFC$ is an isosceles trapezoid. So $MC=MF$, and thus since $M$ is the intersection of the perpendicular bisectors of $CF$ and $AB$, it's the circumcenter.

Solution.Let $O'$ be the midpoint of $AB$, we want to prove that $O'$ is the circumcenter of $(ABC)$. Since, $\angle{CPD}=\angle{CQD}=90^{\circ}\implies CQDP$ is cyclic. Also $\angle{CPD}=\angle{CO'D}=90^{\circ}\implies O'\in (CQDP)$. Also, $CO'$ is the angle bisector of $\angle QCP\implies O'Q=O'P\implies O'$ lies on the perpendicular bisector of $QP$. $F$ lies on the perpendicular bisector of $QP\implies FQ=FP\implies \angle EPQ=\angle CQP\implies EQ=CP$, also $CEQP$ is cyclic$\implies CEQP$ is isosceles trapezium$\implies $ the perpendicular bisectors of $QP$ and $CE$ are the same. But now we have that $O'$ lies on the perpendicular bisector of $AB$ and $CE\implies O'$ is the circumcenter of $(ABC)\implies\angle{ACB}=90^{\circ}.\blacksquare$ [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 = -19.285561891374865, xmax = 13.45609814927643, ymin = -6.738164896172865, ymax = 9.206341425706839; /* image dimensions */ pen qqwwtt = rgb(0,0.4,0.2); pen wwqqcc = rgb(0.4,0,0.8); /* draw figures */ draw(circle((-3.4321388196727844,0.35215619082622873), 4.360283163713907), linewidth(0.8)); draw((-4.0111016113502895,4.673830791443441)--(-7.754326180763476,-0.22296612954991374), linewidth(0.8)); draw((-4.0111016113502895,4.673830791443441)--(0.8890196080282163,0.9349589966495839), linewidth(0.8)); draw((0.8890196080282163,0.9349589966495839)--(-7.754326180763476,-0.22296612954991374), linewidth(0.8)); draw((-5.0682220039537045,0.1368838167464339)--(-2.4912854346628954,3.5141864516634858), linewidth(0.8)); draw((-5.0682220039537045,0.1368838167464339)--(-6.5900055575547745,1.3001702832010966), linewidth(0.8)); draw((-2.4912854346628954,3.5141864516634858)--(-6.5900055575547745,1.3001702832010966), linewidth(0.8) + red); draw(circle((-4.5396618076519974,2.4053573040949376), 2.329237610016973), linewidth(0.8) + qqwwtt); draw((-2.4912854346628954,3.5141864516634858)--(-6.725663595332122,3.2095612758439365), linewidth(0.8) + wwqqcc); draw((-4.0111016113502895,4.673830791443441)--(-3.4321388196727844,0.35215619082622873), linewidth(0.8)); draw((-5.038808964417505,3.3294085937202906)--(-3.4321388196727844,0.35215619082622873), linewidth(0.8)); /* dots and labels */ dot((-4.0111016113502895,4.673830791443441),dotstyle); label("$C$", (-3.9592249322418693,4.83652351663553), NE * labelscalefactor); dot((-7.754326180763476,-0.22296612954991374),dotstyle); label("$A$", (-8.137197177012478,-0.4712065290169389), NE * labelscalefactor); dot((0.8890196080282163,0.9349589966495839),linewidth(4pt) + dotstyle); label("$B$", (0.9648137848092042,1.0635587853885942), NE * labelscalefactor); dot((-5.0682220039537045,0.1368838167464339),dotstyle); label("$D$", (-5.238196027579811,-0.4712065290169389), NE * labelscalefactor); dot((-6.5900055575547745,1.3001702832010966),linewidth(4pt) + dotstyle); label("$Q$", (-7.050071745975227,1.2340882647669866), NE * labelscalefactor); dot((-2.4912854346628954,3.5141864516634858),linewidth(4pt) + dotstyle); label("$P$", (-2.403143432914041,3.6854495308313795), NE * labelscalefactor); dot((-5.038808964417505,3.3294085937202906),linewidth(4pt) + dotstyle); label("$E$", (-5.323460767269006,3.4509714966860896), NE * labelscalefactor); dot((-6.725663595332122,3.2095612758439365),linewidth(4pt) + dotstyle); label("$F$", (-7.135336485664423,3.2804420173076974), NE * labelscalefactor); dot((-3.4321388196727844,0.35215619082622873),linewidth(4pt) + dotstyle); label("$O'$", (-3.490268863951291,-0.08751520041555562), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Let $M$ be the midpoint of $AB$ and $l$ the perpendicular bisector of $PQ$ Claim 1. $M$ lies on $l$ proof. Since $CM \bot AB$, $CQDMP$ is cyclic. Clearly $CM$ bisects $\angle QCP$. Hence, $MQ = MP$. $\blacksquare$ Claim 2. $l$ is also the perpendicular bisector of $CF$ proof. $$\angle FCQ = \angle FPQ = \angle CQP \rightarrow CF || PQ$$Hence, PCFQ is an isosceles trapezium and proves the claim. $\blacksquare$ So, $M$ lies on the perpendicular bisector of both $AB$ and $CF$ which are both chords of $(ABC)$, meaning $M$ is its center and $AB$ is a diameter. This proves $\angle ACB = 90^{\circ}$.

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{PQ}$ respectively. Since $\angle CMD=90^\circ$, and $\angle CPD=\angle CQD=90^\circ$, we have $C,P,M,D,Q,F$ are concyclic, in particular all on $(CD)$. Lemma 1: $PM=QM$. Proof: We have \begin{align*} \angle PQM &= \angle PDM = \angle PDB = 90^\circ- \angle CBA \\ \angle QPM &= 180^\circ-\angle QDM = \angle QDA = 90^\circ - \angle CAB, \end{align*}which are equal since $\triangle CAB$ is isosceles. $\blacksquare$ Lemma 2: $CFQP$ is an isosceles trapezoid. Proof: Since $P,E,F$ are collinear, we have \[ \angle CQP=\angle EQP=\angle EPQ = \angle FPQ = \angle FCQ,\]so $\overline{CF}\parallel \overline{PQ}$, proving the claim. $\blacksquare$ By Claim 2, the perpendicular bisector of $\overline{CF}$ and $\overline{PQ}$ are the same, so by Claim 1, $CM=FM$. The circumcenter of $\triangle ABC$ is the intersection of the perpendicular bisectors of $\overline{CF}$ and $\overline{AB}$, which is $M$. Therefore, $\overline{AB}$ is a diameter of $(ABC)$, so $\angle C=90^\circ$.

[asy][asy] size(8cm); 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 = -8.92, xmax = 8.92, ymin = -7.52, ymax = 2.96; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((-0.48,-2.03), 2.7600181158825747), linewidth(0.8) + ffvvqq); draw((-3.24,-2.02)--(2.28,-2.04), linewidth(0.8)); draw((2.28,-2.04)--(-0.47,0.73), linewidth(0.8)); draw((-3.24,-2.02)--(-0.47,0.73), linewidth(0.8)); draw(circle((-1.1650090578521075,-0.6474999671816959), 1.5429010824035494), linewidth(0.8) + ffvvqq); draw((-2.6820174749235983,-0.36602312512129576)--(0.21749097496619646,0.037509090670412214), linewidth(0.8) + linetype("4 4") + blue); /* dots and labels */ dot((-3.24,-2.02),dotstyle); label("$A$", (-3.8,-1.78), NE * labelscalefactor); dot((2.28,-2.04),dotstyle); label("$B$", (2.6,-1.86), NE * labelscalefactor); dot((-0.48,-2.03),linewidth(4pt) + dotstyle); label("$M$", (-0.4,-1.88), NE * labelscalefactor); dot((-0.47,0.73),linewidth(4pt) + dotstyle); label("$C$", (-0.74,1.1), NE * labelscalefactor); dot((-1.8600181157042153,-2.0249999343633904),dotstyle); label("$D$", (-1.78,-1.82), NE * labelscalefactor); dot((0.21749097496619646,0.037509090670412214),linewidth(4pt) + dotstyle); label("$P$", (0.3,0.2), NE * labelscalefactor); dot((-2.5475090906704123,-1.3325090250338023),linewidth(4pt) + dotstyle); label("$Q$", (-2.46,-1.18), NE * labelscalefactor); dot((-1.3933414138938967,-0.18667468888383276),linewidth(4pt) + dotstyle); label("$E$", (-1.32,-0.02), NE * labelscalefactor); dot((-2.6820174749235983,-0.36602312512129576),linewidth(4pt) + dotstyle); label("$F$", (-2.6,-0.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $M$ be the midpoint of $AB$, obviously $D,M,P,C,F,Q$ are concyclic. Hence $$\angle FCA=\angle FCQ=\angle FPQ=\angle EPA=\angle EQP=90^{\circ}-\angle PQD=90^{\circ}-\angle DCB$$Therefore, $$\angle CFM=\angle CDM=90^{\circ}-\angle DCM=\angle ACM+\angle ACF=\angle FCM$$Hence $CM=FM$, but $MA=MB$, and $C,F,A,B$ are concyclic, hence $M$ is the center of the circle, which implies $\angle ACB=90^{\circ}$.

mathaddiction wrote: $$\angle FPQ=\angle EPA$$ Why is this true? Quote: $$\angle CDM=90^{\circ}-\angle DCM$$ How do we know CMD=90? Quote: $$90^{\circ}-\angle DCM=\angle ACM+\angle ACF$$ pls explain

Let $\angle CBA = \theta$. Claim 01. $CP = FQ$ and $CQ = FP$. Proof. $CF \parallel PQ$ because $\measuredangle CQP = \measuredangle EQP = \measuredangle QPE = \measuredangle QPF = \measuredangle QCF$. Therefore, $CPFQ$ is an isosceles trapezoid, forcing $CP = FQ$ and $CQ = FP$. Claim 02. $BP = CQ$ and $CP = QA$. Proof. Consider the spiral similarity $F : PQ \mapsto BA$. Therefore, \[ \frac{PB}{QA} = \frac{FP}{FQ} = \frac{CQ}{CP} \implies CP \cdot PB = CQ \cdot QA \]Since $BC = CA$, then $BP + PC = QC + QA$. Consider the quadratic equation with $\{ CP, BP \}$ (this denotes length) as a root. Then, \[ (x - BP)(x - CP) = x^2 - (BP + PC)x + (BP \cdot PC) = x^2 - (CQ + QA)x + (CQ \cdot QA) = (x - CQ)(x - QA) \]Thus, $\{ CP, PB \} = \{ QC, QA \}$. However, $CP \not= CQ$ since this would force $D$ to be the midpoint of $AB$ as $DPCQ$ cyclic. Therefore, $CP = QA$ and $CQ = PB$. To finish this, let $BP = CQ = x$ and $CP = AQ = y$. Then, $AB = (x + y) \cdot \cos \theta \cdot 2$ and $AB = BD + DA = \frac{y}{\cos \theta} + \frac{x}{\cos \theta}$. Therefore, $\cos^2 \theta = \frac{1}{2}$, which implies $\theta = 45^{\circ}$ as we know $\theta < 90^{\circ}$. This is enough to force $\angle ACB = 90^{\circ}$.

[asy][asy] size(250); pair A = origin, B = (8, 0), C = (4, 4); pair D = (5/2, 0); pair Q = foot(D, A, C), P = foot(D, B, C); pair EE = extension((A+B)/2, (P+Q)/2, A, C); pair X [] = intersectionpoints(circumcircle(A, B, C), circumcircle(C, P, Q)); pair F = X[1]; draw(A--B--C--cycle, red); draw(Q--D--P, magenta); draw(Q--F--C--P, green); draw(F--EE--P, lightcyan+dashed); draw(circumcircle(A, B, C), orange); draw(circumcircle(C, P, Q), gray); draw(C--D, brown+dashed); draw(P--Q, brown+dashed); pair X3 [] = intersectionpoints(C--D, P--Q); pair R = X3[0]; dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$D$", D, S); dot("$Q$", Q, NW); dot("$P$", P, NE); dot("$F$", F, NW); dot("$E$", EE, SE); dot("$R$", R, SE); [/asy][/asy] First, observe that $P, E, F$ being collinear is equivalent to $CFQP$ being a trapezoid. This is just angle chasing: $$\measuredangle FPQ = \measuredangle EPQ = \measuredangle PQE = \measuredangle PQC,$$from where $CFQP$ is an isosceles trapezoid because it is cyclic. Next, we claim that this collinearity also implies that $$\triangle FQA \sim \triangle FPB.$$This is again angle chasing: we have $$\measuredangle AQF = \measuredangle CQF = \measuredangle CPF = \measuredangle BPF,$$and also $$\measuredangle FAQ = \measuredangle FAC = \measuredangle FBC = \measuredangle FBP.$$In particular, this similarity implies that $$\frac{FP}{PB} = \frac{FQ}{QA} \implies \frac{CQ}{PB} = \frac{CP}{QA} \implies \frac{CQ}{CP } =\frac{PB}{QA}$$From here, we present two finishes: Finish 1 (Synthetic) Observe that $$\frac{PB}{QA} = \frac{PD}{QD}$$because $\triangle AQD \sim \triangle BPD$. We use quadrilateral $CQDP$, which is trivially cyclic. Let the diagonals of the quadrilateral intersect at $R$; then the length condition implies $$\frac{CQ}{PD} = \frac{QR}{PR} \text{ and } \frac{CP}{QD} = \frac{PR}{QR}.$$Combining these two conditions, we obtain $QR=PR$; similarly, $CR=RD$ so $\overline{PQ}$ is a diameter of $(APQ)$. This implies the result. $\square$ Finish 2 (Trigonometric) Let $\angle CAB = \angle CBA = \theta$, and moreover let $AC=BC=x$. We have $$\frac{PD}{QD} = \frac{BD}{AD}$$by similar triangles, while we can write $$CQ=x-AQ = x-AD \cos \theta.$$Therefore, we have $$\frac{BD}{AD} = \frac{x-AD \cos \theta}{x - BD \cos \theta}.$$This expands to $$BDx - BD^2 \cos \theta = ADx - AD^2 \cos \theta \implies (BD-AD)x -\cos \theta(BD^2-AD^2) = 0.$$Factoring, we can write this as $$(BD-AD)(x - \cos \theta (BD+AD)) = 0;$$but $BD+AD = \frac{x\sin(2\theta)}{\sin \theta}$ by the Law of Sines, so $$x(BD-AD)\left(1 - \frac{\sin(2\theta)\cos \theta}{\sin \theta}\right) = 0,$$which implies that $$\sin(2\theta) \cos(\theta) = \sin \theta \implies 2\sin \theta \cos^2 \theta = \sin \theta \implies \sin \theta(1-2\cos^2 \theta)=0.$$This tells us that $\cos^2 \theta = \frac 12$, and since $\theta$ lies in the first quadrant, $\cos \theta = \frac{\sqrt 2}2$, implying $\theta = 45^{\circ}$. This finishes the problem. $\square$ Trigonometry is fun.

Let $M$ denote the midpoint of $AB$, meaning $CM \perp AB$. Since $PDQC$ is cyclic, we haver $\angle CPD = \angle CMD = 90$, thus $M$ lies on the circumcircle of $PQDC$, by inscribed angles. Note, this implies $\angle MCQ = \angle MCP$ as the angle bisector is the perpendicular bisector in an isosceles triangle, so the measure of arcs $MQ$ and $MP$ are equal, implying $MQ = MP$. Then, note that $\angle FPQ = \angle CPQ$ so $FQPC$ is cyclic and for similar reasoning that $MQ = MP$ we have $FQ = PC$, which means $FQPC$ is an isosceles trapezoid or $CF \parallel PQ$. Thus, $M$ lies on the perpendicular bisectors of $CF$ and $AB$ and because they are both chords, the intersection of the perpendicular bisectors of chords is the cirumcenter by a famous lemma (Problem 1.40 of EGMO), and we are done. $\blacksquare$

The collinearity implies that $PCFQ$ is an isosceles trapezoid. Claim: $\triangle FQA$ and $\triangle FPB$ are isosceles. Proof. Define $\angle CAB=\angle ABC=\theta$ and $\tfrac{1}{2}\widehat{FQ}=\tfrac{1}{2}\widehat{CP}=\alpha$. Since $\angle QFP=\angle QFC=\angle ACB=180^\circ-2\theta$ and $\angle EAP=\angle EPQ=\alpha$, we find $\angle FQE=2\theta-2\alpha$. Now, observe that \begin{align*} \angle FAQ&=\angle FAC=\angle FAB-\angle CAB=180^\circ-\angle FCD-\angle CAB \\ &=180^\circ-(\alpha+(180^\circ-2\theta))-\theta=\theta-\alpha. \end{align*}Since this equals one-half of $\angle AQF$'s supplement, the first claim is true. The second claim follows as the two triangles are similar. Now, to finish, observe that \begin{align*} \frac{CD}{QC}=\frac{FQ}{FD}=\frac{QA}{DB}=\frac{QD}{DP}. \end{align*}This implies $QDPC$ is a rectangle, so we are done. edit: dear god, how in the world did i miss perpendicular bisector...

samrocksnature wrote: mathaddiction wrote: $$\angle FPQ=\angle EPA$$ Why is this true? Quote: $$\angle CDM=90^{\circ}-\angle DCM$$ How do we know CMD=90? Quote: $$90^{\circ}-\angle DCM=\angle ACM+\angle ACF$$ pls explain To the first one, he probably meant to write $\angle EPQ$, to the second it's because the median is a perpendicular in an isosceles triangle. To the third, it's a bit more complicated (there's probably an easier way but meh): First, note that $\angle ACF + \angle ACM = \angle FCM$. Also, $90 - \angle DCM = \angle CDM$. Since $CM$ is an angle bisector, we have that $\angle QCM = \angle PCM$, so we can subtract off those arcs. So, all that remains is to prove that $\angle FCQ = \angle CQP$, which is true because $FCPQ$ is an isosceles trapezoid.

Let $M$ is the midpoint of $AB$ and $N$ is the midpoint of $PQ$. Since, $EQ=EP$ and $\angle PCQ=\angle QFP$, so $FCQP$ is a cyclic trapezoid. We have points $D,Q,C,P$ are concyclic. Claim 1: Points $M,N,E$ are collinear. Proof:Let the circumcircle of $\triangle CPQ$ intersects side $AB$ again at $M^\prime$.Since $\angle CM^\prime D=90^\circ$, so $M^\prime$ is the midpoint of $AB$ and also the midpoint of arc $PQ$ not containing $C$, which implies $M^\prime$ lies on the perpendicular bisector of $PQ$. So, points $M,N,E$ are collinear. Claim 2: $F$ is the center of spiral similarity taking $QA$ to $CB$. Proof: Trivial. Claim 3: Points $E,Q,N,F$ are concyclic. Proof: $\angle ENF=180^\circ-\angle FNM=180^\circ-\angle FQA=\angle EQF$ which implies points $E,Q,N,F$ are concyclic. So, $\angle ACB=\angle QFP=180^\circ-\angle EFQ=180^\circ-\angle ENQ=90^\circ$

bruh Let $M$ be the midpoint of $\overline{AB}$, so that points $C$, $D$, $F$, $P$, $Q$, $M$ lie on the circle with diameter $\overline{CD}$. Note that $MP = MQ$. By symmetry, $CPQF$ is an isosceles trapezoid, so $MC = MF$. Since $ABCF$ is cyclic, $MA=MB$, and $MC=MF$, we conclude the $M$ is the circumcenter.

rocketsri wrote: samrocksnature wrote: mathaddiction wrote: $$\angle FPQ=\angle EPA$$ Why is this true? Quote: $$\angle CDM=90^{\circ}-\angle DCM$$ How do we know CMD=90? Quote: $$90^{\circ}-\angle DCM=\angle ACM+\angle ACF$$ pls explain To the first one, he probably meant to write $\angle EPQ$, to the second it's because the median is a perpendicular in an isosceles triangle. To the third, it's a bit more complicated (there's probably an easier way but meh): First, note that $\angle ACF + \angle ACM = \angle FCM$. Also, $90 - \angle DCM = \angle CDM$. Since $CM$ is an angle bisector, we have that $\angle QCM = \angle PCM$, so we can subtract off those arcs. So, all that remains is to prove that $\angle FCQ = \angle CQP$, which is true because $FCPQ$ is an isosceles trapezoid. That's right. For the third point notice that from the first equality we have $$90^{\circ}-\angle DCM=90^{\circ}-\angle DCB+\angle BCM=\angle FCA+\angle BCM=\angle FCA+\angle ACM$$Sorry for making a lot of typos in my solutions

Quick. Let $X$ be the midpoint of $PQ$. Observe that $$\frac{FQ}{FP}=\frac{QA}{PB}=\frac{DQ}{DP}$$Thus $(FD ; QP)=-1$. It's well-known that $CD$ must be the $C$-median in $\triangle CPQ$ and so $X\in CD$. Consequently $X$ is the circumcenter of $\triangle CPQ$ and we are done. $\mathcal{Q.E.D.}$

Not-so elegant algebraic calculation, sorry for my bad Latex. First, it's easy to deduce that $CPQF$ is an isosceles trapezium. Moreover, $ABCF$ is cyclic hence angles $CQP$ and $FBA$ are equal. As $AFB$ and $CQD$ are both right angles hence $BAF=DCP$ (angles). Notice that $BAF= 180- ACB- CQP$ hence q.e.d

Define N = midpoint of PQ and M = midpoint of AB. Claim: M lies on (CPQ) and M lies on NE. Proof: $\angle{CQD} = \angle{CPD}$ so D $\in$ (CPQ). Now $\angle{CMD} = 90^{o} = \angle{CPD}$ so M $\in$ (CPQ). Now as $\angle{QCM} = \angle{PCM}$ we have $\hat{PM} = \hat{QM}$ which gives M $\in$ NE, as required. Claim: $\triangle CEP \cong \triangle FEQ$ Proof: We have $\triangle CEP \sim \triangle FEQ$, since $CPQF$ is cyclic and $E$ is the intersection point of the diagonals. Since $EP=EQ$, we have the required congruence. Claim: CF $\parallel$ PQ Proof: $\angle{CFP} = \angle{CQP} = \angle{EQP} = \angle{EPQ} = \angle{FPQ}$, and the claim follows. Since a trapezium inscribed in a circle is isosceles, CPQF is an isosceles trapezium and so CF and PQ have the same perpendicular bisector which is NE. Since the perpendicular bisector of CF contains circumcenter of $\triangle ABC$, and the circumcenter also lies on $CM$, The circumcenter of $\triangle ABC$ is $M$. Thus, the $\triangle$ ABC is right angled.

I'm on mobile, so I can't post full solution, but there are steps to solve problem with 10 minute's angle chasing, that I did during TST. Step $1$: Show that $QF=QA$. Step $2$: Show that $\angle FQA=2 \angle FDA$. From Step $1$and Step $2$ we get $Q$ is circumcentre of $\triangle AFD$ $\implies$ $\angle A=45$. So we are done.

oh i didn’t say EQP is isosceles

Note that $CPQF$ is an isoceles trapezoid. Since $F$ is the Miquel point of $ABPQ$, \[\frac{AQ}{BP}=\frac{FQ}{FP}=\frac{CP}{CQ}\]hence $AQ\cdot CQ = BP \cdot CP$, so $P$ and $Q$ have the same power with respect to $(ABC)$. This implies that the circumcenter $O$ of $(ABC)$ is on the perpendicular bisector of $PQ$. However, an angle chase gives us that the midpoint $M$ of $AB$ is on $(CPDQ)$ and is the midpoint of arc $PQ$. Hence, if $O\ne M$, $OM$ is the perpendicular bisector of both $AB$ and $PQ$, implying that $AD = DB$, a contradiction.

ez complex

Let $M$ be the midpoint of $AB$. Note that $\omega = (CPQDFM)$ with diameter $CD$. As $CM$ bisects $\angle ACB$, it bisects $\angle QCP$ and hence is the midpoint of arc $PQ$ of $\omega$ not containing $C$. As $EP=EQ$ and $P,E,F$ collinear, we have that $E$ and $C$ are reflections across the perpendicular bisector of $PQ$, which is $EM$. Hence $FM=CM$. Now $M$ is the intersection of perpendicular bisectors of $CF$ and $AB$ of $(ABCF)$, so it must be the center of $(ABCF)$. It follows that $\angle ACB = 90^{\circ}$.

Note that since $$\measuredangle FPQ = \measuredangle EPQ = \measuredangle PQE = \measuredangle PQC,$$quadrilateral $CPQF$ is an isosceles trapezoid. Claim: $AQ = FQ$. Proof: Just angle chase: $$\begin{aligned} \measuredangle AFQ &= \measuredangle AQF + \measuredangle FAQ \\ &= \measuredangle CQF + \measuredangle FAC \\ &= \measuredangle CDF + \measuredangle FAC \\ &= (\measuredangle PDB + \measuredangle ADQ) + (\measuredangle CDP + \measuredangle QDF) + \measuredangle FAC \\ &= 2 \cdot \measuredangle CBA + 2 \cdot \measuredangle QCF + \measuredangle FAC \\ &= 2 \cdot (\measuredangle CBA + \measuredangle ABF) + \measuredangle FBC \\ &= \measuredangle CBF \\ &= \measuredangle QAF, \end{aligned}$$which gives what we wanted. Now, note that we also have $FQ = CP$ from isosceles trapezoid $CPQF$, so $AQ = CP$. Thus, $QC = PB$ as well. To finish, denote $\alpha = \angle CAB = \angle ABC$, and let $AD = x, BD = y$. Straightforward computations gives $AQ = x \cos \alpha$ and $QC = PB = y \cos \alpha$, as well as $AC = \tfrac{x + y}{2 \cos \alpha}$. Then since $AQ + QC = AC$ we have $\tfrac{1}{2 \cos \alpha} = \cos \alpha$ or $\cos \alpha = \tfrac{\sqrt{2}}{2}$, so $\alpha = 45^{\circ}$ and $\angle ACB = 90^{\circ}$, as desired.

OK, a slightly harder G1 ig. We start off with the following observation. Claim : Points $C,P,M,D,Q,F$ lie on the same circle with $CPQF$ infact being an isoceles trapezoid. Proof : Note that $D$ clearly lies on $(CPQ)$ due to the right angles. Further, $CM \perp AB$ which gives $M$ also on $(CPQ$. Further, we are given that $F$ also lies on this circle. Now, note that \[\measuredangle QCF = \measuredangle QPF = \measuredangle QPE = \measuredangle EQP = \measuredangle CQP\]Thus, $CF \parallel PQ$ which means $CPQF$ is indeed an isosceles trapezoid. Let $O$ and $O_1$ be the centers of $(ABC)$ and $(CPQ)$). Let $M$ be the midpoint of $AB$ and $M_1$ the midpoint of $PQ$. Now, we have the following key claim. Claim : Points $E,O,O_1,M$ and $M_1$ lie on the same line. Proof : Now, note that by the isosceles trapezoid, $CF\parallel PQ$ and we have $E-O_1-O$. Clearly, the midpoint of $PQ$ also lies on its perpendicular bisector and thus $E-O_1-M_1-O$. Now, also note that \[\measuredangle MPQ = \measuredangle MDQ = \measuredangle ADQ = \measuredangle PDB = \measuredangle PDM = \measuredangle PQM\]Thus, $MP=MQ$. This means $M$ also lies on the perpendicular bisector of $PQ$. Thus, $E-O_1-M_1-O-M$. But, clearly $O, M$ also lie on the perpendicular bisector of $AB$. This means both points $O$ and $M$ lie on the perpendicular bisector of $PQ$ and the perpendicular bisector of $AB$. Since two distinct lines can intersect at most 1 point we have two possibilities. Either, $EO_1$ and $OM$ are the same line. But then, $E=C$ which is a degenerate case. Or, we must have the two given intersection points being the same or $O=M$. This happens if and only if $\angle ACB=90^\circ$ which was the desired conclusion.

By angle chasing,we get $PB=PF$, and since $CFPQ$ is an isosceles trapezoid $PF=CQ$,so $PC=QA$.$PC*PB=CQ*QA$,so $OP=OQ$.This means $O$ and $O1$ are on the perpendicular bisector of $PQ$,and since $O1$ is also on the line $CD$ we get $PQ$ is diameter which gives us $\angle{PCQ}=90$.

Note the spiral similarity from $F$ mapping $\overline{PQ} \mapsto \overline{BA}$. Then from some simple angle chasing we may conclude, \begin{align*} \angle FCA &= \angle FBA\\ &= \angle FPQ\\ &= \angle EQP \end{align*}Then $\overline{PQ} \parallel \overline{CF}$ and $FCPQ$ is an isosceles trapezoid. Now let $M$ be the midpoint of $\overline{AB}$ which lies on $(FCPQD)$ as $ABC$ is isosceles. Now we claim $MP = MQ$ which is visible after some simple trig outlined below, \begin{align*} MQ &= \frac{AM \cdot \sin A }{\sin(\angle AQM)}\\ &= \frac{BC \cdot \sin B}{\sin(\angle AQD + \angle DQM)}\\ &= \frac{BC \cdot \sin B}{\sin(\angle BPD + \angle DPM)}\\ &= \frac{BC \cdot \sin B}{\sin(\angle MPB)}\\ &= MP \end{align*}Thus $\overline{EM}$ is the perpendicular bisector of $\overline{PQ}$. However the $M$ is the intersection of two of the perpendicular bisectors of the cyclic quad $ABCF$ namely of $\overline{FC}$ and $\overline{AB}$ so $M$ is the center and thus $\angle ACB = 90$. $\square$

Let $G$ be the midpoint of $AB$. Note that $CPQD$ is cyclic. Claim: $G$ lies on $(CPQ)$ Proof: We angle chase: \[\angle CGD=90=\angle CQD\]$\square$ Claim: $GE$ is the perpendicular bisector of $PQ$ Proof: Using LoS, we can say: \[GQ=\frac{\sin{\angle CAB}}{\sin{\angle AQG}}\cdot GA,\]\[GP=\frac{\sin{\angle CBA}}{\sin{\angle GPC}}\cdot BG.\]By simple-angle chasing, the RHS of the above equations are the same, so $GP=GQ$, as desired $\square$ Claim: $FCPQ$ is an isosceles trapezoid Proof: It is clear that: \[\angle EPQ=\angle EQP,\]so we are done by the fact that $FCPQ$ is cyclic $\square$ Therefore, $FC\parallel PQ$, so $GE$ is the perpendicular bisector of $FC$. Therefore, since $G$ is the midpoint of $AB$, $G$ is the circumcenter of $(ABC)$, so $\angle ACB=90$ $\blacksquare$.

Denote by $M$ the midpoint of $AB$, and by $O$ the centre of the circumcircle of $\Delta ABC$. Claim 1: $MP = MQ$. Proof: Notice that $CQDMP$ is a cyclic polygon; as such, $\sin CPM = \sin BPM$. But now the Law of Sines on triangles $AQM$, $BPM$ finishes. $\square$ Claim 2: $CF \parallel PQ$. Proof: Since $P, E, F$ are collinear, $\angle CFP = \angle CQP = \angle FPQ$ (This last part is where we use $P, E, F$ collinear). $\square$ Note that due to Claim 2, for any point $X$, $XC = XF \iff XP = XQ$. But now $OC = OF \implies OP = OQ$. Also $OA = OB$. But note that the conditions $XA = XB, XP = XQ$ are both satisfied by $O, M$. So we break into two cases: Case 1: $PQ \parallel AB$ (i.e. the perpendicular bisectors of $AB$, $PQ$ coincide) This isn't possible, as it would imply $AQ = BP$, but $AQ/AD = BP/BD \implies AD = DB$, contradiction. Case 2: The aforementioned perpendicular bisectors do not coincide. Then we get that they meet at either no point or at a unique point. But now $O, M$ lie on both, so we must have $O = M$, i. e. The midpoint of $AB$ is the circumcentre of $\Delta ABC$, giving us $\angle ACB = 90^\circ$. $\square$

Let $M$ be the midpoint of $AB$. $BC=CA=> \angle CMD = \angle CMA = 90 = \angle DPB = \angle DQA$, hence $C,P,M,D,Q$ are concylic in a circle with diameter $CD$. Hence, $F$ also lies on this circle. From the collinearity of $P,E,F$ we get (using $EQ=EP$) that: $\angle CFP = \angle CQP = \angle EQP = \angle EPQ = \angle FPQ => CF \parallel PQ$ and so $FCPQ$ is an isosceles trapezoid, as it is cyclic with $CF \parallel PQ$, and hence $PQ,CF$ have the same perpendicular bisector. Now we have that: $BC=CA=> \angle ACM = \angle BCM => \angle QCM = \angle PCM => MQ=MP$, as $QMPC$ is cyclic. Using that $PQ,CD$ have the same perpendicular bisector, we get that $MC=MF$. Hence, $M$ is the point where the perpendicular bisectors of $AB,CF$ meet, so it is the center of the circle $(AFCB)$, from where we conclude that $\angle ACB =90$, as needed.

After reading everyone else's solutions I can't help but think I made some error... Well it doesn't hurt to post... We use directed angles $\measuredangle$ modulo $\pi$. Let $M$ be the midpoint of $\overline{PQ}$. Make the observation that $CPDQ$ is cyclic as $\measuredangle CQD = \measuredangle CPD = 90^{\circ}$. Now observe $\triangle QEM \sim \triangle DCP.$ Indeed, we have that $\measuredangle MQE = \measuredangle PQC = \measuredangle PDC,$ and $\measuredangle EMQ = \measuredangle CPD = 90^{\circ}$. We make the following claim: Claim: $ECPM$ is cyclic. Proof: We find that by angle chasing \[\measuredangle QEM = \measuredangle DCP = \measuredangle MCP. \]However, $\measuredangle QEM = \measuredangle MEP,$ as $EM$ is the perpendicular bisector of $\overline{PQ}$. $\square$ Hence we finish the problem by a final angle chase: \begin{align*} \measuredangle QDP &= \measuredangle QDC + \measuredangle CDP \\ &= \measuredangle QPC + \measuredangle CQP \\ &= \measuredangle QEM + \measuredangle EQM \\ &= 90^{\circ}. \end{align*}Therefore $\measuredangle QDP = \measuredangle QCP = \measuredangle ACB = 90^{\circ}$, as desired. $\blacksquare$

Since $C,E,Q$ and $P,E,F$ are collinear, we see that $CF$ is parallel to $PQ$. Since $CF$ is the radical axis of $(ABC), (CPQ)$, by LinPoP we know that the power difference is the same at $P,Q$, and since $P,Q$ lie on $(CPQ)$ we just get that $AQ \cdot QC = CP \cdot PB$. Let $M$ be the midpoint of $AB$, then let $DM = x$. Compute $AM = AC \cos BAC, AD = AC \cos BAC - x$, symmetrically for $BM,BD$. We then have that $(AC \cos BAC - x) (\cos BAC) (AC - (AC \cos BAC - x) \cos BAC) = (AC \cos BAC + x) (\cos BAC) (AC - (AC \cos BAC + x) \cos BAC)$. This then forces $(AC \cos BAC - x)(AC \sin^2 BAC + x \cos BAC) = (AC \cos BAC + x)(AC \sin^2 BAC - x \cos BAC) $, which resolves to $-\sin^2 BAC + \cos^2 BAC = \sin^2 BAC - \cos^2 BAC$, which is just $\sin BAC = \cos BAC$, forcing $\angle BAC = \frac{\pi}{4}$ as desired.

Heres my solution: Let $M$ denote the midpoint of $AB$. Then, notice that: $M$ lies on $(CQDP)$ as $\angle CMB = \angle CQD = 90^\circ$. Note that: $\angle BCM=\angle PCM = \angle ACM = \angle QCM$ which implies $MP=MQ$. Thus, $MO \perp PQ$. Also, note that: $\angle CQP=\angle FPQ$ which implies $CP=FQ$ or $CPQF$ is a cyclic trapezoid. Therefore, the perpendicular bisector of $CF$ passes through $M$, which implies it's the circum-center of $\triangle ABC$ and we are done.

Since $\angle FPQ = \angle CQP$ and $F$ lies on $(CPQ)$, it follows that $\overline{CF} \parallel \overline{PQ}$. Then, since $\angle CFD = 90^{\circ}$, it follows that line $FD$ passes through the antipode of $C$ WRT $(ABC)$ – call this point $M$. We also have from $\angle CFM = 90^{\circ}$ that line $DM$ is perpendicular to line $PQ$. From this, we now have \begin{align*}\angle BDM &= \angle PDM - \angle PDB = \angle DPQ + \angle CBA - 180^{\circ} \\ &= \left( 180^{\circ} - 2 \angle CBA - \angle BCD \right) + \angle CBA - 180^{\circ} = \angle CDB, \end{align*}implying by symmetry (and the fact that $D$ is not the midpoint of $\overline{AB}$) that $\angle BCA = 90^{\circ}$. edit: wow #3 is beautiful. its a shame that you can solve the whole problem without constructing the midpoint of AB tho