Claim: We have $\overline{CX}\parallel \overline{AB}$ and $\overline{BY}\parallel \overline{AC}$. Proof: We have \begin{align*} \angle DXC &= 180^\circ - \angle DQC = \angle QDC+\angle ACB \\ &= \angle DAC+\angle ACB = 180^\circ-\angle ADC=\angle ADB. \end{align*}Hence \begin{align*} \angle DCX &= 180^\circ-\angle CDX-\angle DXC \\ &= 180^\circ - \angle PDB - \angle ADB \\ &= 180^\circ-\angle PAD-\angle ADB \\ &=\angle ABC. \end{align*}This proves the claim. $\blacksquare$ The key is to now define $A'=\overline{CX}\cap \overline{BY}$. By the claim, $ACA'B$ is a parallelogram. Hence $M\in \overline{AA'}$. Now think of the problem in terms of $\triangle A'BC$. Notice $X$ is the point on $\overline{A'C}$ such that $\angle DXC=\angle ADB=\angle A'EC$, where the last equality follows by reflection over $M$. Hence $\triangle DXC\sim \triangle A'EC$, so $DXA'E$ cyclic. Similarly, $DEYA'$ is cyclic. Combining, in particular $DEXY$ is cyclic.

Let $A'$ be the reflection of $A$ over $M$. We know that $ABA'C$ and $ADA'E$ are parallelograms. \[\angle AQD = \angle BPD = 180 - \angle BYD \Longrightarrow BY\parallel AC\]\[\angle APD = \angle CQD = 180 - \angle DXC \Longrightarrow CX \parallel AB\]Hence $BY\cap CX = A'$. \[\angle BAC = 180-\angle PDQ = 180-\angle YDX=\angle BA'C \Longrightarrow A' \text{ lies on } (DYX).\]\[\angle DEA' = \angle ADE = \angle APD = \angle BYD = 180-\angle DYA' \Longrightarrow E \text{ lies on } (DYA').\]Hence $E$ lies on $(DYX).$

Essentially the same compared to the solutions above (and probably many below as well - in the future) - posting for storage. By Reim's Theorem on the circumcircles of $PDQA$ and $DQCX$ and using that $A,Q,C$ and $P,D,X$ are collinear, we get that $PA \parallel$ $XC$ and therefore $AB \parallel CX$ and similarly, $BY \parallel AC$. Let $M$ be the midpoint of $BC$ and let $S$ be the point such that $ABSC$ is a parallelogram. Notice that there for $S \in BY$ and $S \in CX$ as $BY \parallel AC$ and $AB \parallel CX$. Notice that $M$ lies on $AS$ and $ME = MD$ as $BD = EC$ meaning that $SEAD$ is also a paralleogram. Then $$\angle SED = \angle EDA = \angle DAC + \angle DCA = \angle DQC + \angle DCQ = \angle DXC$$meaning that $S,E,D,X$ are conyclic. Analogously , $S,Y,E,D$ are also conyclic meaning that in conclusion $D,E,X,Y$ are conyclic. $\square$

Note that after completing the parallelogram, you can also finish by showing (with some length chasing) that $\operatorname{Pow}_{(XDY)}(B)=\operatorname{Pow}_{(XDY)}(C)$.

Imayormaynotknowcalculus wrote: Note that after completing the parallelogram, you can also finish by showing (with some length chasing) that $\operatorname{Pow}_{(XDY)}(B)=\operatorname{Pow}_{(XDY)}(C)$. Exactly my approach:(probably?)

Mine is an Inversive Solution - We start with a claim - Claim - $(BPD)$ & $(CDQ)$ are tangent at $D$. Proof -Consider a tangent $l_a$ at $D$ to $(BPD)$ and let $\measuredangle(l_a , MN)$ deontes the angle between the tangent $l_a$ and the chord $MN$. Now by Reims Theorem , $PY \parallel QX$ , so $$ \measuredangle(l_a , DX) =\measuredangle(l_a ,PD) = \measuredangle PBD = \measuredangle PYD = \measuredangle XQD$$$\blacksquare$ Redefine $Y$ as $ DQ \cap (DEX)$ , so we have to prove that $Y$ lies on $(DPB)$ Now we invert around $D$ with arbitrary radius , call this mapping as $\Psi$ , and denote the image of object $K$ by $K'$. Note that \begin{align*} \Psi: (APQD) \to \overline {A'P'Q'} \\ \Psi: (PBD) \to \overline {P'B'} \\ \Psi: (CQXD) \to \overline {C'X'Q'} \\ \Psi: (EXYD) \to \overline {E'X'Y'} \\ \Psi: Y \to \overline{P'B'} \cap \overline{DQ'} \\ \end{align*} In the inverted sketch note that $B'P' \parallel C'Q'$ as $(BPD)$ & $(CDQ)$ are tangent at $D$ , also $P'Q' \parallel C'B'$ , due to tangency given in question . So we got a rectangle $P'Q'C'B'$. So in the inverted picture we have to prove that $Y'$ lies on $P'D'$ , which is equivalently to prove that $ \triangle B'E'Y' \sim C'E'X'$ , but note that $\measuredangle E'B'Y' = \measuredangle E'C'X' = 90$ , hence we have to prove that $$\frac{B'E'}{C'E'} = \frac{B'Y'}{C'X'}$$, but we have $BD = CE$ which is equivalently having $C'D. E'D = B'D.C'E'$ , and we also have $D'E' = D'B' + B'E'$ , by putting this value and doing some computations we get $$\frac{B'E'}{C'E'} = \frac{B'D'.D'E'}{C'E'.C'D'}$$, but from $C'D. E'D = B'D.C'E'$ we got this equal to ${\frac{D'E'}{C'E'}}^2$ , so now we have to prove that $$ ({\frac{D'E'}{C'E'}})^2 = \frac{B'Y'}{C'X'} $$ but from $\triangle D'B'Y' \sim D'C'Q' $ and $\triangle D'B'P' \sim D'C'X'$ , we get $\frac{B'Y'}{C'X'} = \frac{C'Q'. B'P'}{{C'X'}^2}$ , but as $C'Q' = B'P'$ from rectangle $P'Q'C'B'$ , we got this equal to $({\frac{B'P'}{C'X'}})^2$ , so we have to prove - $$ \frac{D'E'}{C'E'} = \frac{B'P'}{C'X'}$$, now using Inversion distance formula this is equivalently to prove $$ \frac{CX}{CE} = \frac{DX.BP}{DP.BD}$$, in the reinverted picture. Now note that as $BD = CE$ , this is equivalently to prove that $CX . DP = DX. BP$ , but this follows from the fact $\triangle BDP \sim CXD$. $\blacksquare$ Original Diagram - [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 = -7.05, xmax = 12.51, ymin = -11.07, ymax = 1.97; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen bfffqq = rgb(0.7490196078431373,1,0); pen qqffff = rgb(0,1,1); draw((-0.2,1.88)--(-5.42,-4.26)--(7.98,-4.76)--cycle, linewidth(0.5) + ffvvqq); /* draw figures */ draw((-0.2,1.88)--(-5.42,-4.26), linewidth(0.5) + ffvvqq); draw((-5.42,-4.26)--(7.98,-4.76), linewidth(0.5) + ffvvqq); draw((7.98,-4.76)--(-0.2,1.88), linewidth(0.5) + ffvvqq); draw(circle((0.19930503825233759,-1.28120070027316), 3.1863198805175252), linewidth(0.5) + blue); draw(circle((4.007759537127925,-4.948044404971607), 3.9766889484996537), linewidth(0.5) + qqwuqq); draw(circle((-2.6710397474259597,-4.129865231015716), 2.7520387948447462), linewidth(0.5) + qqwuqq); draw((-2.9849537476134755,-1.3957885077292604)--(4.461364456383953,-8.898778206475576), linewidth(0.5) + bfffqq); draw((-2.233188374984234,-6.846849534242625)--(3.3750654351639153,-1.022009106294425), linewidth(0.5) + bfffqq); draw(circle((1.1608410608336417,-7.7734595696584154), 3.4870900072901767), linewidth(0.5) + linetype("4 4") + qqffff); /* dots and labels */ dot((-0.2,1.88),dotstyle); label("$A$", (-0.13,1.65), NE * labelscalefactor); dot((-5.42,-4.26),dotstyle); label("$B$", (-5.35,-4.05), NE * labelscalefactor); dot((7.98,-4.76),dotstyle); label("$C$", (8.05,-4.55), NE * labelscalefactor); dot((-2.9849537476134755,-1.3957885077292604),dotstyle); label("$P$", (-2.91,-1.19), NE * labelscalefactor); dot((0.06057838829875983,-4.464499193593237),dotstyle); label("$D$", (0.15,-4.27), NE * labelscalefactor); dot((3.3750654351639153,-1.022009106294425),linewidth(1pt) + dotstyle); label("$Q$", (3.45,-0.87), NE * labelscalefactor); dot((2.504638229241978,-4.555695456314999),dotstyle); label("$E$", (2.59,-4.35), NE * labelscalefactor); dot((4.461364456383953,-8.898778206475576),linewidth(1pt) + dotstyle); label("$X$", (4.55,-8.73), NE * labelscalefactor); dot((-2.233188374984234,-6.846849534242625),linewidth(1pt) + dotstyle); label("$Y$", (-2.15,-6.69), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Inverted Picture- [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -14.67, xmax = 14.67, ymin = -9.06, ymax = 9.06; /* image dimensions */ pen ttzzqq = rgb(0.2,0.6,0); draw((-3.06,2.94)--(3.84,2.96)--(3.8518842941878244,-1.1400814947994555)--(-3.02,-1.16)--cycle, linewidth(0.5) + red); /* draw figures */ draw((-3.06,2.94)--(3.84,2.96), linewidth(0.5)); draw((-3.06,2.94)--(3.84,2.96), linewidth(0.5) + red); draw((3.84,2.96)--(3.8518842941878244,-1.1400814947994555), linewidth(0.5) + red); draw((3.8518842941878244,-1.1400814947994555)--(-3.02,-1.16), linewidth(0.5) + red); draw((-3.02,-1.16)--(-3.06,2.94), linewidth(0.5) + red); draw((-9.480003696671316,-1.1787246483961487)--(3.8704057936921346,-7.529998823786401), linewidth(0.5) + ttzzqq); draw((-9.480003696671316,-1.1787246483961487)--(3.8518842941878244,-1.1400814947994555), linewidth(0.5) + red); draw((3.8518842941878244,-1.1400814947994555)--(3.8704057936921346,-7.529998823786401), linewidth(0.5) + blue); draw((-3.02,-1.16)--(-3.039180121421116,-4.242858109714994), linewidth(0.5) + blue); /* dots and labels */ dot((-3.06,2.94),dotstyle); label("$P'$", (-2.94,3.24), NE * labelscalefactor); dot((3.84,2.96),dotstyle); label("$Q'$", (3.96,3.27), NE * labelscalefactor); dot((-3.02,-1.16),dotstyle); label("$B'$", (-2.91,-0.87), NE * labelscalefactor); dot((3.8518842941878244,-1.1400814947994555),linewidth(4pt) + dotstyle); label("$C'$", (3.96,-0.9), NE * labelscalefactor); dot((-9.480003696671316,-1.1787246483961487),dotstyle); label("$E'$", (-9.36,-0.87), NE * labelscalefactor); dot((3.8704057936921346,-7.529998823786401),dotstyle); label("$X'$", (3.99,-7.23), NE * labelscalefactor); dot((-3.039180121421116,-4.242858109714994),linewidth(4pt) + dotstyle); label("$Y'$", (-2.91,-3.99), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

A cleaner length finish, with the diagram as in #2: Since $BD$ is tangent to $(APD)$, we have that $BP\cdot BA=BD^2$, so $BP = \frac{BD^2}{BA} = \frac{BD^2}{CF}$ since $CF=AB$. Then as $AB\parallel CX$, we note that $\triangle CXD\sim \triangle BPD$ and hence $CX=\frac{CD}{BD} \cdot BP$. Combining, \[ CX\cdot CF = \frac{CD}{BD} \cdot \frac{BD^2}{CF} \cdot CF = CD\cdot BD\]which implies $E$ lies on $(FXD)$.

This is what I submitted. Define $K \equiv \overline{BY} \cap \overline{CX}$. We begin with the following claim. Claim. We have, $ABKC$ a paralleogram. Proof. We show that opposite sides are parallel. Indeed, $$\angle DBY = \angle PBY - \angle PBD = PDQ - \angle ABC = 180^\circ - \angle BAC - \angle ABC = \angle C$$so $$\angle ABK + \angle BAC = \angle PBD + \angle DBY + \angle BAC = \angle B + \angle C + \angle A = 180^\circ$$implying that $\overline{AC} \parallel \overline{BK}$. Similarly, $\overline{AB} \parallel \overline{CK}$, which establishes our claim. $\square$ Now since $AB=KC$ and $BD=CE$, it is easy to see that $\triangle ABD \sim \triangle KCE$. Also, $$\angle YKX + \angle YDX = \angle BKC + \angle PDQ = \angle A + 180^\circ - \angle A = 180^\circ$$implying that $YDXK$ is cyclic. It suffice to show that $E$ lie on this circle. Indeed, $$\angle CEK = \angle BDA = \angle BPD = \angle DYK$$where first equality follows from similarity, while second follows from the similarity of $\triangle BDA$ and $\triangle BPD$. This implies that $E \in \odot(YDXK)$, which finishes the problem. $\blacksquare$

By Reim's theorem, we obtain that $CX\parallel AB$ and $BY\parallel AC$. Let $A'$ be the reflection of $A$ over $M$, the midpoint of $BC$. Thus, $X,Y$ lie on $A'C$, $A'B$ respectively. By Reim again, we obtain that $A'DXY$ is cyclic. $E$ also lies on that circle, since $\measuredangle DEA'=\measuredangle EDA=\measuredangle DPA=\measuredangle DXA'$, where we used tangent-chord theorem.

The same solution: Let $BY\cap CX=T$. We will show that $X$ and $Y$ are on the circumcircle of $DET$. Because of the circles we have $\angle A=\angle QDX=180-\angle QCX$ and $\angle A=\angle PDY=180-\angle PBY$. These equalities show that $ABTC$ is a parallelogram. $\textbf{Claim:}$ $\triangle DXC \sim \triangle ADB$. $\textbf{Proof:}$ $\angle ABD=\angle ECT=\angle DCX$ because $ABTC$ is a parallelogram. $\angle ADB=180-\angle ADC=180-\angle DQC=\angle DXC$. Second equality comes from tangecy condition and the third one comes form the fact that $DQCX$ is cyclic. $\angle DAB=\angle CDX$ and $\angle ADB=\angle DXC$ finishes the proof. $\blacksquare$ Note that $\angle BAD=\angle CDX$ from the claim. Because $ABTC$ is a parallelogram we have ,$AB=CT$. With $\angle ABD=\angle ECT$ and $BD=CE$, we can say that $\triangle BDA \cong \triangle CET \implies \angle ETC=\angle BAD$. Thus we have $\angle ETC=\angle DAB=\angle CDX$ which implies $DEXT$ is a cyclic quadrilateral. By symmetry $DEYT$ is also a cyclic quadrilateral. In conclusion; $D$, $E$, $X$, $Y$ are concyclic.$\square$

Imayormaynotknowcalculus wrote: Note that after completing the parallelogram, you can also finish by showing (with some length chasing) that $\operatorname{Pow}_{(XDY)}(B)=\operatorname{Pow}_{(XDY)}(C)$. But why length bash such an elegant problem :woozy_face:, my solution is essentially the same parallelogram + angle chase thingy : D

Here's what I submitted. Nice problem . Let $E$ be the intersection point of $(DXY)$ with $BC$. We will prove that $CE=BD$, which is sufficient. Let $Z$ be the intersection point of $BY$ with $CX$. We have that $\angle BYD=180-\angle BPD =180- \angle DQA$, and thus $BY$ is parallel to $AC$ (we used the cyclic quadrilaterals $BYDP$ and $PDQA$). Similarly, $CX$ is parallel to $AB$. Therefore $ABZC$ is a parallelogram. In addition, we have that $\angle DYZ =\angle BPD =\angle DQA = \angle DXC$, so $YDXZ$ is cyclic (and also we have assumed that $E$ also lies on that circle). Let $M$ be the intersection point of $AZ$ and $BC$. Obviously $M$ is the midpoint of $AZ$ and $BC$. Note that it’s sufficient to prove that $ADZE$ is parallelogram, which will give that $M$ is midpoint of $DE$ and $BC$ at the same time. We have $MA = MZ$, so we need only $AD$ and $ZE$ to be parallel, which is equivalent to $\angle ADC =\angle DEZ$. Indeed, we have that $\angle ADC =\angle BAD+ \angle ABD = \angle PDB+ \angle ABD=180- \angle BPD= \angle BYD=\angle DEZ$ (we used that $BC$ is tangent to $(APDQ)$, and $BPDY$ and $YDEZ$ are cyclic), so we are done.

Also one question, will we get our marks deducted if we don't use directed angles?

My solution: From $\angle PDB =\angle PQY$ and $\angle PYQ = \angle PBD$, we get $ \triangle PBD \sim \triangle PYQ$, then we obtains: \[ \frac{{BD}}{{QY}} = \frac{{PB}}{{PY}}\]From Reim Theorem, we obtains $ \triangle PBY \sim \triangle XCQ$, then: \[\frac{{CX}}{{QX}} = \frac{{PB}}{{PY}} = \frac{{BD}}{{QY}} = \frac{{CE}}{{QY}}\]Combine with $\angle XCE= \angle XQY$, we get $ \triangle XCE \sim \triangle XQY$, leads to $\angle XEC = \angle DYX$, thus $D$, $E$, $X$, and $Y$ are concyclic. (Q.E.D)

Nice problem. Here's a sketch: Extend $BY$ and $CX$ to meet at $T$, and note that $ABTC$ is a parallelogram. Moreover, by applying the sine law on triangles $BDY,ABD,CDX,ACD$, it's easy to see that $\frac{BY}{AB}=\frac{CX}{AC} \implies BY \cdot BT = CX \cdot CT$. Through angle chasing, observe that $DYTX$ is a cyclic quadrilateral, hence $\text{Pow}(B)=\text{Pow}(C)$ with respect to the circumcircle of $DYTX$. Let this circle now intersect $BC$ at $E'$—clearly, $BD=E'C$ and, as $E'$ lies on the interior of $BC$, $E=E'$. Hence proved.

My problem. Since \(\measuredangle BYD=\measuredangle BPD=\measuredangle APD=\measuredangle AQD\), we have \(\overline{BY}\parallel\overline{AC}\) and analogously \(\overline{CX}\parallel\overline{AB}\). Construct \(A'=\overline{BY}\cap\overline{CX}\), so \(ABA'C\) is a parallelogram. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=lightblue; pen sec=heavygreen; pen tri=lightred; pen fil=invisible; pen fil2=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,Ap,D,EE,Op,X,Y,Xp,Yp,P,Q; A=dir(120); B=dir(200); C=dir(340); Ap=B+C-A; D=(3B+2C)/5; EE=B+C-D; Op=extension(D,D+(0,1),(A+D)/2,(A+D)/2+rotate(90)*(A-D)); X=2*foot(Op,A,B)-A; Y=2*foot(Op,A,C)-A; Xp=extension(X,D,C,Ap); Yp=extension(Y,D,B,Ap); filldraw(circumcircle(B,D,X),tfil,tri); filldraw(circumcircle(C,D,Y),tfil,tri); filldraw(circumcircle(D,Xp,Yp),sfil,sec); draw(X--Xp,gray); draw(Y--Yp,gray); filldraw(circumcircle(A,X,Y),fil2,pri2); fill(B--Ap--C--cycle,fil2); draw(B--Ap--C,pri2); filldraw(A--B--C--cycle,fil,pri); dot("\(A\)",A,N); dot("\(B\)",B,W); dot("\(C\)",C,E); dot("\(A'\)",Ap,SE); dot("\(D\)",D,dir(285)); dot("\(E\)",EE,NE); dot("\(P\)",X,W); dot("\(Q\)",Y,dir(30)); dot("\(X\)",Xp,dir(-30)); dot("\(Y\)",Yp,dir(220)); [/asy][/asy] Note that \(\measuredangle XDY=\measuredangle PQD=\measuredangle PAQ=\measuredangle XA'Y\), so \(D\), \(X\), \(Y\), \(A'\) are concyclic. Now observe that \(\measuredangle A'ED=\measuredangle ADE=\measuredangle APD=\measuredangle BPD=\measuredangle BYD=\measuredangle A'YD\), so \(E\) lies on \((DXA'Y)\) as well.

coming back and solving problems that i failed before is kind of satisfying Note that $PD$ is antiparallel to $BY$ and $AQ$ is antiparallel to $PD$, so by Reim's, we have that $AC\parallel BY$. Similarly, $CX$ is antiparallel to $DQ$ which is antiparallel to $AP$, so $AB\parallel CX.$ Extend $CX$ and $BY$ to meet at $F$. Note that $ACFB$ is a parallelogram. We have $$\angle PQD=\angle PDB=\angle CDX.$$Similarly, $\angle QPD=\angle QDC=\angle BDY.$ Additionally, from parallelogram, $$\angle YFX=\angle PAQ=\angle PQD+\angle QPD,$$so $$\angle YFX+\angle YDX=\angle PQD+\angle QPD+(180-\angle PQD+\angle QPD)=180,$$so $DXFY$ is cyclic. Note that $ADFE$ is also a parallelogram. We then have $$\angle DEF=\angle EDA=\angle CDA.$$Note that $\angle CDA$ intercepts arc $AD$ of $(APDQ)$, and $\angle APD$ also does, so $$\angle CDA=\angle APD.$$Then, from cyclic $PDYB$, $\angle APD=\angle BYD$ Since $\angle BYD=\angle DEF$, $DEFY$ is cyclic. Since $DEFY$ and $DXFY$ are both cyclic, $DEXY$ is cyclic, so we are done.

Let $F = BY \cap CX$. From \[\angle AQY = \angle AQD = 180^\circ - \angle APD = \angle BPD = 180^\circ - \angle BYD = \angle DYF = \angle QYF,\]we have $AC \parallel BF$, and similarly $AB \parallel CF$. Therefore $ACBF$ is a parallelogram, and $\triangle AED \cong \triangle FDE$ by symmetry. Now, \[\angle FYD = \angle AQD = \angle ADE = \angle FED\]gives $(DEFY)$ cyclic. Furthermore, \[\angle DXF = \angle APD = 180^\circ - \angle BPD = \angle BYD = 180^\circ - \angle DYF\]gives $(DFXY)$ cyclic. This implies $(DEFXY)$ cyclic, as desired.

All geo skills are lost. Took an eternity to solve this. What is happening? How will I survive. Solution: Redefine $E \coloneqq \odot(DXY) \cap BC \ne D$. We will show that $\overline{EC} = \overline{BD}$. Define $A'$ to be intersection of $CX$ and $BY$. [asy][asy] import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(14); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.1, xmax =4.5, ymin = -6.510333635931812, ymax = 2.0219760482168474; /* image dimensions */ pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen dcrutc = rgb(0.8627450980392157,0.0784313725490196,0.23529411764705882); pen xfqqff = rgb(0.4980392156862745,0.,1.); pen ffqqff = rgb(1.,0.,1.); draw((0.,1.5120028456075445)--(-1.3,-2.1)--(3.8599599134652864,-2.0974705678136822)--cycle, linewidth(0.6) + yqqqyq); /* draw figures */ draw((0.,1.5120028456075445)--(-1.3,-2.1), linewidth(0.6) + yqqqyq); draw((-1.3,-2.1)--(3.8599599134652864,-2.0974705678136822), linewidth(0.6) + yqqqyq); draw((3.8599599134652864,-2.0974705678136822)--(0.,1.5120028456075445), linewidth(0.6) + yqqqyq); draw(circle((0.7144658148831527,-0.22282356535170195), 1.876188710337632), linewidth(0.6) + dcrutc); draw(circle((2.2878250220842684,-2.4089294192579134), 1.602689842995383), linewidth(0.6) + dcrutc); draw(circle((-0.29240484589548915,-1.9003832679125319), 1.027178092788282), linewidth(0.6) + dcrutc); draw((-0.4240698709768342,-2.919087913761613)--(2.493259892490037,-0.8194605624186455), linewidth(0.6) + xfqqff); draw((3.3008885931981005,-3.6508299522003296)--(-0.941686247033682,-1.1044382343579542), linewidth(0.6) + xfqqff); draw((0.,1.5120028456075445)--(0.7153855296911991,-2.0990120502654803), linewidth(0.6) + xfqqff); draw(circle((1.2809543613056267,-4.086489121818744), 2.0663816717256385), linewidth(0.6) + linetype("4 4") + blue); draw((3.3008885931981005,-3.6508299522003296)--(1.8445743837740893,-2.098458517548202), linewidth(0.6)); /* special point */ draw((3.8599599134652864,-2.0974705678136822)--(2.559959913465289,-5.7094734134212235), linewidth(0.6) + yqqqyq); draw((-1.3,-2.1)--(2.559959913465289,-5.7094734134212235), linewidth(0.6) + yqqqyq); draw((-0.4240698709768342,-2.919087913761613)--(3.3008885931981005,-3.6508299522003296), linewidth(0.6) + xfqqff); draw((1.8445743837740893,-2.098458517548202)--(2.559959913465289,-5.7094734134212235), linewidth(0.6) + ffqqff); /* dots and labels */ dot((0.,1.5120028456075445),dotstyle); label("$A$", (0.060387493516161274,1.6625564095072887), NE * labelscalefactor); dot((-1.3,-2.1),dotstyle); label("$B$", (-1.7,-2.2754326754844003), NE * labelscalefactor); dot((3.8599599134652864,-2.0974705678136822),dotstyle); label("$C$", (3.9202418744405523,-1.9472669184017597), NE * labelscalefactor); dot((0.7153855296911991,-2.0990120502654803),dotstyle); label("$D$", (0.7635998301218195,-2.4785829060593683), NE * labelscalefactor); dot((-0.941686247033682,-1.1044382343579542),dotstyle); label("$P$", (-1.33,-1.1502929369153463), NE * labelscalefactor); dot((2.493259892490037,-0.8194605624186455),dotstyle); label("$Q$", (2.716967431804204,-0.6658577716981148), NE * labelscalefactor); dot((3.3008885931981005,-3.6508299522003296),dotstyle); label("$X$", (3.420179768409862,-3.7599920527630135), NE * labelscalefactor); dot((-0.4240698709768342,-2.919087913761613),dotstyle); label("$Y$", (-0.4865554349549063,-3.353691591613077), NE * labelscalefactor); dot((1.8445743837740893,-2.098458517548202),dotstyle); label("$E$", (1.9043665095043318,-1.9785208000286778), NE * labelscalefactor); dot((2.559959913465289,-5.7094734134212235),linewidth(1.pt) + dotstyle + invisible,UnFill(0)); label("$A'$", (2.6700866093638265,-5.963390707460744), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] By ``Reim's Theorem'', we know that $CX \parallel AB$ and $BY \parallel AC$. This shows that $ABA'C$ is a parallelogram. Also observe that $A' \in \odot(DYEX)$ since \[\measuredangle YDX = \measuredangle QDP = \measuredangle QAP = \measuredangle CAB = \measuredangle CA'B. \]We will now prove that $\triangle EA'B \cong \triangle DAC$. Since $A'B \parallel AC$, we already have that $\measuredangle EBA' = \measuredangle DCA$ and $\overline{BA'} = \overline{AC}$. Finally note that \[\measuredangle CAD = \measuredangle QDC = \measuredangle YDE = \measuredangle YA'E\]which proves the congruence with the ASA rule. To finish off, just note that $\triangle EA'B \cong \triangle DAC \implies \overline{BD} = \overline{EC}$ as desired. $\blacksquare$

Nice problem! Let $\omega$ be the circumcircle of triangle $DXY$. Let $BY$ and $CX$ be extended to meet at $T$. We have $$\angle TXD=\angle CXD=\angle CQD=\angle AQD=\angle APD=\angle BPD=\angle BYD=\angle TYD\Longrightarrow T\in\omega$$Moreover, $$\angle BAC=\angle PAQ=\angle PDQ=\angle PDQ=\angle PDY=\angle PBY=\angle ABT\Longrightarrow AC\parallel BT$$and similarly $AB\parallel CT$. It follows that $ABTC$ is a parallelogram. Claim. $Pow_{\omega}(B)=Pow_{\omega}(C)$ Proof. Note that $\angle DAC=\angle DAQ=\angle CDQ=\angle BDY$ and $\angle BAD=\angle PAD=\angle PDB=\angle XDC$. We have $$\frac{BD/AB}{DC/CA}=\frac{BD}{DC}\cdot\frac{CA}{AB}=\frac{\sin(\angle BAD)}{\sin(\angle DAC)}=\frac{\sin(\angle XDC)}{\sin(\angle BDY)}=\frac{CX}{BY}\cdot\frac{\frac{BD}{\sin(\angle BPD)}}{\frac{DC}{\sin(\angle DQC)}}=\frac{CX}{BY}\cdot\frac{BD}{DC}$$$$\Longrightarrow\frac{CA}{AB}=\frac{CX}{BY}=\frac{BT}{CT}\Longrightarrow BT\cdot BY=CT\cdot CX,$$from where it follows that $Pow_{\omega}(B)=Pow_{\omega}(C)$ Let $E'\ne D$ be the point of intersection of $BC$ with $\omega$. Then, from our claim we have $$BE'\cdot BD=CE'\cdot CD=E'C\cdot DC=(BC-BE')\cdot(BC-BD)=BC^2-(BD+BE')\cdot BC+BE'\cdot BD$$$$\Longrightarrow BD+BE'=BC=BD+BE\Longrightarrow BE'=BE\Longrightarrow E'=E,$$as desired. $\square$

Claim: there exists a negative homothety centered at $D$ that sends $P$ to $X$, $B$ to $C$ and $Y$ to $Q$. Proof: due to angle chasing, we have that $\angle PDY = \angle QDX = \angle A$, that $\angle PYD = \angle B$ and that $\angle DXQ = \angle C$. Hence, $\triangle BPD \sim \triangle CXD$. As a result, we have that $\angle YBD = \angle YPD = \angle C$, hence $\overline{BY} \parallel \overline{QC}$; similarly, $\overline{CX} \parallel \overline{PB}$. This is sufficient to prove the cliam. In fact, this means that $(PDY)$ is externally tangent to $(QDX)$ at point $D$, although we won't use this fact. Let $A'$ be the intersection of lines $BY$ and $CX$. Due to the angle chasing we did in the previous claim, we have that $\angle DBY = \angle C$ and $\angle DCX = \angle B$, hence $A'$ is just the point such that $ABA'C$ is a parallelogram. Claim: $BY \cdot A'B = CX \cdot A'C$. Proof: We begin by using our previous claim, that "everything's similar": \begin{align*} \frac{CD^2}{CQ \cdot CX} &= \frac{BD^2}{BP \cdot BY}\\ \frac{CD^2 \cdot BY}{CQ} &= \frac{BD^2 \cdot CX}{BP}. \\ \left( \frac{AC \cdot CQ}{CD^2} \right) \left( \frac{CD^2 \cdot BY}{CQ} \right) &= \left( \frac{BP \cdot BA}{BD^2} \right) \left( \frac{BD^2 \cdot CX}{BP} \right) \\ AC \cdot BY &= CX \cdot AB \\ A'B \cdot BY &= CX \cdot A'C. \end{align*} Note that $\angle YDX = \angle PDQ = 180^{\circ} - \angle BAC - 180^{\circ} - \angle BA'C$, hence $DXYA'$ is cyclic. Therefore, our previous claim just establishes that $B$ and $C$ have equal powers WRT $(DXYA')$, so the two points at which $(DXYA')$ intersects $\overline{BC}$ are equidistant from $B$ and $C$, respectively.

By Reim's, we get that $BY\parallel AC$, and $CX\parallel AB$. Let $A' = CX\cap BY$. We see that $ACA'B$ is a parallelogram. It is easy to see now that $XA'YD$ is cyclic. Let $P = (A'YX)\cap BC\neq D$. Note that, $\triangle{BYD}\sim \triangle{CQD}$, and that \[BP\cdot BD = POW_{(A'YX)}(B) = BY\cdot BA' = \frac{QC\cdot BD}{CD}\cdot AC = CD\cdot BD \implies P = E\]Thus, $DEXY$ is cyclic.

Let intersection point of $BY$ and $CX$ be $A'$. $\angle PAQ=\angle PDY=180-\angle PBY$ which means that $BY//AC$, similarly $CX//AB$, which implies that $ABA'C$ is parallelogram . Then $\angle YA'X=\angle PAQ=180-\angle PDQ=180-\angle XDY$, then $(Y D X A')$ is cyclic $(1)$. From $ABA'C$ parallelogram we know that $BA'C$ and $CAB$ triangles are similar and from $EB=DC$, we have $\angle BA'E=\angle DAC=\angle QDC=\angle EDY$, then $(Y A' D E)$ is cyclic $(2)$. Then from $(1)$ and $(2)$ we have $(E Y A' X D)$ is cyclic, so we are done:).

No auxiliary constructions for this method.

$ \angle{BPY} = \angle{BDY} = \angle{CEQ} = \angle{CXQ}$,$\angle{PBD} = \angle{PYQ}$. So $\triangle{BDP}$ and $\triangle{YQP}$ are similar by AAA so $QY : DY = BD : BP$. Also $\angle{BYP} = \angle{BDP} = \angle{CEX} = \angle{CQX}$ so $\triangle{BPY}$ and $\triangle{CXQ}$ are similar by AAA and $BP : CX = YP : QX$. Moreover $BD = CE$ so $CX : CE = QX : QY$.And $\angle{XCE} = \angle{XQY}$ so $\triangle{CXE}$ so $\triangle{QXY}$ are similar by SAS and $\angle{CEX} = \angle{DYX}$. Then we are done.

Used the 10% hint on ARCH. Assume WLOG that $B,D,E,C$ are collinear in that order. Note that by Reim's theorem, $BY\parallel AC$ and $CX\parallel AB$. Thus if $BY$ and $CX$ meet at $Z$, quadrilateral $ABZC$ is a parallelogram. We have \[ \angle CDX=\angle PDB=\angle PAD. \]Similarly, $\angle BDY=\angle QAD$, so \[ \angle YDX=180^{\circ}-\angle BDY-\angle CDX =180^{\circ}-\angle QAD-\angle PAD =180^{\circ}-\angle BAC =180^{\circ}-\angle BZC. \]Therefore, quadrilateral $YDXZ$ is cyclic. But \[ \angle EZX=\angle EZC=\angle DAB=\angle PAD=\angle CDX=\angle EDX, \]so quadrilateral $EDZX$ is cyclic. Therefore, $E$ and $Y$ are both on $(DXZ)$, so quadrilateral $DEXY$ is cyclic. $\blacksquare$

Bit hard for a p1? First, note that since $(APDQ)$ is cyclic, \[ \angle BYD = 180^{\circ} - \angle BPD = \angle APD = 180^{\circ} - \angle AQD = \angle DQC, \]which implies $BY \parallel AC$. Analogously, $CX \parallel AB$. Let $A' = BY \cap CX$, and realize by our prior observation that $A'=B+C-A$. By easy angle chasing using the fact that $ABA'C$ is a parallelogram, $(A'XDY)$ is cyclic. Let $N$ denote the intersection of $(DXY)$ and $BC$. Now using the tangency condition, we have $\angle BAD = \angle BDP = 180^{\circ} - \angle BDX = \angle NA'C$. Similarly, $\angle NA'B = \angle CAD$. Thus, upon reflection about $M$, we have $N=E$, as desired.

reaganchoi wrote: In $\triangle ABC$, points $P$ and $Q$ lie on sides $AB$ and $AC$, respectively, such that the circumcircle of $\triangle APQ$ is tangent to $BC$ at $D$. Let $E$ lie on side $BC$ such that $BD = EC$. Line $DP$ intersects the circumcircle of $\triangle CDQ$ again at $X$, and line $DQ$ intersects the circumcircle of $\triangle BDP$ again at $Y$. Prove that $D$, $E$, $X$, and $Y$ are concyclic. $<YBP=<PDQ=180-<A$ so $YB//AC$ similar $XC//AB$ let $A'$ be the symmetric poin of $A$ we respect the midpoint of $BC$ then we have: $<XMY=<A=180-<XDY$ so $X,D,Y,A'$ are concyclic. $<DEA'=<ADC=<APD=<BYD$ so $E$ belongs to $(XDYA')$ We use that $ADA'E$ is parallhlogram

Solution using Ptolemy's sine lemma (no need to any syntehtic constructions). Let $\angle BAD=\angle PQD=\angle PDB=\angle PYB=\angle CDX=\alpha$ and $\angle CAD=\angle QPD=\angle QDC=\angle PDY=\beta$. Ptolemy sine to quadrilaterals $PBYD$ and $QDXC$ from pencil $D$. $(....1)$ $DP\sin(\beta)+DY\sin(\alpha)=DB\sin(\alpha+\beta)$. $(....2)$ $DQ\sin(\alpha)+DX\sin(\beta)=DC\sin(\alpha+\beta)$. If $DEXY$ is cyclic, then $DE\sin(\alpha+\beta)+DY\sin(\alpha)=DX\sin(\beta)(?)$. Let's write $(....2)$ as $DQ\sin(\alpha)+DX\sin(\beta)=DE\sin(\alpha+\beta)+DB\sin(\alpha+\beta)$ more precisely $DQ\sin(\alpha)+DX\sin(\beta)=DE\sin(\alpha+\beta)+DP\sin(\beta)+DY\sin(\alpha)$ if we prove $DQ\sin(\alpha)=DP\sin(\beta)$ we will finish the problem. So,sine thrm to $\triangle DPQ$ gives us $DQ\sin(\alpha)=DP\sin(\beta)$.

Angle-Chase: By Reims, $BY \parallel AC$ and $CX \parallel AB$. Extend it to point $F$ such that $ABFC$ is a parallelogram. Notice that: $$\angle YDX = \angle PDQ = 180^\circ - \angle PAQ=180^\circ \angle YFX$$Thus, $YDXF$ is cyclic. Note that: $\triangle ABD \cong \triangle FCE$. Thus, we have: $AD \parallel BF$. Therefore: $$\angle FBD = \angle ADE = \angle AQD = \angle DXC$$which implies $FEDX$ is also cyclic and we are done.

Solved with adrian_042