Circumcircles intersect on AO

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Source: InfinityDots MO Problem 3

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276 posts

talkon

#1 h Mar 28, 2017, 1:21 PM • 12 Y

Y by artsolver, jonyj1005, Ankoganit, aopser123, anantmudgal09, monkey8, don2001, GeoMetrix, ApraTrip, Adventure10, Mango247, Funcshun840

Let $\triangle ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$ . The line through $O$ parallel to $BC$ intersect $AB$ at $D$ and $AC$ at $E$ . $X$ is the midpoint of $AH$ . Prove that the circumcircles of $\triangle BDX$ and $\triangle CEX$ intersect again at a point on line $AO$ .

Proposed by TacH

This post has been edited 1 time. Last edited by talkon, Jan 3, 2019, 6:16 PM

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#2 h Mar 28, 2017, 1:31 PM • 5 Y

Y by artsolver, mihajlon, don2001, Adventure10, MS_asdfgzxcvb

diagram

solution

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#3 h Mar 28, 2017, 2:06 PM • 5 Y

Y by anantmudgal09, don2001, fuzimiao2013, Adventure10, Mango247

Let $AP, BQ, CR$ be the altitude from $A, B, C$ , wrt $\triangle ABC$ . Let $M$ be the midpoint of $BC$ . Let $O, H$ be the circumcenter and orthocenter of $\triangle ABC$ . Let $H'=QR \cap AP$ . Let $O'$ be the circumcenter of $\triangle BXC$ . Let $Y'=XO' \cap AO$ . Let $O_1, O_2$ be the circumcenters of $\triangle BXD$ and $\triangle CXE$ , and $\omega_1$ , $\omega_2$ be the corresponding circumcircle. Let $\omega$ be the circumcircle of $\triangle ABC$ .
Let $R$ be the circumradius of $\triangle ABC$ .

Lemma. $\angle BXE = \angle CXD = 90$ . Also, $H'$ is the orthocenter of $\triangle XBC$ .

Proof of Lemma. Let $AO \cap BC = K$ and $QR \cap BC = T$ .
Now, harmonic chasing, we have $(T,H',R,Q)=(T,P;B,C)=-1$ by projecting at $A$ .
Also, we have $(P,H';H,A)=-1$ , so since $X$ is the midpoint of $AH$ , we have $PB \cdot PC = PH \cdot PA = PX \cdot PH'$ , which forces $H'$ to be the orthocenter of $\triangle XBC$ . This proves the second claim.

It now suffices to show that $CH' \parallel XE$ , which will show that $\angle BXE=90$ .
It will similarly follow that $BH' \parallel XD$ , which will show that $\angle CXD = 90$ .
Let's show that $\frac{AX}{AH'}=\frac{AE}{AC}$ .

We have $\frac{AE}{AC} = \frac{AO}{AK} = 1-\frac{OK}{AK} = 1-\frac{OM}{AP} = 1-\frac{AX}{AP}$ .
(Note that $OM = \frac{1}{2}AH = AX$ by a well known lemma)
It now suffices to show that $\frac{AX}{AH'} + \frac{AX}{AP} = 1$ .
$(A,H;H',P)=-1$ shows $\frac{1}{AH'}+\frac{1}{AP} = \frac{2}{AH} = \frac{1}{AX}$ , which implies the desired.
The Lemma is now proved $\blacksquare$ .

I claim that $\triangle AXB \sim \triangle AEY'$ and $\triangle AXC \sim \triangle ADY'$ .
I will just show the first one, the second will follow similarly.

Since $Y' \in AO$ , we have $\angle XAB = \angle HAB = \angle OAC = \angle Y'AE$ . We will show $\frac{AX}{AB} = \frac{AE}{AY'}$ .

Notice the similar triangles $\triangle AXY' \sim \triangle OO'Y'$ . This shows that $\frac{AY'}{AO} = \frac{AX}{AX-OO'} = \frac{AH}{AH-2OO'} = \frac{AH}{AH-2OM+2O'M} = \frac{AH}{AH-AH+XH'} = \frac{AH}{XH'}$ .
Therefore, $AY' = \frac{R \cdot AH}{XH'}$ .

Now note that $X$ is the center of $(ARHQ)$ so $RX = AX$ . Therefore, from sine law in $\triangle XH'R$ , we have $XH' = \frac{RX}{ \sin \angle AH'R} \cdot \sin \angle XRQ = \frac{AH \cdot \sin (90-A)}{2 \sin (90+C-B)}$ . Also, we have, from sine law in $\triangle AOE$ , $AE= \frac{R}{\sin \angle AEO} \cdot \angle AOE = \frac{R \sin (90+C-B)}{\sin C}$ . Now $AX \cdot AY' = AX \cdot \frac{R \cdot AH \cdot 2 \cdot \sin (90+C-B)}{AH \cdot \sin (90-A)} = AX \cdot \frac{2R \cdot \sin (90+C-B)}{\sin (90-A)} = AX \cdot \frac{2 \cdot AE \cdot \sin C}{\sin (90-A)} = \frac{AH \cdot AE \cdot \sin C}{\cos A} = 2R \cdot AE \cdot \sin C = AB \cdot AE$ .

This implies $\frac{AX}{AB} = \frac{AE}{AY'}$ . Therefore, we have $\triangle AXB \sim \triangle AEY'$ as desired.
Similarly, we have $\triangle AXC \sim \triangle ADY'$ . This also shows $\triangle AXD \sim \triangle ACY'$ and $\triangle AXE \sim \triangle ABY'$ .

Now let's angle chase. I claim that $Y'$ lies on both circles. This will show that $Y' \equiv Y$ , which will finish the problem.

I will show $\angle DBY' + \angle DXY' = 180$ . Note that from $\angle BXE = \angle CXD = 90$ , we have $\angle BXD = \angle CXE$ . We have $\angle ABY' + \angle DXO' = \angle AXE + \angle DXO'$ . Since $O'$ and $H'$ are isogonal wrt $\angle BXC$ , we have $\angle AXE + \angle DXO' = \angle AXE + \angle BXD + \angle BXO' = \angle AXE + \angle CXE + \angle H'AC = 180$ . Therefore, we have $D, B, X, Y'$ cyclic, which implies $Y' \in \omega_1$ .

Similarly, we have $Y' \in \omega_2$ , so $Y' \equiv Y$ . Since $Y' \in AO$ , so is $Y$ . We are done. $\blacksquare$ .

This post has been edited 1 time. Last edited by rkm0959, Mar 28, 2017, 2:07 PM

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#4 h Mar 28, 2017, 3:53 PM • 4 Y

Y by GGPiku, TacH, Adventure10, Mango247

My solution : Let $B',H',B_1,F$ be the antipode of $B$ in the circle $\odot O$ ,the intersection of $AH, \odot O$ , the foot of $B$ in $AC$ , the intersection of $(BDX),(CEX)$ . $\angle B'AE=\angle B'BC=\angle B'OE$ so $A,O,E,B'$ are concyclic. $\angle OB'E=\angle EAO=\angle BAH'$ so $B,H',B_1,E$ are concyclic. by simple angle chasing : $X,B,H',B_1,E$ are concyclic. thus $\angle BXE=\angle CXD=90$ so $\angle EFC=\angle CXE=\angle XBD=\angle DFB=\alpha$ , $\angle BFC=\angle XFC+\angle XFB=\angle XEA+\angle XDA=180-\angle A-(180-(90+\alpha))=\alpha+90-\angle A$ so $\angle BFE=90-\angle A$ $\rightarrow$ $B,F,O,E$ are concyclic . similarly $C,O,D,F$ are concyclic. $AB$ cuts $(BOE),(COD)$ at $E',D'$ res. $\angle EE'B=\angle EFD=\angle DFC=\angle CD'B$ $\rightarrow$ $E'E \parallel CD'$ so $\tfrac {AE'}{AD'}=\tfrac {AE}{AC}=\tfrac {AD}{AB}$ thus $AD.AD'=AE.AE'$ so $A$ lies on radical axis of $(BOE),(COD)$ or line $OF$ . done . $\blacksquare$ .

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#5 h Mar 28, 2017, 5:48 PM • 1 Y

Y by Adventure10

Is it secret who is the author of this amazing problem?

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#6 h Mar 28, 2017, 8:45 PM • 2 Y

Y by Adventure10, Mango247

Let $C_1,B_1$ be the midpoints of $AB,AC$ respectively. Note that $X$ is the orthocenter of $AB_1C_1$ and that $X_1$ , the reflection of $X$ over $B_1C_1$ lies on $DE$ . Let $C_1X$ cut $AO$ at $R_1$ . By sine theorems in $\triangle AR_1C_1, \triangle DX_1R_1$ we get $\frac{C_1R_1}{C_1A}=\frac{cos \angle ACB}{sin \angle ABC}=\frac{C_1D}{C_1X_1}$ so $BDXR_1$ is cyclic. Because of same reason $CER_2X$ is cyclic, where $XB_1 \cap AO = R_2$ . By sine theorems in $\triangle AR_1C_1$ and $AR_2B_1$ we get $\frac{AR_1}{AR_2}=\frac{AB^2}{AC^2}=\frac{AC_1 \cdot AB}{AB_1 \cdot AC}$ , so our two circles must intersect on $AO$ .

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2312 posts

#7 h Mar 29, 2017, 4:01 PM • 12 Y

Y by rkm0959, talkon, anantmudgal09, kapilpavase, Saro00, don2001, enhanced, AmirKhusrau, fuzimiao2013, Adventure10, Mango247, ehuseyinyigit

Let $S$ be the projection of $D$ on $BC.$ Obviously $ADSX$ and $DXHS$ are parallelogram, so $H$ is the orthocenter of $\triangle CXS$ $\Longrightarrow$ $CX$ is perpendicular to $DX$ $( \bigstar ).$ Let the perpendicular from $B,$ $C$ to $AB,$ $AC$ cut $AC,$ $AB$ at $Y,$ $Z,$ respectively and let $W$ be the second intersection of $AO$ with $\odot (AYZ).$ Notice $(B,E,O,W,Y)$ and $(C,D,O,W,Z)$ are concyclic, so $\measuredangle BWD = \underbrace{ \measuredangle (\perp AB, OY) + \measuredangle (OZ,AB) = \measuredangle (BX,\perp CA) + \measuredangle (CA, CX) }_{\because \ \triangle ABC \cup X \text{ and } \triangle AYZ \cup O \ \text{are inversely similar}} \stackrel{ (\bigstar )}{=} \measuredangle BXD$ $\Longrightarrow$ $W$ $\in$ $\odot (BDX).$ Similarly, $W$ lies on $\odot (CEX),$ so the second intersection of $\odot (BDX),$ $\odot (CEX)$ lies on $AO.$

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225 posts

baopbc

#8 h Mar 30, 2017, 6:13 AM • 3 Y

Y by don2001, Adventure10, Mango247

Let the line passes through $X$ and perpendicular to $AB$ intersects $AO$ at $K$ and $AC$ at $M$ . $F$ is the reflection of $E$ through $M$ , since $M$ is the midpoint of $AC$ so $ME\cdot MC$ $=$ $MF\cdot MA$ . Note that $\angle AKX$ $=$ $\angle ACB$ $=$ $\angle AEO$ $=$ $\angle OFM$ so $KFMO$ is a cyclic quadrilateral. Thus $\angle FKM$ $=$ $\angle FOM$ $=$ $\angle EOM$ $=$ $\angle XAF$ which imply $AFKX$ is a cyclic quadrilateral i.e $MF\cdot MA$ $=$ $MK\cdot MX$ . Hence the circle $(CEX)$ passes through $K$ . Define $L$ similarly and we get $(BDX)$ passes through $L$ . The tangent at $A$ of $(O)$ meets $BC$ at $S$ , then we get $\triangle ABC$ $\cup$ $S$ $\sim$ $\triangle XLK$ $\cup$ $A$ which imply $\tfrac{AK}{AL}$ $=$ $\tfrac{SC}{SB}$ $=$ $\tfrac{AC^2}{AB^2}$ $=$ $\tfrac{AE\cdot AC}{AD\cdot AB}$ . Hence $(CEX)$ and $(BDX)$ intersects on $AO$ . $\square$

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andria

#10 h Apr 1, 2017, 8:52 AM • 3 Y

Y by mijail, Adventure10, Mango247

From Two parallel lines $\angle BXE=\angle CXD=90^{\circ}$ hence if $Y=\odot(CEX)\cap \odot(BDX)$ then:
$90^{\circ}=\angle BXE=\angle A+\angle ABX+\angle AEX=\angle A+\angle DYX+\angle CYX=\angle A+\angle DYC$ $\Longrightarrow \angle DYC=90^{\circ}-\angle A\ \clubsuit$ On the other hand $\angle EOC=\angle OCB=90^{\circ}-\angle A$ ; So combining this with $\clubsuit\Longrightarrow YCOD$ is cyclic. Let $AB\cap \odot(YCOD)=C'$ then $\angle AC'C=\angle DYC=90^{\circ}-\angle A\Longrightarrow \angle C'CA=90^{\circ}$ thus we conclude that $C'DOY$ is cyclic ; similarly $B'EOY$ is cyclic so because $DEB'C'$ is cyclic $\Longrightarrow AD.AC'=AE.AB'\Longrightarrow A,O,Y$ are collinear.
Q.E.D

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#11 h Apr 9, 2017, 3:52 AM • 3 Y

Y by don2001, Adventure10, Mango247

Fairly long solution, but it exhibits some interesting properties of the configuration that haven't been discussed in other solutions.

First, prove that $\angle CXD = \angle BXE = 90^{\circ}$ as has been done numerous times in this thread.

Let $\Omega$ be the circumcircle of $ABC$ and let $AH$ cut $\Omega$ at $A, F$ . Let $M$ be the midpoint of $BC$ and let the line through $F$ parallel to $HM$ cut $AO$ at $P$ . We shall prove that $BDXP$ and $CEXP$ are cyclic which is of course sufficient.

Notice that $DE$ is the perpendicular bisector of $AF$ and thus $\angle DFO = \angle DAO = \angle DBO$ . Thus $B, D, O, F$ are concyclic. Analogously $C, E, O, F$ are concyclic. Using this fact we find that

$\angle BFE = \angle BFO + \angle EFO = \angle ECO + \angle ABE = \angle CAO + \angle ABC = \angle BAH + \angle ABC = 90^{\circ}$
Thus $F, B, X, E$ are concyclic and analogously $F, C, X, D$ are concyclic.

Now, let $Q = BE \cap CD, Y = (BDO) \cap BE, Z = (CEO) \cap CD$ , then $\angle QYD = \angle BOD = \angle COE = \angle QZE$ which implies $D, E, Y, Z$ are concylic. Thus

$\frac{QY}{QZ} = \frac{QD}{QE} = \frac{QC}{QB} \implies QY \cdot QB = QZ \cdot QC$
Which implies $Q$ lies on radical axis $OF$ of $(BDO), (CEO) \implies O, Q, F$ , are collinear.

Now, by Ceva, $A, Q, M$ are collinear, and since $OM$ is parallel to $AF$ we get $\frac{OQ}{OF} = \frac{QM}{QA} = \frac{OM}{AF} = \frac{AX}{AF} = \frac{OX}{FP}$ using the well known fact that $AH = 2OM$ ( $XOMH$ is a parallelogram). It follows that $X, Q, P$ are collinear.

Now let $\mathcal{I}$ be the composition of the inversion with center $A$ and radius $\sqrt{AD \cdot AC}$ with the reflection about the angle bisector of $\angle BAC$ , and let $\mathcal{I}(W) = W'$ for any point $W$ . Then it's clear that $D = C', E = B'$ , so $Q'$ is the intersection of the circumcircles of $ABE, ACD$ . Now notice that rays $AH$ and $AO$ are swapped by $\mathcal{I}$ since $O, H$ are isogonal conjugates. Moreover, as triangles $ADF$ and $AOC$ are isosceles with $\angle FAD = \angle CAO$ we have:

$\frac{AD}{AF} = \frac{AO}{AC} \implies AD \cdot AC = AO \cdot AF = AX \cdot AP$
Where the last equality follows from $OX$ being parallel to $PF$ . Thus $X = P'$ . Now notice that

$\angle BQC = \angle BXC + \angle XBQ + \angle XCQ = (\angle CXE + \angle CXD - \angle BXE) + \angle XFE + \angle XFD = 180^{\circ} - \angle DXE + \angle BAC$
On the other hand, as $Q' = (ABE) \cap (ACD)$ , it must also lie on $(DQB)$ and $EQC$ . It follows that

$\angle DQ'E = \angle DQ'Q + \angle EQ'Q = \angle DBQ + \angle ECQ = \angle BQC - \angle BAC = 180^{\circ} - \angle DXE$
From which it follows that $X, D, Q', E$ are concyclic and hence $Q, B, P, C$ are concyclic.

Now, let $\angle ABC = \beta, \angle PBC = \alpha$ , then $\angle DBP = \alpha + \beta$ and it suffices to prove that $\angle DXP = 180^{\circ} - \alpha - \beta$ . But $\angle XDQ = \angle XFC = \angle ABC = \beta$ and $\angle DQX = \angle PQC = \angle PBC = \alpha$ , so we are finally done.

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#12 h Apr 11, 2017, 7:29 AM • 3 Y

Y by anantmudgal09, don2001, Adventure10

WLOG $AB<AC$ , and define $\{A,H'\}\equiv AH\cap \odot(ABC)$ , $P\equiv AH\cap DE$ , $Q\equiv CH\cap AB$ , $V\equiv BE\cap CD$ , $U \equiv AO\cap XV$ . Note that by reflection about $O$ , $H'D$ passes through the antipode of $C$ with respect to $\odot(ABC)$ , so $\angle CH'D=\angle CQD=\tfrac{\pi}{2}$ . Then $AX\cdot AH'=AH\cdot AP=AQ\cdot AD$ , so pentagon $CXQDH'$ is cyclic. Then $\angle AQX=\angle AH'D=\tfrac{\pi}{2}-\angle AH'C=\angle BCH$ , so $\{CH,CD\}$ are isogonal with respect to $\angle XCB$ and consequently $\angle BCV=\angle XCQ=\angle XDQ$ .

I now claim that $U\in\odot(BVC)$ ; note that $\{H,V\}$ are isogonal conjugates in $\triangle XBC$ . Let $V'$ be the point on $\odot(BVC)$ with $VV'\parallel BC$ ; notice that $\overline{AOV'}$ are collinear from negative homothety through $V$ . Then
$\begin{align*} \angle AUX&=\pi-\angle AXV-\angle XAO\\ &=\angle HXV-\angle XAO\\ &=\angle BXC-2\angle BXH-\angle B+\angle C\\ \end{align*}$ and $\angle VCV'=\pi-\angle VV'C-\angle CVV'=\angle VBC-\angle VCB=\angle XBH-\angle XCH$ so we want to show that $\angle BXC-2\angle BXH-\angle B+\angle C=\angle XBH-\angle XCH$ $\Longrightarrow \angle BXC-\angle BXH=\angle B-\angle XCH$ $\Longrightarrow \angle CXH+\angle XCH=\angle B$ which is true, as desired. Then $\angle BUX=\angle BUV=\angle BCV=\angle XDQ=\pi-\angle BDX$ , and we're done.

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#13 h Apr 19, 2017, 10:13 PM • 1 Y

Y by Adventure10

Is there a good complex solution to this? I tried it on the contest but got too messy to do by hand.

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#14 h Jul 18, 2018, 5:29 PM • 3 Y

Y by Adventure10, Mango247, Lcz

Let $P$ be the second intersection of $(BDX)$ and $(CEX)$ , and perform an inversion about $A$ with radius $\sqrt{AB\cdot AE}=\sqrt{AC\cdot AD}$ and reflect across the angle bisector of $\angle BAC$ . We see that $B\to E$ and $D\to C$ , while lines $AX$ and $AO$ swap. We look at this and say it would be really nice if $X$ and $P$ are inverses of each other. In fact, this is true. We show this using complex numbers.

Let $(ABC)$ be the unit circle so $A=a,B=b,C=c$ . Then

$d+ab\bar{d}=a+b,d+bc\bar{d}=0 \implies d=\frac{c(a+b)}{c-a},$
while

$x=a+\frac{b+c}{2}.$
Let $Q$ be the inverse of $X$ under this inversion: we have

$(q-a)(x-a)=(c-a)(d-a) \implies (q-a)\left(\frac{b+c}{2}\right)=(c-a)\frac{ac+bc-ac+a^2}{c-a}=a^2+bc$
$q-a=\frac{2(a^2+bc)}{b+c}.$
We want to show that $Q,B,X,D$ are concyclic; this is equivalent to $q-a,b-a,x-a,d-a$ being concyclic. These are

$\frac{2(a^2+bc)}{b+c},b-a,\frac{b+c}{2},\frac{a^2+bc}{c-a}.$
The cross ratio is

$\begin{align*} \frac{\frac{ \frac{2(a^2+bc)}{b+c}-\frac{b+c}{2} }{ \frac{2(a^2+bc)}{b+c}-\frac{a^2+bc}{c-a} }}{\frac{ (b-a)-\frac{b+c}{2} }{ (b-a)-\frac{a^2+bc}{c-a} }} &= \frac{\frac{ \frac{4(a^2+bc)-(b+c)^2}{2(b+c)} }{ \frac{(a^2+bc)(2(c-a)-(b+c)}{(b+c)(c-a)} }}{\frac{ \frac{2(b-a)-(b+c)}{2} }{ \frac{(b-a)(c-a)-(a^2+bc)}{c-a} }}\\ &= \frac{\frac{ \frac{4a^2-(b-c)^2}{2(b+c)} }{ -\frac{(a^2+bc)(2a+(b-c)}{(b+c)(c-a)} }}{\frac{ -\frac{2a-(b-c)}{2} }{ -\frac{a(b+c)}{c-a} }}\\ &= \frac{\frac{ \frac{(2a+(b-c))(2a-(b-c))}{2(b+c)} }{ -\frac{(a^2+bc)(2a+(b-c))}{(b+c)(c-a)} }}{\frac{ -\frac{2a-(b-c)}{2} }{ -\frac{a(b+c)}{c-a} }}\\ &= -\frac{(2a+(b-c))(2a-(b-c))(b+c)(c-a)(2)(a)(b+c)}{(2)(b+c)(a^2+bc)(2a+(b-c))(2a-(b-c))(c-a)}\\ &= -\frac{a(b+c)}{(a^2+bc)}, \end{align*}$
the conjugate of which is

$-\frac{\frac{1}{a}\frac{b+c}{bc}}{\frac{a^2+bc}{a^2bc}}=-\frac{a(b+c)}{a^2+bc},$
so it is real, and the inverse of $X$ lies on $(BDX)$ and hence $(CEX)$ . This point is obviously on $AO$ , finishing the proof.

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#15 h Apr 6, 2019, 10:35 AM • 2 Y

Y by Adventure10, Mango247

After establishing that $\angle{BXE}=\angle{DXC}=90^{\circ}$ it is easy to finish with inversion. Let $Y=(CEX) \cap (BDX).$ Let $Z'$ be the image of $Z$ under an inversion with pole $A,$ power $\sqrt{AC\cdot AD},$ followed by a reflection over the bisector of angle ${A}.$ Clearly $Y'$ lies inside triangle ${ADE}.$
$360^{\circ}-\angle{BY'E}=\angle{AY'B}+\angle{AY'E}=\angle{AEY}+\angle{ABY}=360^{\circ}-\angle{BAC}-\angle{BYE} \implies \angle{BY'E}=\angle{BAC}+\angle{BYE}$ $\angle{BYE}=\angle{BYX}+\angle{EYX}=\angle{ADX}+\angle{ACD}=90^{\circ}=\angle{BAC} \implies \angle{BY'E}=90^{\circ}$ Analogously, $\angle{CY'D}=90^{\circ},$ which means that $Y'$ is the intersection of circles $(DXC)$ and $(BXE)$ which lies inside $\triangle{ADE},$ but this is exactly $X,$ hence the conclusion.

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#16 h Mar 11, 2020, 7:46 AM • 3 Y

Y by sameer_chahar12, mueller.25, amar_04

This was a really nice prblem. One of the nicest configurations ive seen in a while. Here goes my solution.

Proof: Define $F$ as the required point of intersection. We begin with a few claims.

Claim 1: If $I=\overline{DH'} \cap \overline{BC}$ then $\angle IHC=90^\circ$ .

Proof: For this notice that as $H'$ is the reflection of $H$ over $\overline{BC}$ $\implies$ $\angle CHI=\angle CH'D$ . But now notice that trivially we have that $DA=DH'$ hence $\angle CH'D=\angle DH'A+\angle AH'C=\angle B+90^\circ-\angle B=90^\circ$ . $\square$ .

Now we bring the point $X$ in the picture.
Claim 2: $\angle CXD=90^\circ$ .

Proof: For this notice that by claim1 it just suffices to show that $(DXCH')$ is cyclic. Now we use phantom points. Define $D=\odot(CXH') \cap \overline{AB}$ . Then we need to show that $\overline{OD}\parallel \overline{BC}$ . Now i present a proof told to me by amar_04.

Now observe that it suffices to show that $\overline{AD}=\overline{DH'}$ . Alter the labelling a bit. Let $\overline{CE}\perp \overline{AB}$ and $\overline{AQ}\perp \overline{BC}$ where $E\in \overline{AB}$ and $Q\in \overline{BC}$ . Note That $\overline{HX}\cdot \overline{HH'}=\overline{HA}\cdot \overline{HQ}=\overline{HE}\cdot \overline{HC}$ . So, $E\in\odot(CH'X)$ . So now draw a line $\ell$ to $BC$ though $H'$ meeting $\overline{AB}$ at $T$ . So, $\angle HTE=\angle EH'X=\angle XDA\implies \overline{AD}=\overline{DX}=\overline{DH'}$ $\square$ .

Now back to the main problem. Observe by claim 2 we have that $(CXDH')$ is cyclic. But also $(CHIH')$ is cyclic by claim 1. Hence we have that there exists a spiral similirity at $C$ such that $\overline{HI} \mapsto \overline{XD}$ $\implies$ $\triangle {XCD} \sim \triangle {HCI}$ $\implies$ $\angle XCD=\angle HCI=\angle ACO$ and as we also know that $\angle DAX=\angle OAC$ so we can directly conclude that $(X,O)$ are isogonal conjugates w.r.t $\triangle {ADC}$ . So we may have that $\angle EFB=\angle XFB+\angle XFE=\angle ADX+\angle XCE=\angle EDC+\angle OCD=\angle DCB+\angle OCD=\angle OCB=\angle OBC=\angle DOB$ . Hence we have that $(OEFB)$ is cyclic. Similiarly conclude that $(ODCF)$ is cyclic. Now we state a lemma which finishes the problem.

Lemma: If $\odot(OEB)\cap \odot(OCD)=F$ then $O \in \overline{AF}$ .

Proof: Observe that $\angle EFB=\angle DOB=\angle EOC=\angle DFC$ . So we have that $(\overline{FD},\overline{FE})$ are isogonal w.r.t $\angle BFC$ . Now applying the isogonal line lemma to $(\overline{FD},\overline{FE})$ w.r.t $\angle BFC$ we have that if $G=\overline{CD}\cap \overline{BE}$ then $(\overline{FG},\overline{FA})$ are isogonal. But notice that $\angle BFC=\angle OFB+\angle OFC=\angle EBC+\angle DCB=180^\circ -\angle BGC$ . So we have that $(BGCF)$ is cyclic. Now just notice that $\angle GFC=\angle GBC=\angle EBC=\angle OEB=\angle OFB$ with which we have that $(\overline{FG},\overline{FO})$ are isogonal w.r.t $\angle BFC$ . Now earlier we already had $(\overline{FG},\overline{FA})$ are isogonal with which we are done. $\blacksquare$ .
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draw(circle((-4.974307769629735,-6.808894958451459), 13.748640203456471), linewidth(0.8)); draw((-4.7801760376472116,13.950725144137508)--(8.336947610659198,-10.249183848185663), linewidth(1.2) + linetype("4 4")); draw(circle((1.3557146108993223,-4.335137577706486), 9.149511325000091), linewidth(0.4) + linetype("4 4")); draw((-4.771162566203684,6.938244361073334)--(8.336947610659198,-10.249183848185663), linewidth(0.4)); draw((8.336947610659198,-10.249183848185663)--(-7.56,-2.28), linewidth(0.4)); draw((2.477103169504549,4.745393529389301)--(8.336947610659198,-10.249183848185663), linewidth(0.4)); draw((-4.771162566203684,6.938244361073334)--(-6.3587113083741285,4.734036441371462), linewidth(0.4) + wrwrwr); draw((-4.771162566203684,6.938244361073334)--(8,-2.26), linewidth(0.4)); draw((0.21098652855647299,4.742480783064174)--(8,-2.26), linewidth(0.4)); draw((-4.771162566203684,6.938244361073334)--(2.477103169504549,4.745393529389301), linewidth(0.4)); draw(circle((-3.068492691278137,0.5606330775406088), 5.314398760492078), linewidth(0.4)); 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label("$D$", (-6.171059813296446,5.09852905289869), NE * labelscalefactor); dot((2.477103169504549,4.745393529389301),linewidth(4pt) + dotstyle); label("$E$", (2.665951167836507,5.09852905289869), NE * labelscalefactor); dot((8.336947610659198,-10.249183848185663),linewidth(4pt) + dotstyle); label("$F$", (8.52812676918213,-9.906890583381548), NE * labelscalefactor); dot((-4.7564880054789525,-4.478563882765253),linewidth(4pt) + dotstyle); label("$H'$", (-4.596146965173741,-4.132210140264955), NE * labelscalefactor); dot((-5.1393956738612,-2.276888683385425),linewidth(4pt) + dotstyle); label("$I$", (-4.946127598089897,-1.9448311845389725), NE * labelscalefactor); dot((-1.1581917939609276,2.200896647923807),linewidth(4pt) + dotstyle); label("$G$", (-0.965097898668617,2.561169464256551), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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#17 h Feb 10, 2021, 6:37 PM • 1 Y

Y by p_square

Claim 1. $\angle BXD=\angle CXE$

We present two proofs of this claim but before that let's see how this claim finishes.
Perform $\sqrt{AB\cdot AE}$ inversion with center $A$ and reflect across the angle bisector of $\angle BAC$ . Observe that this transformation maps $B$ to $E$ and $C$ to $F$ . Let $X$ go to a point $X'$ under the inversion. Now, note that $X'$ lies on $AO$ as $AO$ and $AH$ are isogonal. Also, $\angle CXE=\angle BXD=\angle BXA-\angle DXA=\angle AEX'-\angle ACX'=\angle EX'C$ . Thus, $CEXX'$ are cyclic. Similarly, $BDXX'$ are collinear and we get the desired result.
Now, we prove claim 1.
Let $D',E'$ be the feet of perpendiculars from $D,E$ respectively to $BC$ and let $H$ be the orthocenter and $T$ be the foot of perpendicular from $A$ to $BC$ .

Remember that $AX=DD'=EE'=XH$ so $EE'AX, DD'AX, EE'XH, DD'XH$ are all parallelograms.

Proof 1(by p_square):
We claim that $H$ is the orthocenter of $E'XC'$ .
Proof: We have that $CH\perp AB=AE\parallel XE'\implies E'X\perp CH$ and we have $XH\perp CE'$ . Thus, we get that $EX\parallel E'H\perp XC\implies \angle EXC=90$ . Similarly, $\angle BXD=90\implies \angle BXD=\angle CXE$ .
Proof 2(Mine):Now, let $U$ be a point on $AH$ such that $\frac{TA}{TX}=\frac{TU}{TH}$ .
$\angle DXB=\angle DXT-\angle BXT=\angle D'HT-\angle BXT=\angle BUT-\angle BXT=\angle XBU$ . Similarly, we have $\angle CXE=\angle XCU$ . Thus, we want $\angle XBU\angle XCU$ . Let the reflection of $B$ in $T$ be $B'$ . Now, $TX\cdot TU=TA\cdot TH=BT\cdot TC=TB'\cdot TC$ . Thus, we have that $XUB'C$ is cyclic. So, $\angle BXD=\angle XBU=\angle XB'U=XCU=\angle CXE$ and thus, we are done.

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#18 h Apr 11, 2021, 7:28 AM • 2 Y

Y by tapir1729, math_comb01

Very cool. The key idea is that the problem inverts to itself with the two intersections of $(BDX)$ and $(CEX)$ swapped. First, a claim:

Claim: $\angle BXE=\angle CXD=90^\circ$ .
Proof

$[asy] size(6cm); defaultpen(fontsize(10pt)); pair A,B,C,H,O,D,EE,X,Y; A=dir(110); B=dir(220); C=dir(320); H=A+B+C; O=origin; D=extension(A,B,O,O+C-B); EE=extension(A,C,O,O+C-B); X=(A+H)/2; Y=reflect(circumcenter(B,D,X),circumcenter(C,EE,X))*X; draw(A--H); draw(A--Y,dashed); draw(circumcircle(B,D,X),gray); draw(circumcircle(C,EE,X),gray); draw(unitcircle); draw(D--EE); draw(A--B--C--cycle,linewidth(1)); dot("\(A\)",A,A); dot("\(B\)",B,dir(210)); dot("\(C\)",C,dir(-30)); dot("\(H\)",H,S); dot("\(O\)",O,SW); dot("\(D\)",D,W); dot("\(E\)",EE,NE); dot("\(X\)",X,dir(305)); dot("\(Y\)",Y,dir(300)); [/asy]$

From the claim, we deduce $\measuredangle BXD=\measuredangle EXC$ . Denote by $\Psi$ inversion at $A$ with radius $\sqrt{AB\cdot AE}=\sqrt{AC\cdot AD}$ followed by reflection over the bisector of $\angle BAC$ , so $\Psi$ swaps $(B,E)$ and $(C,D)$ .

Let $Y=\Psi(X)$ lie on $\overline{AO}$ . Then $\measuredangle BYD=\measuredangle BYA+\measuredangle AYD=\measuredangle XEA+\measuredangle ACX=\measuredangle EXC=\measuredangle BXD$ so $Y$ lies on $(BDX)$ . Similarly $Y$ lies on $(CEX)$ , so $Y$ is the desired intersection.

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#19 h Apr 11, 2021, 12:45 PM

Y by

This is routine with complex numbers once you guess the intersection (which is not easy).

Solution

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#20 h Sep 28, 2022, 9:55 PM

Y by

Let $AH,BH,CH$ hit $BC,AC,AB$ at $A_1,B_1,C_1$ , now let $AH$ hit $(ABC)$ again at $H_A$ now by PoP we have that $XH \cdot HA_1=AH\cdot HA_1=BH \cdot HB_1=CH \cdot HC_1$ so $XBH_AB_1, XCH_AC_1$ are cyclic. Now consider an inversion with reflection that sends $D$ to $C$ and $B$ to $E$ , this gives that $O,H_A$ are inverses so $DBH_AO, ECH_AO$ are cyclic and by angle chase.
$\angle BH_AD=\angle BOD=\angle OBC=90-\angle BAC=\angle OCB=\angle EOC=\angle CH_AE$ Now note that $H_A,A$ are symetric w.r.t $DE$ so $\angle BAC=\angle DH_AE$ so $\angle BH_AE=\angle CH_AD=90$ and that means $CH_AC_1DX, BH_AEB_1X$ are cyclic so $\angle BXE=90=\angle CXD$ so $\angle BXD=\angle CXE$ and since $DE \parallel BC$ we get $(DXE), (BXC)$ tangent and also inverting gives $(DOC) \cap (BOE)=X'$ and $(DX'E), (BX'D)$ are tangent at $X'$ and $A,O,X'$ colinear and $\triangle BDX \sim \triangle ECX'$ so now $\angle DXB=\angle EX'C=\angle DX'B$ hence $DBX'X$ is cyclic but inverting gives $CEXX'$ cyclic so $(BDX),(CEX)$ meet at $X'$ which lies in $AO$ thus we are done

Note: I wrote this at the hospital with one hand because yesterday i had finger surgery, it went good so i still alive lol.

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#21 h Dec 20, 2023, 4:24 PM

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Claim: $\angle BXE = \angle CXD = 90^\circ$ .

Proof: Let $P$ be the foot from $E$ to $BC$ , so that $AXPE$ is a parallelogram. Let $Q$ be the reflection of $C$ over $P$ , so that $HQ\parallel AC$ . Let $R$ be the reflection of $Q$ over $E$ , so that $AHCR$ is a parallelogram. Since $AR\parallel HC\perp AB$ and $CR\perp CB$ , $R$ is actually the antipode of $B$ in $(ABC)$ . Then we have $\angle QBR = 90^\circ - \angle A = \angle HBA$ and $\angle BQR = 180^\circ - \angle C = \angle BHA$ , so $\triangle BQR\sim\triangle BHA$ . Hence, we also have $\triangle BQE\sim\triangle BHX$ , inducing the second spiral similarity $\triangle BHQ\sim\triangle BXE$ , so $\angle BXE = \angle BHQ = 90^\circ$ as desired, and $\angle CXD = 90^\circ$ follows by symmetry.

Note that the claim implies $\angle DXB = \angle CXE$ . Now, if $T$ is the point on $AO$ satisfying $\triangle CAX\sim\triangle BAT$ , we have $\angle BTD = \angle CXE = \angle BXD$ , so $T$ lies on $(BDX)$ . Hence by PoP, the other intersection $T'$ of $(BXD)$ with $AO$ satisfies

$AT' = \frac{AB\cdot AD}{AT} = \frac{AB\cdot AD}{AX\cdot \frac{AB}{AC}} = \frac{AD\cdot AC}{AX} = \frac{\sqrt{AB\cdot AC\cdot AD\cdot AE}}{AX}$
which is symmetric in $B$ and $C$ , as desired.

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#22 h Dec 25, 2023, 6:18 PM

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Very Nice Problem!
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15 cm); 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 = -3.582269769522469, xmax =9.3495599740257218, ymin = -6.750263899423706, ymax = 4.296190961229847; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen ffvvqq = rgb(1,0.3333333333333333,0); pen ccqqqq = rgb(0.8,0,0); pen ttffqq = rgb(0.2,1,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw((-0.9162004708649716,4.002019967362306)--(-1.54,-2.25)--(6.4,-1.69)--cycle, linewidth(0.4) + zzttff); /* draw figures */ draw((-0.9162004708649716,4.002019967362306)--(-1.54,-2.25), linewidth(0.4) + zzttff); draw((-1.54,-2.25)--(6.4,-1.69), linewidth(0.4) + zzttff); draw((6.4,-1.69)--(-0.9162004708649716,4.002019967362306), linewidth(0.4) + zzttff); draw(circle((2.2537760815942054,0.5286034145393064), 4.702485926886521), linewidth(0.8) + ffvvqq); draw((xmin, 0.07052896725440806*xmin + 0.3696469150817805)--(xmax, 0.07052896725440806*xmax + 0.3696469150817805), linewidth(0.4)); /* line */ draw(circle((4.501891557302407,-1.575832419914503), 6.079387758338865), linewidth(0.4) + blue); draw(circle((-0.4654900211539337,-7.461514690475687), 8.969132346717876), linewidth(0.4) + ccqqqq); draw((-0.9162004708649716,4.002019967362306)--(8.418109611952028,-6.225799265460478), linewidth(0.4)); draw((-0.7399765524591768,1.503416552822999)--(2.2537760815942054,0.5286034145393064), linewidth(0.4)); draw((-0.5637526340533822,-0.9951868617163078)--(8.418109611952028,-6.225799265460478), linewidth(0.4)); draw((2.2537760815942054,0.5286034145393064)--(2.43,-1.97), linewidth(0.4)); draw((-0.5637526340533822,-0.9951868617163078)--(5.423752634053378,-2.9448131382836977), linewidth(0.4)); draw(circle((3.2243161541048657,-4.775751493570609), 5.392414025604744), linewidth(0.4) + ttffqq); draw(circle((0.9920346646962718,-6.8130193744306435), 7.449256109078176), linewidth(0.4) + ttffqq); draw(circle((0.8450117237704114,-0.23329172358850042), 2.35124296344326), linewidth(0.4) + fuqqzz); draw((-1.54,-2.25)--(3.4407305698579895,0.6123180887745355), linewidth(0.4)); draw((-1.2876844867144508,0.2788278580842878)--(6.4,-1.69), linewidth(0.4)); draw((-1.54,-2.25)--(8.418109611952028,-6.225799265460478), linewidth(0.4)); draw((8.418109611952028,-6.225799265460478)--(6.4,-1.69), linewidth(0.4)); draw((8.418109611952028,-6.225799265460478)--(3.4407305698579895,0.6123180887745355), linewidth(0.4)); draw((8.418109611952028,-6.225799265460478)--(-1.2876844867144508,0.2788278580842878), linewidth(0.4)); draw((-0.7399765524591768,1.503416552822999)--(8.418109611952028,-6.225799265460478), linewidth(0.4)); draw(circle((3.103865690898049,0.5190026237808545), 3.9678954088420353), linewidth(0.4) + linetype("4 4") + blue); draw(circle((-3.230341837388172,0.0722575084357314), 2.8723049042034083), linewidth(0.4) + linetype("4 4") + ccqqqq); draw((xmin, -14.178571428571441*xmin-8.98839385168748)--(xmax, -14.178571428571441*xmax-8.98839385168748), linewidth(0.4)); /* line */ draw((xmin, 1.2853434304193623*xmin-0.27057111715418225)--(xmax, 1.2853434304193623*xmax-0.27057111715418225), linewidth(0.4)); /* line */ draw(circle((0.9503652849289946,-0.818840955612733), 2.872304904203409), linewidth(0.4) + blue); /* dots and labels */ dot((-0.9162004708649716,4.002019967362306),dotstyle); label("$A$", (-0.8552137275571697,3.913076919819897), NE * labelscalefactor); dot((-1.54,-2.25),dotstyle); label("$B$", (-1.477774044848339,-2.0890430622693166), NE * labelscalefactor); dot((6.4,-1.69),dotstyle); label("$C$", (6.455879229349382,-1.5303350852131397), NE * labelscalefactor); dot((2.2537760815942054,0.5286034145393064),linewidth(4pt) + dotstyle); label("$O$", (2.321440199133668,0.6566075678353237), NE * labelscalefactor); dot((-0.5637526340533822,-0.9951868617163078),linewidth(4pt) + dotstyle); label("$H$", (-0.5040258562647153,-0.8598855127457276), NE * labelscalefactor); dot((-1.2876844867144508,0.2788278580842878),linewidth(4pt) + dotstyle); label("$D$", (-1.2223646839083722,0.4011982068953572), NE * labelscalefactor); dot((3.4407305698579895,0.6123180887745355),linewidth(4pt) + dotstyle); label("$E$", (3.502708493481015,0.7364229931290632), NE * labelscalefactor); dot((-0.7399765524591768,1.503416552822999),linewidth(4pt) + dotstyle); label("$X$", (-0.6796197919109425,1.6303557564189461), NE * labelscalefactor); dot((8.418109611952028,-6.225799265460478),linewidth(4pt) + dotstyle); label("$F$", (8.48319103181037,-6.095777412015042), NE * labelscalefactor); dot((5.423752634053378,-2.9448131382836977),linewidth(4pt) + dotstyle); label("$A'$", (5.482131040765758,-2.8233449749717203), NE * labelscalefactor); dot((2.43,-1.97),linewidth(4pt) + dotstyle); label("$M$", (2.4970341347798954,-1.8495967863880978), NE * labelscalefactor); dot((-1.228100235432486,0.8760099836811528),linewidth(4pt) + dotstyle); label("$P$", (-1.1585123436733804,1.0077954391277777), NE * labelscalefactor); dot((2.7418997645675143,1.1560099836811528),linewidth(4pt) + dotstyle); label("$N$", (2.800332750896106,1.279167885126492), NE * labelscalefactor); dot((1.5817007453631895,-0.4560261269164034),linewidth(4pt) + dotstyle); label("$J$", (1.650990626666255,-0.33310370580704657), NE * labelscalefactor); dot((-0.0448438436610229,-8.352572211207967),dotstyle); label("$A'_{1}$", (-8.182269769522469,4.296190961229847), NE * labelscalefactor); dot((-0.397291680472613,-3.3553653821293556),linewidth(4pt) + dotstyle); label("$K$", (-0.3284319206184881,-3.2224221014404177), NE * labelscalefactor); dot((1.725249073262489,1.9469664450008537),linewidth(4pt) + dotstyle); label("$L$", (1.7946583921949864,2.0773221380638875), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
We first give 2 solutions
Solution 1:
Let $K$ be the reflection of orthocenter in $BC$ , and Let $BL \perp AC$ , Let $F$ be the intersection of the circles.
Claim 1: $\measuredangle BXE = \measuredangle CXD = 90^\circ$
Proof
Now just invert at $A$ with radius $\sqrt{AB \cdot AE}$ , then do some angle chasing to get the inverse of $X$ is $F$ . Hence we're done.
$\quad$
$\quad$
Solution 2:
Let $(COD)$ and $(BOE)$ intersect at $F'$ , $(FBD),(FCE)$ intersect at $X'$
Claim: $FA,FD$ are isogonal in $BFC$
Claim 2 $A-O-F$
CLaim 3 $FO$ and $FJ$ are isogonal where $J = BE \cap CD$
Claim 4: $F-J-X$
Claim 5: $AX \perp BC$
ALL the proofs of claims are angle chase or DDIT.
Hence we're done
Fact: While playing around with this problem, I found out that under $\sqrt{XB \cdot XC}$ inversion $H$ and $F$ are interchanged. I will be interested in any solution using that.
Remark

This post has been edited 2 times. Last edited by math_comb01, Dec 25, 2023, 6:21 PM

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#23 h Mar 26, 2025, 11:23 AM

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Let $AH\cap (ABC)=F$ and $(BOE)\cap (COD)=P,(BOE)\cap (ABC)=Q,QC\cap AF=W$ . Let $EF$ be the altitude from $E$ to $BC$ . Let $C'$ be the antipode of $C$ on $(ABC)$ . We will show that $P\in AO,O\in (CEX)$ and similarily $P\in (CDX)$ .
Claim: $A,O,P$ are collinear.
Proof: If $D^*,E^*,P^*$ are the images of $D,E,P$ under the inversion around $(ABC)$ , then $F,B,D^*$ and $F,C,E^*$ are collinear. DDIT at $P^*BFC$ gives $(\overline{AD^*},\overline{AE^*}),(\overline{AB},\overline{AC}),(\overline{AF},\overline{AP^*})$ is an involution which must be reflection over the angle bisector of $\measuredangle CAB$ thus, $P^*$ lies on $AO$ .
Claim: $E,C,P,W$ are concyclic.
Proof: First notice that $\measuredangle BQE=90-\measuredangle A=\measuredangle BQC'$ hence $Q,E,C'$ are collinear and $\measuredangle CQE=90$ .
$\measuredangle APQ=\measuredangle OPQ=90-\measuredangle QAB=\measuredangle AWQ$ Thus, $A,W,Q,P$ are concyclic. Since $E$ lies on the perpendicular bisector of $BC'$ , we have $EC'=EB$ and since $C'B\perp BC$ we observe $\measuredangle OBQ=180-\measuredangle QEO=\measuredangle (BC,EQ)=\measuredangle EBC$ so $\measuredangle OBE=\measuredangle QBC$ .
$\measuredangle WPE=\measuredangle WPO+\measuredangle OPE=\measuredangle WQA+\measuredangle OBE=180-\measuredangle B+\measuredangle QBC=180-\measuredangle ACQ$ Which proves the claim.
Claim: $X$ lies on $(ECPW)$ .
Proof: Let $A'$ be the reflection of $A$ over $E$ . Note that $\measuredangle AFA'=90$ and $VH=VF=VA'$ . Since $E,V,C,Q$ lie on the circle with diameter $EC$ , we get $\measuredangle FQC'=\measuredangle C=\measuredangle VQC'$ hence $F,V,Q$ are collinear. We have $\measuredangle BCQ=\measuredangle EVQ=90-\measuredangle VFA'=\measuredangle A'HF$ thus,
$\measuredangle XEA=\measuredangle HA'A=\measuredangle C-\measuredangle FA'H=\measuredangle BCQ-90=\measuredangle AWC$ As desired. $\blacksquare$

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