it's nice problem! Here my solution:(some projectivity) Let $k_1,k_2$ be circles passing through $A,C$ and tangents to $BE$ at the one point $P$, respectively. (1) Suppose that $CX$ be the perpendicular to $AB$. Let $AZ$ is altitude of the triangle $ABC$, $k$ be circumcircle of quadrilateral $AXZC$ and $BX'$, $BZ'$ are tangent to $k$ and $X',Y' \in k$, respectively $X'$ nearer $X$. So, we have \[ BP^2=BX\cdot BA=BZ\cdot BC=BX'^2=BZ'^2 \] and \[ BP=BX'=BZ' \] and $ \angle AX'C=90^\circ $, $A,X',Q$ are collinear. Easily get that $ \angle BX'C=\angle X'AC=\angle BX'P $ and $X',P,C$ are collinear $ \Rightarrow $ $X'=Y$. Hence the quadrilateral $AYXC$ is cyclic. (2) Suppose that $AYXC$ is cyclic. Let $k$ is circumcenter $AYXC$, $BY'$ is tangent to $k$, which $Y'$ is nearer $Y$. \[ BP^2=BX\cdot BA =BY'^2 \] and \[ BP=BY'=BQ \] and $ \angle PY'Q=90^\circ $. We have that $ \angle Y'CA=\angle Y'XA=\angle Y'QB$, because, $ BY'^2=BX\cdot BA=BP^2 $ $ \Rightarrow $ $ \angle BY'X=\angle BAQ=\angle BQX$ and $BXY'Q$ is cyclic. So $CEY'Q$ is cyclic and $ \angle CY'Q=90^\circ =PY'Q$ and $Y'=Y$. Hence, $ \angle CYA=90^\circ$ $ \Rightarrow $ $ \angle CXA=90^\circ $.

First part:(If $XYAC$ is cyclic then $AXC=90$) I assume that $BK$ is tangent to the circumcircle of $XYAC$ at $K$.Clearly,$BK=BP=BQ$ and thus angle $PKB=90$So quadrilateral $PKAE$ is cyclic.So angles $KAE=KPB$But as $BP=BK$ so angles $BPK=BKP$ So angles $BKP=KAE$.So $BKP-KAX=KAE-KAX$ .Now as $BK$ is tangent so angles $KAX=BKX$.So $BKP-BKX=KAE-KAX$ so angle $PKX=XAC$ But angles $XAC=XKC$.So $PKX=CKX$.Thus $P$,$C$ and $K$ are collinear.That means $K=Y$So angle $PYQ=90$So $AXC=AKC=90$ (Proved) Second part:(If $AXC=90$ then $XYAC$ is cyclic) I extend $AB$ upto $S$ such that $BX=BS$So $PXQS$ is a parellelogram.So $XPQ=PQS$.Now as $BP$ is a tangent.So $XPQ=XAP$.So $XAP=SAP=SQP$Thus $SQAP$ is a cyclic quadrilateral.We already have $CYQ=PYQ=90$Thus $BP=BY=BQ$.Thus $BYQ=BQY$.$BYQ=YAB+YBA$As $APSQ$ is cyclic so $YAB=QAS=QPS=XQP$.So $BYQ=YBA+XQP$.$YBA=BYQ-XPQ=BQY-XPQ=YQX$So quadrilateral $YXBQ$ is cyclic.So $XYA=180-YBE$So $XYA=90+A$As $ AXC=90$ so $XCA=90-A$Thus in quadrilateral $XYCA$ $XYA+XCA=180$.Thus $XYCA$ is cyclic.(Proved)

I think that there are gaps in above two solutions in proving math=inline]$AYXC$[/math] cyclic [math=inline]$\Rightarrow$[/math] [math=inline]$AX\perp XC$[/math part. mathuz wrote: We have that $ \angle Y'CA=\angle Y'XA=\angle Y'QB$, because, $ BY'^2=BX\cdot BA=BP^2 $ $ \Rightarrow $ $ \angle BY'X=\angle BAQ=\angle BQX$ and $BXY'Q$ is cyclic. As I understood, the last equality should be $\angle BY'X=\angle BAY'$, and I think that mathuz assumed $Y'\in QA$ here, which seems to be not clear from the rest of the solution. mathdebam wrote: First part:(If $XYAC$ is cyclic then $AXC=90$) I assume that $BK$ is tangent to the circumcircle of $XYAC$ at $K$.Clearly,$BK=BP=BQ$ and thus angle $PKB=90$So quadrilateral $PKAE$ is cyclic.So angles $KAE=KPB$But as $BP=BK$ so angles $BPK=BKP$ So angles $BKP=KAE$. I think that mathdebam did the similar mistake here, assuming $K\in QA$. Here is a solution for the part math=inline]$AYXC$[/math] cyclic [math=inline]$\Rightarrow$[/math] [math=inline]$AX\perp XC$[/math: If we fix the points $A,P,X$, then $B,Q$ are determined uniquely, and the whole figure is determined by the line $l=CY$ passing through $P$. We will invert with a circle centered at $A$. Denote the image of a point by lowercase letters of the name of the point. Let a circle $\omega=(o)$ be a image of the line $PQ$, then $a$ is a point on $\omega$. We can easily check from $BP^2=BQ^2=BX\cdot BA$ that $p$ and $q$ are tangency points of two tangents from $x$ to $\omega$. Clearly $c\in ao, y\in aq$. As $C,P,Y$ lie on a line $l$, $acpy$ is cyclic. By the assumption [$AYXC$ is cyclic], $y,x,c$ are collinear. The desired conclusion is $ac\perp cx$. From the configuration, it is well-known that $ax$ is a symmedian of triangle $aqp$. Let $q'$ be a point on $aq$ satisfying $xq=xq'$, and $p'$ be a point on $ap$ satisfying $xp=xp'$. By easy angle chasing, $\triangle aqp\sim\triangle ap'q'$, $p',x,q'$ lie on a line perpendicular to $ao$. Let $p'q'\cap ao=c'$. If $y=q'$, then $c=c'$ and the conclusion follows. If $y\ne q'$, $\angle pyx=\angle pyc=\angle pac=\angle pac'=\angle pq'x$, so $pxq'y$ is cyclic, $\angle ayp=180^{\circ}-\angle pxq'=\cdots=2\angle aqp-180^{\circ}$. This means $\angle APY=2\angle APQ-180^{\circ}$ in the original figure. Then $BP$ is an angle bisector of $\angle APC$, so $BP$ is an perpendicular bisector of $AC$. Therefore, $AB=BC$. This contradicts to the assumption.

semisimplicity wrote: In a triangle $ABC$, with $AB \ne BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \ne B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. Let $Y$ be the point of intersection of the lines $CP$ and $AQ$. Prove that the points $C, X, Y, A$ are concyclic if and only if $CX$ is perpendicular to $AB$. Suppose $CXYA$ is cyclic. Let $Z=AP \cap CQ$. Observe that $X$ is the $A$-HM point in $\triangle APQ$ so $\angle XCP=\angle XAY=\angle XQP$, hence $CPXQ$ is also cyclic. Now observe \begin{align*}\angle CYA=\angle AXP+\angle CXP=\angle APE+\angle CQP=\angle CZA\end{align*}hence $CZYA$ is cyclic. So $\angle QAP=\angle QCP$. But $PQ \perp AC$ hence $P$ is the orthocenter of $\triangle AQC$. Consequently, $$EA \cdot EC=EP \cdot EQ=EB^2-BX \cdot BA$$proving $CX \perp AB$. Conversely, suppose $CX \perp AB$. Then equate $EA \cdot EC=EP \cdot EQ$ as we did earlier, hence $P$ is the orthocenter in $\triangle AQC$. Consequently, $\angle CYA=\angle CXA=90^{\circ}$ and $CXYA$ cyclic follows.

I have a question, couldn't we just rephrase the problem to make it more simple? The following is seen when we notice that under the condtions $P$ must be the orthocenter of $ACQ$. We rephrase it into the following: IMOTC 2013, Rephrased wrote: Given the triangle $ACQ$, let $P$ be the orthocenter of the aforementioned triangle.Define point $B$ to be the midpoint of $QP$. Point $Y$ is defined as the intersection between $QA$ and the circle $C(B,BP)$, while $X$ is the intersection of the segment $BA$ and the circle defined by the points $ACY$. Let $E$ be the intersection of $QB$ with $AC$. Prove that $BE$ is tangent to $(APX)$. [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 = -22.615563337902902, xmax = 12.428618664251017, ymin = -12.497845536357767, ymax = 9.345837415657604; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffzztt = rgb(1,0.6,0.2); /* draw figures */ draw((-3.255978015520864,6.154157407886398)--(-19.307548531452394,-2.1753062111916175), linewidth(1.2) + blue); draw((-19.307548531452394,-2.1753062111916175)--(0.8653711710020958,-10.201091469157413), linewidth(1.2) + blue); draw((0.8653711710020958,-10.201091469157413)--(-3.255978015520864,6.154157407886398), linewidth(1.2) + blue); draw(circle((-12.229053225604495,-0.391600497516424), 7.299774097046311), linewidth(1.2) + linetype("2 2") + red); draw((-12.229053225604495,-0.391600497516424)--(-3.255978015520864,6.154157407886398), linewidth(1.2) + qqwuqq); draw((-12.229053225604495,-0.391600497516424)--(0.8653711710020958,-10.201091469157413), linewidth(1.2) + qqwuqq); draw((-19.307548531452394,-2.1753062111916175)--(-2.240759658011324,2.12534350492747), linewidth(1.2) + ffzztt); draw((-6.695494696710308,4.369327130079976)--(0.8653711710020958,-10.201091469157413), linewidth(1.2) + ffzztt); draw((-3.255978015520864,6.154157407886398)--(-8.310508295971845,-6.550472756490348), linewidth(1.2) + ffzztt); draw(circle((-1.1953034222593835,-2.023467030635507), 8.433262787131921), linewidth(1.2) + linetype("2 2") + red); draw((-8.353130253829232,2.4358419165089753)--(0.8653711710020958,-10.201091469157413), linewidth(1.2) + ffzztt); draw(circle((-5.9229566308024015,4.457308522098457), 3.161023738532947), linewidth(1.2) + red); draw((-8.353130253829232,2.4358419165089753)--(-5.150557919756596,1.3921052161587695), linewidth(1.2) + qqwuqq); /* dots and labels */ dot((-3.255978015520864,6.154157407886398),dotstyle); label("$A$", (-3.0898251595727477,6.556446247324706), NE * labelscalefactor); dot((-19.307548531452394,-2.1753062111916175),dotstyle); label("$Q$", (-19.15828977715229,-1.772440058120004), NE * labelscalefactor); dot((0.8653711710020958,-10.201091469157413),dotstyle); label("$C$", (1.0353307935955949,-9.826315966686822), NE * labelscalefactor); dot((-5.150557919756596,1.3921052161587695),linewidth(4pt) + dotstyle); label("$P$", (-4.97561073816399,1.7241207021846146), NE * labelscalefactor); dot((-12.229053225604495,-0.391600497516424),linewidth(4pt) + dotstyle); label("$B$", (-12.086593857435133,-0.0830904772986714), NE * labelscalefactor); dot((-2.240759658011324,2.12534350492747),linewidth(4pt) + dotstyle); label("$E$", (-2.068357971169158,2.431290294156335), NE * labelscalefactor); dot((-6.695494696710308,4.369327130079976),linewidth(4pt) + dotstyle); label("$Y$", (-6.547098720323358,4.670660668733451), NE * labelscalefactor); dot((-8.310508295971845,-6.550472756490348),linewidth(4pt) + dotstyle); label("$Z$", (-8.15787390203671,-6.251180807274235), NE * labelscalefactor); dot((-8.353130253829232,2.4358419165089753),linewidth(4pt) + dotstyle); label("$X$", (-8.197161101590696,2.745587890588211), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] To see this we just do an inversion around $(QPY)$, call it $\varphi$. Notice that $(QYP)$ and $(AYC)$ are orthogonal under the conditions, this implies that $X \xrightarrow{\varphi} A$. Meaning that $\angle BPX = \angle BAP = \angle XAP$, which gives us the tangency.