Hard geometry problem This is geometry problem in my textbook for contests, but without a solution. Also my professor can't solve it, so it is quite a challenge:
The triangle $ABC$ is given. The points $D$ and $E$ on the line $AB$ are such that $AD = AC$ and $BE = BC$, with the arrangement $D-A-B-E$. The circumscribed circles around triangles $DCB$ and $ECA$ intersect at the point $X \neq C$, and the circumscribed circles around triangles $DEC$ and $ABC$ intersect at $Y \neq C$. If $DY + EY = 2XY$ is true, determine the $\measuredangle ACB.$

So far, I chased the angles and tried with trigonometry, but soon it becomes too complicated. Textbook suggests that this problem is as hard as third on olympiad. I know fundamental trigonometric theorems(law of sin, law of cosine, etc) and I can understand a lot of solutions for other problems with similar difficulty, so feel free to use whatever you need to solve this.
Can someone help?
Thanks in advance :)
 A: Here is an alternative solution using inversion. It has two phases.
(1) First we show by inversion that $X$ is the center of $\odot(DCYE)$, and the ex-center $I_C$ of $\Delta ABC$.
(2) Using this information, with the given relation, $Y$, and $X$, and its reflection $X^r$ in $AB$ are on an ellipse with focars in $D,E$. We obtain a contradiction by a convexity argument, showing that $YX^r$ passes through an explicit point $Z$ (on the circumcircle of $\Delta ABC$ and its $C$-angle bisector) inside the ellipse, but in the opposite half-plane w.r.t. $AB$.
I will give all details and the related pictures.
The established geometrical properties are of interest also if we remove the given metric condition.

(1) Quick proof: Consider the point $x=I_C$, the ex-center of $\Delta ABC$ w.r.t. $C$. Let $\alpha,\beta,\gamma$ be the angle bisectors in this triangle, drawn below, intersecting in the incenter $I$. Then we have the angle situation:

The angle between the prolongation of $xA=I_CA$ and the side $AC$ in the isosceles triangle $\Delta ACD$ is $\hat A/2$, so it is also the $A$-angle bisector in it. Thus also the perpendicular bisector of $CD$, giving $xC=xD$. Similarly $xC=xE$.
Then the angle $\widehat BIX$ (exterior w.r.t. $\Delta BIC$) is $\hat B/2+\hat C/2$, so the angle in $x$ in $\Delta xIB=\Delta I_CB$ is its complement, $\hat A/2$, equal to $\widehat BDC$, so $BCDx$ is a cyclic quadrilateral. Similarly $ACEx$ cyclic. This gives $x=X$.
$\square$

(1) Proof by inversion. This is given, since some points constructed on the road ($(D^*)'$, $(E^*)'$ are related to the position of $Y$. (The exposition may seem long, but this is only because of details in defining the inversion and exhibiting bonus properties. It may be skipped, consider only the first proof instead.)
We denote by $*$ the inversion centered in $C$ with power $k^2=CA\cdot CB$. The image of a point $Z$ is thus denoted by $Z^*$. We will consider only the points $A,B,C,D$ first, and locate their $*$-values. First of all, $A^*$, $B^*$ are obtained easily, we have $CA^*=CB$, $CB^*=CA$, so the points $A^*,B^*$ are the reflections of $B,A$ w.r.t. the bisector $\gamma$ of the angle $\hat C$ in $\Delta ABC$. This reflection is also later helpful, let us denote it by a prime sign, or by $R:Z\to Z'$. (So $A^*=B'$ and $B^*=A'$.)

Let us find now the position of $D^*$. It is the image of $D$. Where is $D$ located? It is on the line $CD$, on the line $AB$, and on the circle $\odot(A,AC)$ centered in $A$ of radius $AC$. This circle hits the ray $(CA$ in $A_1$, so that $\Delta CDA_1$ has a right angle in $D$. The image of $A_1$ is $A_1^*$, the mid point of $CA^*$. Then the image $D^*$ is a/the point on $CD$, on the circle $\odot(CA^*B^*)=(AB)^*$, and on the perpendicular in $A^*_1$ on $CA$.
The $*$-image of the circle $\odot(CDB)$ is the line $D^*B^*$, and angle chasing shows:
$$
\begin{aligned}
\widehat{D^*B^*C} 
&= \widehat{D^*A^*C} &&\text{ since $CD^*A^*B^*$ cyclic, the image of $DAB$}\\
&= \widehat{DAC} &&\text{ since $\Delta D^*A^*C\sim\Delta DAC$ by inversion}\\
&= \widehat{DCA} = \frac 12\widehat{CAB} =\frac 12 \hat A\ .
\end{aligned}
$$
So $B^*D^*$ is the reflection $R:Z\to Z'$ of the angle bisector $\alpha$ in $A$. Because
$B^*=A'$, and the angles correspond.
Similarly, $A^*E^*$ is the reflection of the angle bisector $\beta$ in $B$. We can now localize $X^*$:
$$
\begin{aligned}
X^*
&=(\ \odot(DCB)\cap \odot(ECA)\ )^*\\
&=\odot(DCB)^*\cap \odot(ECA)^*\\
&=B^*D^*\cap C^*E^*\\
&=\alpha'\cap\beta'\\
&=(\alpha\cap\beta)'\\
&=I'\\
&=I\ ,\text{ the incenter $I=\alpha\cap\beta\cap\gamma$ of $\Delta ABC$.}
\end{aligned}
$$
In particular $X^*\in \gamma$, so $X\in\gamma$.
Recall now that $D^*$ is on $\alpha'$ and on $\odot(B^*A^*C)=\odot(A'B'C)=\odot(A'B'C')=\odot(ABC)'$.
So $(D^*)'=\alpha\cap \odot(ABC)$ is the intersection of the circumcircle of $\Delta ABC$ with its angle bisector in $A$. Similarly:
$(E^*)'=\beta\cap \odot(ABC)$.
This implies that $(D^*E^*)'$ intersects $\gamma'=\gamma$ perpendicularly in a point $X_1^*$, say, which is the mit point of $CI$.

(The angle is a right angle, since it is related to the sum of the measures of the arcs $(CD^*)'$, $(E^*)'A$, and the arc from $A$ to the circle intersection with the angle bisector $\gamma$. The intersection is the mid point, since seen from $(E^*)'$ the two angles against the catheti $CX_1^*$ and $IX_1^*$ correspond to two equal arcs on the drawn circle.)
So $D^*E^*$ intersects $\gamma$ in the mid point $X_1^*$ of $CI$. Its image by inversion is a point $X_1$ on the circle $\odot(CDE)$, also orthogonal to $\gamma$, so $CX_1$ is a diameter, and $X$ is the mid point on it. So $X$ is the center of $\odot(CDE)$.

For the convenience of the reader, here is a concluding picture for the arguments used so far, i give it since i have it, it may help to better figure out the idea of proof.

From the picture we get immediately:
$$
\begin{aligned}
\widehat{AXC} &=\widehat{AEC} =\frac 12\hat B\ ,\\ 
\widehat{CXB} &=\widehat{CDB} =\frac 12\hat A\ ,\\ 
\widehat{AXB} &=\frac 12(\hat A+\hat B)=\frac 12(180^\circ-\hat C)\ . 
\end{aligned}
$$

(2)
Now let $Y$ enter the stage. Recall that
$X$ is the center of $\odot(CDE)$, so:
$$XC=XD=XE=XY\ .$$
We show first (claim) that in the case of $\hat C=60^\circ$ we have
$$
XC=XD=XE=XY=YD=YE\ ,
$$
and $X,Y$ correspond via the reflection $r$ in the line $DABE$. In notation $X=Y^r$ and $Y=X^r$.
(So this case leads to a solution for the relation $YD+YE=2YX$.)
In a picture:

Proof of the claim:
Indeed, if $\hat C=60^\circ$, then the angle in $X$ in $\Delta XAB$ is
$\frac 12(180^\circ-\hat C)=\frac 12(180^\circ-60^\circ)=60^\circ$.
So $X$ is on the reflected circle $\odot(ABC)^r=\odot(ABC^r)$.
We also have $\widehat {DXB}=\widehat{DCB}$, for instance (using $\hat A+\hat B=120^\circ$):
$$
\begin{aligned}
\widehat {DCE} 
&=
\frac 12\hat A+\hat C+\frac 12\hat B=120^\circ\ ,
\\[2mm]
\widehat {DXE} 
&=
\widehat {DXB}+
\widehat {AXE}-
\widehat {AXB}
\\
&=
(180^\circ-\widehat {DCB})+
(180^\circ-\widehat {ACE})-
60^\circ
\\
&=
180^\circ+180^\circ-60^\circ-(\widehat {DCE}+\hat C)
\\
&=120^\circ\ .
\end{aligned}
$$
So $X$ is also on the reflected circle $\odot(DCE)$, so $X^r$ has the defining properties of $Y$, i.e.
$$
X=Y^r\ ,\ Y=X^r\ .
$$
The claimed properties follow by reflection.
$\square$

We show now that $\hat C=60^\circ$ is also a necessary condition for $2XY=YD+YE$.
We assume this condition.
Let $r$ be again the reflection in the line $DABE$.
We want to show $Y=X^r$. Assume that this is not the case. (And get a contradiction.)
Claim: The circles $\odot(DAX^r)$ and $\odot(EBX^r)$ intersect for the second time in $Y$.

Proof: Let $y\ne X^r$ be the second point of intersection of the two circles.
(We show $Y=y$.)
Then:
$$
\begin{aligned}
\widehat{DyA}
&
=
\widehat{DX^rA}
=
\widehat{DXA}
\\
&=
\widehat{DXC}-\widehat{AXC}
=
\hat B-\hat B/2=\hat B/2\ ,
\\
\widehat{DyX^r}
&
=
\widehat{DAX^r}
=
\widehat{DAX}
=
\widehat{DXC}-\widehat{AXC}
\\
&=
180^\circ-\widehat{AXD}-\widehat{ADX}
=
180^\circ-\hat B/2-\hat C/2\ ,
\\
\widehat{AyX^r}
&
=
\widehat{AyD}
+\widehat{DyX^r}
=
180^\circ-\hat B/2-\hat C/2\ .
\end{aligned}
$$
We used the circle $\odot(DAyX^r)$. Using the other circle
$\odot(EByX^r)$ we obtain the similar equalities. We know thus "all angles" around $y$. In particular:
$$
\begin{aligned}
\widehat{AyB}
&=360^\circ-
\widehat{AyX^r}-\widehat{ByX^r}
=360^\circ
-
2(180^\circ-\hat C/2)
\\
&=\hat C\ ,
\\
\widehat{DyE}
&=\widehat{DyA}+\widehat{AyB}+\widehat{ByE}
=\frac 12\hat B+C+\frac 12\hat A/2
\\
&=\widehat{DCB} \ .
\end{aligned}
$$
So $y$ satisfies the properties of $Y$, giving
$y=Y$.
$\square$

Bonus 1: $D,(E^*)',Y$ are colinear since
$$
\widehat{(E^*)'YA} =
\widehat{(E^*)'BA} =\hat B/2=\widehat{DYA}\ ,
$$
and similarly $E,(D^*)',Y$ are colinear. (We use the notations from the first step.)
$\square$

Bonus 2: Let $Z$ be the intersection of the bisector $\gamma$ of $\hat C$
with the circles $\odot(ABC)$ and its $\gamma$-reflection $\odot(A'B'C')=\odot(B^*A^*C)$.
Then $X^r,Y,Z$ are colinear and their common line is the angle bisector in $Y$ in
$\Delta AYB$.
Proof: From the above computations $X^ry$ bilds with $Ay$ inside the cyclic polygon $AYX^rD$ the angle $180^\circ-\widehat{ADX^r}=180^\circ-\widehat{ADX}
=180^\circ-\widehat{BDX}=180^\circ-\widehat{BCX}=180^\circ-\hat C/2$.
The exterior/suplement angle to this angle is thus $\hat C/2$. The similar comuptations done "on the other side" show that $X^ry$ is the angle bisector of
$\widehat{AyB}$. Since $\overset\frown{AZ}=\overset\frown{ZB}$ by the definition of $Z$, this angle bisector also passes through $Z$.
$\square$

So far we did not use  the given property of $Y$.
(The above claims+proofs still hold in the non-degenerate case $Y\ne X^r$.)
The above arguments "depend on the picture". But the other picture is

so $Z,X^r,Y$ are (in the final) in this order on the common line and the above arguments can be transposed to obtain their coliniarity.

(2) getting the contradiction.
Let us now finally suppose $Y\ne X^r$ and $2YX=YD+YE$. This implies
$$
YD+YE=2YX=2XY=XD+XE\ ,
$$
so $Y,X,X^r$ are points on the corresponding ellipse with focal points in $D,E$.
Both $X^r,Y$ are in the same half-plane w.r.t. the focal line $DE$, but the line $X^rY$ also passes through the point $Z$ in the interior of the ellipse.
(Since $Z$ is between the intersection $\gamma\cap DE$ and $X\in \gamma$.)
Contradiction. The assumption $Y\ne X^r$ is false. So $Y=X^r$ and (as mentioned) then $\hat C=60^\circ$.
$\square$
A: Here is my approach to the problem. Let $Bx$ be the bisector of $\angle CBA$ and cut $CX$ at the point $I$. Since $ACEX$ is a cyclic quadrilateral, we have $\angle AXC = \angle AEC = \frac{\angle ABC}{2}=\angle ABI$. Hence we have, $AXBI$ is a cyclic quadrilateral as well. By that means we obtain: $\angle CXB=\angle CXI = \angle CAI$ and we've already have $\angle CXB = \angle BDC = \frac{\angle CAB}{2}$. We infer that $I$ is an incenter of $\triangle ABC$ (because it is an intersection point of two bisector lines of $ \angle ABC$ and $ \angle CAB$).
We have $IAXB$ is a cyclic quadrilateral and $C, X, I$ are colinear, it follows that $X$ is excenter with respect to $\angle ACB$. Moreover, we can obtain $\angle ECX = \angle ECB + \angle BCX = \angle CEB + \angle XCA = \angle CEB + \angle XEA = \angle CEX$ or $XC=XE$. Similarly, we have $XC = XD$ so $X$ is a center of $(EXD) $.
Here, let $V$ is the intersection point of $YX$ and $(EXD)$ or $YV$ is the diameter of $(EXD)$. Furthermore, let $G$ be the point on $(EXD)$ that $\triangle GDE$ is equilateral triangle. By Ptolemy's theorem of $DYEG$, we get: $YD + YE = YG$. Therefore: $YG = 2XY=YV$. Since $YV$ is a diameter and $YG$ is just an chord of $(EXD)$, we must have $G \equiv V$ or $\angle DYE = 120^o$. Lastly, we have $\angle ACB = 60^o$ due to $\angle DYE = \frac{\angle ACB}{2} +90^o$.
