Let line $AD$ intersect $(BDX)$ and $(CDY)$ at $P$ and $Q$ respectively, where $P$ and $Q$ are different from $D$. If $A$ lies on the radical axis of $(BDX)$ and $(CDY)$, $AP \cdot AD = AQ \cdot AD \iff AP = AQ$. Note that $\angle EPB = \angle DPB = \angle DXB = \angle DXE = \angle EDX = \angle PDX = \angle PBX = \angle PBE \iff EP = EB$. Similarly, $EQ = EC$. Now we have $AP = AQ \iff EA + AP = EA + AQ \iff EP = EQ \iff EB = EC \iff EB^2 = EC^2$ $ \iff ED^2 + BD^2 = ED^2 + CD^2 \iff BD^2 = CD^2 \iff BD = CD \iff AB = AC$.

This problem was proposed by Macedonia

This was my problem. The shortest solution I found was this: Note that the condition of the problem is that $E$ lies on the radical axis of the circumcircles of $\triangle BDX$ and $\triangle CDY$. This gives us $EB \cdot EX = EC \cdot EY$. However, $EX = EY$ because $E$ is the center of $\omega$ and this means that $BE = CE$. Now using Pythagoras' theorem we have the following: $$AB^2 = AD^2 + BD^2 = AD^2 + (BE^2-DE^2) = AD^2 + (CE^2 - DE^2)= AD^2 + CD^2 = AC^2$$From here we obtain $AB = AC$.

Let intersection point be S. we have BE.EX = DE.ES = AE.EY and EY = EX so BE = CE so E lies on perpendicular bisector of BC. In triangle ABC perpendicular bisector of BC and altitude of A are same so ABC is isosceles.

Let circumcircs of triangles BDX and CDY intersects at L other than D. Since CDYL is concyclic, <YCL = <YDL = <CYD = < DLC => LE = EC Since BDXL concyclic, <LBX = <LDX = <DXB = <BLD => LE = BE. Therefore BE = EC and in triangle BEC, BD = DC. In triangle ABC, BD = DC and AD altitude, so AB = AC.

Because both intersection points on $AD$, so $\overline{AD}$ is the radical axis of $(BDX)$ and $(CDY)$. Thus, because $E\in AD$, $EB\cdot EX = EC\cdot EY$. Furthermore, because $E$ is the center of $(AYDX)$, $EY=EX \Rightarrow EB=EC$, and so $AD$ is the perpendicular bisector of $BC$ - $AB=AC$.

Let the second intersection point of $(BDX)$ and $(CDY)$ be $R$,different than $A$.By power of point we get that $RE\cdot ED = YE\cdot EC = BE\cdot EX$ . Since $EX$ and $EY$ are both radii, we get $BE=EC$, so $AD$ is the perpendicular bisector of $BC$ ,which implies $AB=AC$ .

First of all please tell me if the following is incorrect ( I am not that good with inversion and have to practice hence this post) Denote $\omega_1$ as the cline $(YAX)$, $\omega_2$ as the cline $\overline{BE}$, $\omega_3$ as the cline $\overline{CE}$, $\omega_4$ as the cline $(BDX)$, $\omega_5$ as the cline $(YDC)$ Let $\Psi_D$ be a map sending every point to its inverse with respect to the circle centered at $D$ with radius $\overline{DE}$ After applying $\Psi_D$ we get the following diagram: Note that $\omega_1$ is tangent to $\overline{BC}$ The new problem is quite easy to prove so I decided to present the stupidest prove ever Throw the problem into the coordinate plane. Let $D = (0,0)$, $B = (b,0)$, $ C = (c,0)$, $E = (1,0)$ Let $t = X_y = Y_y$ $\omega_2$ and $\omega_3$ have equations: $(x-A)^2+(y-B)^2 = A^2 + B^2 \Longleftrightarrow x^2 -2Ax + y^2 - 2By=0$ And since $y = t$ and $B=\frac{1}{2}$ we get the following: $x^2 -2Ax + t^2 - t =0$ So the equation of $\omega_2$ is given by : $x^2 - 2(\frac{b}{2}) + t^2 - t =0 \Longleftrightarrow x^2 - b + T=0$ Similarly the equation of $\omega_3$ is given by : $x^2 - 2(\frac{c}{2}) + t^2 - t =0 \Longleftrightarrow x^2 - c + T=0$ Solving for $x$ in both we get $Y = (\frac{b-\sqrt{b^2-4T}}{2},t)$ and $X = (\frac{c+\sqrt{c^2-4T}}{2},t)$ (In the problem statement we are given where $Y$ and $X$ are with respect to $\overline{AD}$ and other points) We also know that $\overline{CY}$, $\overline{BX}$, $\overline{DE}$ are concurrent. Let their intersection point be $F = (0,f)$ $YC_{slope} = FC_{slope} \Longleftrightarrow \frac{2t}{2c-b+\sqrt{b^2-4T}}=\frac{f}{c} $ $XB_{slope} = FB_{slope} \Longleftrightarrow -\frac{2t}{c-2b+\sqrt{c^2-4T}}=\frac{f}{b} $ We can equate the expressions for $f$ which results in: $c=-b$ hence establishing $\overline{AB} = \overline{AC}$