rip got sniped lol. I think this is pretty much the same. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); dotfactor *= 1.2; pair A = dir(120), B = dir(210), C = dir(330), D = extension(A,incenter(A,B,C),B,C), E = foot(B,C,A), F = foot(C,A,B), M = (B+C)/2, K = extension(M,(E+F)/2,A,B), X = extension(A,D,M,K); draw(A--B--C--A^^F--E--K, heavyblue); draw(circumcircle(C,E,D), heavygreen); draw(circumcircle(A,E,F), purple); draw(circumcircle(B,M,E), orange); draw(E--M--F^^M--K, purple); draw(A--D, red); dot("$A$", A, dir(90)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$D$", D, dir(270)); dot("$E$", E, dir(45)); dot("$F$", F, dir(200)); dot("$X$", X, dir(180)); dot("$K$", K, dir(135)); dot("$M$", M, dir(270)); [/asy][/asy] Let $F$ be the foot from $C$ to $\overline{AB}$, $K = \overline{AB} \cap (BME)$, and redefine $X = \overline{KM} \cap \overline{AD}$. It is well known that $M = \overline{EE} \cap \overline{FF}$ in $(AEF)$. Then $$\angle EKB = 180^\circ - \angle EMB = 180^\circ - 2\angle ECB = 180^\circ - 2\angle EFK,$$so $KE = KF$. This means $\overline{KM}$ is the perpendicular bisector of $\overline{EF}$, so $X$ is the midpoint of minor arc $\overline{EF}$ in $(AEF)$. Now from $\angle ECD = \angle EFA = \angle EXA$, we have $ECDX$ cyclic, so our $X$ is indeed the $X$ defined in the problem. Also $\angle XME = \angle KME = \angle KBE = 90^\circ - \angle BAC$, so we are done.

Let $F=AD\cap BE$. We claim that $\triangle ABF \sim \triangle EMX$. Firstly, we have $\frac{1}{2}\angle BAC + \angle ACB = \angle ADB = \angle XEC$ but as $\angle MEC = \angle ACB$, we have $\angle XEM = \frac{1}{2}\angle BAC$. Also, from $\triangle AEX \sim \triangle ADC$, we have $\frac{EX}{DC} = \frac{AE}{AD} \implies EX=\frac{AE\cdot DC}{AD}$. So now it is equivalent to prove that $\frac{AE\cdot DC}{AD\cdot AF} = \frac{a}{2c}$ where $a=BC$ and $c=AB$. Let the foot from $D$ to $AB$ be $G$. Then we know $\frac{AE}{AF} = \frac{AG}{AD}$. Replacing $AG$ with $AD\cdot \cos{\frac{A}{2}}$, it is equivalent to show $\frac{DC\cdot \cos{\frac{A}{2}}}{AD} = \frac{a}{2c}$. But we know $\cos{\frac{A}{2}} = \frac{\sqrt{(b+c-a)(b+c+a)}}{2\sqrt{bc}}$ and $DC = \frac{ab}{b+c}$ and $AD = \frac{\sqrt{bc} \cdot \sqrt{(b+c-a)(b+c+a)}}{b+c}$. After dividing, we see that $\frac{DC\cdot \cos{\frac{A}{2}}}{AD} = \frac{a}{2c}$ and thus we have $\triangle ABF \sim \triangle EMX \implies \angle XME = \angle ABE = 90 - \angle BAC$. $\square$