$F \equiv AM \cap BC$ is the foot of the A-external bisector. Thus circle $\Gamma_a$ with diameter $EF$ is the A-Apollonius circle of $\triangle ABC,$ which is orthogonal to its circumcircle $\Gamma.$ Hence, it is enough to show that $D \in \Gamma_a.$ $\angle AFE= |\angle B- \angle FAB|=| \angle B-90^{\circ}+\frac{_1}{^2}\angle A|=\frac{_1}{^2}| \angle B-\angle C|$ $\angle ADE=\angle ACM=|\angle MCB-\angle C|=|90^{\circ}-\frac{_1}{^2}\angle A-\angle C|=\frac{_1}{^2}| \angle B-\angle C|$ $\Longrightarrow \angle AFE=\angle ADE$ $\Longrightarrow$ $D \in \Gamma_a.$

Dear Mathlinkers, 1. F is the point of intersection of AM and BC. 2. Remark that the tangent Tm to "Gamma" at M is parallel to EF. 3. According to a converse of Reim's theorem applied to Gamma with the basic points A and D, the borning monians MAF and MDF, the parallels Tm and FE, the points A, D, E and F are concyclic on the A-Apollonius circle of ABC. 4. This last circle being orthogonal to "Gamma", we are done. Sincerely Jean-Louis

Note that $\angle EDK= \angle MDK= \frac{1}{2} \hat MAD= \frac{1}{2}( \hat DB+ \hat BM)= \frac{1}{2} ( \hat DB+ \hat MC)= \angle KED$. Thus, $K$ lies on the perpendicular bisector of $ED$. Also, $\angle DAE= \frac{1}{2} (2 \angle DAE)= \frac{1}{2}( \angle DAE+ \angle NAC + \angle DAE - \angle BAN)= \frac{1}{2}( \angle DAC- \angle BAD)=\frac{1}{2}[\frac{1}{2} \hat CD- \frac{1}{2} \hat BD]= \frac{1}{2} \angle DKE$. Thus, $K$ is the circumcenter of $\triangle AED$.

As $DE$ meets $(DBC)$ at $M$ which is the midpoint of arc $BC$, by Fact 5, $DE$ is the angle bisector of $\angle BDC$. This means, $E$ is the feet of perpendicular from both $A$ in $\angle BCA$ and from $D$ in $\angle BDC$. Thus $(AED)$ is the A-Apollonius circle of $\Delta ABC$. Thus tangents at $D$ to $(ADE)$ and $(ABC)$ are perpendicular. We know that circumcenter of $(AED)$ lies on $BC$, thus it is the intersection of $BC$ and tangent at $D$ to $(ABC)$.

A different approach perhaps: Note that it suffices to show that $AD$ is the $A$-symmedian, in which case $\odot (ADE)$ will be the $A$-Appolonius circle, and its center will be the point where the tangents to $\Gamma$ at $A$ and $D$ meet $BC$. But this easily follows from the following harmonic bundle:- $$-1=(B,C;M,AE \cap \Gamma) \overset{E}{=} (C,B;D,A)$$