A advanced geometry question

Question

In triangle $\triangle ABC, \overline{AB}>\overline{AC}$. Mis foot of perpendicular from $C$ on external bisector of angle $\widehat{A}$. $D$ is midpoint of $\overline{BC}$ .Prove that $\overline{AB}+\overline{AC}=2\overline{DM}$.

Answer 1

I will present a slightly unusual solution to this problem... Assume an ellipse with $B,C$ as foci and $A$ as a point on the ellipse.

Then what the question asks us to show is that the length $DM$ is half the length of the major axis of this ellipse.

For this, simply observe that the external angle bisector is actually the tangent to this ellipse at $A$. And it is again well known that the foot of perpendicular from foci to a tangent lies on the auxilliary circle.

Hence $M$ lies on the auxilliary circle, which is exactly what was intended.