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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 $\overline{AB}+\overline{AC}=2\overline{DM}$.

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    $BA$ extended meets $CM$ extended at $E$. Triangle $CAE$ is isosceles. Show $BE = 2DM$ – 2017-02-25

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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.