Note that $MC=\frac{1}{2}AB$ so we need to prove that $AD=\frac{1}{2}\left(BC+CA\right)$. Now if we define the point $D'$ as the point on side $AB$ such that $AD'=\frac{1}{2}\left(AB+CA\right)$ (and hence $D'B=\frac{1}{2}\left(AB-CA\right)$) then we wish to prove that $D=D'$ i.e. that $D'M\perp AL$. Thus we wish to prove that $\angle{AD'M}=90-\angle{LAD'}=90-\frac{1}{2}\angle{CAB}$. Define $E$ as the point such that $D'$ is the midpoint of $BE$. Then $D'M\parallel EC$ ($D'$ midpoint of $BE$, $M$ midpoint of $BC$). Thus we wish to prove that $\angle{AEC}=90-\frac{1}{2}\angle{CAB}$. Now $BE=2BD'=AB-CA$ so $AE=AB-\left(AB-CA\right)=CA$ hence $\triangle{AEC}$ is isosceles. Thus, $\angle{AEC}=\frac{1}{2}\left(180-\angle{CAB}\right)=90-\frac{1}{2}\angle{CAB}$ as required.

Dear Mathlinkers, another proof consist to think to the Simson's line.... which will avoid all calculation. Who want to consider this idea? Sincerely Jean-Louis

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Highway%20to%20Geometry%201.pdf p. 45... Sincerely Jean-Louis

Let the perpendiculars to $AL$ through $B,C$ cut resp. $AC,AB$ at $B',C'$ it s clear that $ABB',ACC'$ are isoceles and $BC\parallel B'C' \parallel DE$ where $E$ is the second intersection with $AC$ .Since $DE$ pass through the midpoint of $BC$ then $D,E$ are the midpoints of $BC',CB'$ hence $2AD=AB+AC'=AB+AC$

Let $E$ be the reflection of $D$ in $\overline{AL}.$ Then, by LoS and $\triangle AED$ isosceles, $$BD=\frac{BM\sin\angle DMB}{\sin\angle BDM}=\frac{CM\sin\angle EMC}{\sin\angle AED}=CE.$$Hence, $$AB-AD=AD-AC\implies AD=\frac{AB+AC}{2}.$$$\square$