Equilateral triangle ABC has side points M, D and E. Given AM = MB and $\angle$ DME $=60^o$, prove AD + BE = DE + $\frac{1}{2}$AB

A nice path without trigonometry

Consider the Figure below, where point $F$ has been taken so that $DF \cong DE$, point $L$ so that $BL \cong BE$, and point $G$ so that $AG\cong AF$.

1. $\angle AMD \cong \angle MEB$, because they are both supplementary to $\angle EMB + 60^\circ$.
2. Conclude therefore that $\triangle ADM \sim \triangle EMB$. In particular we have that $\frac{DM}{EM}=\frac{AM}{EB}$.
3. From the latter proportionality, the congruence $AM\cong MB$, and SAS cryterion, conclude that $\triangle DEM \sim \triangle ADM$, and in particular $\angle ADM \cong\angle MDE$.
4. Hence, by SAS cryterion, $\triangle DFM \cong\triangle DEM$, and in particular $FM\cong ME$, and $\angle FMD \cong \angle DME = 60^\circ$.
5. Use the latter congruences plus the fact $\angle FGM \cong \angle MLE = 120^\circ$ and $\angle GFM \cong \angle EMB$ (why?), to show that $\triangle FGM \cong \triangle EML$.
6. The thesis follows from the fact that $AD-DE\cong AF \cong AG \cong ML\cong MB-LB \cong MB-BE$.