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 I would like to ask if someone could help me with solving the following probelm.
The triangle $ABC$ is equilateral. $M$ is the midpoint of $AB$. The points $D$ and $E$ are on the sides $CA$ and $CB$, respectively, such that $\angle DME=60^o$

Prove that $AD+BE = DE + \frac{1}{2}AB$.
Thank you in advance.
 A: 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 $AG\cong AF$.



*

*$\angle AMD \cong \angle MEB$, because they are both supplementary to $\angle EMB + 60^\circ$.

*Conclude therefore that $\triangle ADM \sim \triangle EMB$. In particular we have that $\frac{DM}{EM}=\frac{AM}{EB}$.

*From the latter proportionality, the congruence $AM\cong MB$, and SAS cryterion, conclude that $\triangle DEM \sim \triangle ADM$, and in particular $\angle ADM \cong\angle MDE$. 

*Hence, by SAS cryterion, $\triangle DFM \cong\triangle DEM$, and in particular $FM\cong ME$, and $\angle FMD \cong \angle DME = 60^\circ$.

*Use the latter congruences plus the fact $\angle FGM \cong \angle MLE = 120^\circ$ and $\angle GFM \cong \angle EMB$ (why?), to show that $\triangle FGM \cong \triangle EML$.

*The thesis follows from the fact that $AD-DE\cong AF \cong AG \cong ML\cong MB-LB \cong MB-BE$.

A: To simplify, assume all lengths relative to $|AM|$. In other words, $AM=1$
Let $x = CD, \quad y = CE,\quad z=DE$ 
Assume the statement to be proved is true, this follows:
$AD + BE = (2-x)+(2-y) = z + 1 \quad → x+y+z=3$
$ΔCDE:\; z^2 = x^2 + y^2 - x y$ 
$$(3-x-y)^2 = x^2+y^2-xy$$
$$x^2+y^2+2xy-6(x+y)+9 = x^2+y^2-xy$$
$$3xy - 6(x+y) + 9 = 0$$
$$xy - 2(x+y) + 3 = 0$$
Let $T = xy-2(x+y)+3$. If $T=0$, we proved $AD+BE=DE+{1\over2}AB$

Laws of Cosines: 
$ΔADM:\; DM^2 = 1 + (2-x)^2 - (2-x) = x^2-3x+3$
$ΔBEM:\; EM^2 = 1 + (2-y)^2 - (2-y) = y^2-3y+3$ 
$ΔDEM:\; z^2 = x^2+y^2-xy = DM^2 + EM^2 - DM·EM$ 
$$DM·EM = xy-3(x+y)+6 = T+(3-(x+y))$$
$$(DM·EM)^2 = (T+(3-(x+y)))^2$$
$$T^2 + (x+y)T + z^2 = T^2+2(3-(x+y))T+(3T+z^2) $$
$$3(x+y-3)\;T = 0$$

All is left is to show $(x+y-3)$ is non-zero.
If $x+y=3$, we have $DM·EM = xy - 3(x+y) + 6 = xy-3$
Using AGM inequality $\large \sqrt{xy} ≤ {x+y \over 2} = 1.5$
This implied $DM·EM ≤ 1.5^2 - 3 < 0$, thus $x+y≠3$
A: Let $a$ be the side length of the equilateral triangle ABC. Apply the cosine rule to the triangle CDE,
$$DE^2 = (a-AD)^2+(a-BE)^2-(a-AD)(a-BE)$$
$$= \left(AD + BE -\frac a2\right)^2 + 3\left(\frac {a^2}{4} -AD\cdot BE \right)\tag{1}$$
Now, recognize $\angle DMB = 60 + \angle ADM$ to conclude $\angle EMB = \angle ADM$ and, likewise, $\angle DMA = \angle MEB$. Thus, the triangles ADM and BEM are similar, which yields,
$$AD\cdot BE = AM\cdot BM = \frac 14 a^2$$
As a result, (1) simplifies to $ DE^2= \left(AD + BE -\frac a2\right)^2$, which is,
$$AD+BE = DE + \frac{1}{2}AB$$
