Hellenic Mathematical Olympiad

We extend median $AM$ by $M\Theta =AM$. Then $AB\Theta \Gamma$ is a parallelogram. Hence $AB\parallel \Gamma \Theta$ and ${\hat{A}}_1={\hat{O}}_1$. But from $A\Delta =\Delta E$ we get ${\hat{A}}_1={\hat{E}}_1$. Figure 1

and since ${\hat{E}}_1={\hat{E}}_2$, we find that ${\hat{O}}_1={\hat{E}}_2$. Therefore the triangle $E\Theta$ is isosceles with $\Gamma E=\Gamma \Theta$. Finally from the parallelogram $AB\Theta \Gamma$ we have $\overrightarrow{AB}=\Gamma \Theta$, from which we have $AB=\Gamma E$.

---

Alternative solution.

From the midpoint $M$ of $B\Gamma$ we draw line $\delta \parallel AB$. Then $\delta \parallel B\Delta$. Let $B\Delta$ intersect the segment $\Gamma \Delta$ at $Z$. Then $Z$ is the midpoint of $\Gamma \Delta$, that is $\Gamma Z=Z\Delta$ (1) and moreover $B\Delta =2\cdot MZ$. (2) Figure 2

Also we have ${\hat{A}}_1={\hat{M}}_1$. However from $A\Delta =\Delta E$ we get that ${\hat{A}}_1={\hat{E}}_1$ and ${\hat{E}}_1={\hat{E}}_2$. Hence ${\hat{M}}_1={\hat{E}}_2$, and $EMZ$ is isosceles with $ZM=EZ$. (3)

Hence $$\begin{array}{rl}\Gamma E& =\Gamma Z+ZE=\Delta Z+ZE(\text{from (1)})\\ & =\Delta E+2\cdot ZM(\text{from (3)})\\ & =A\Delta +\Delta B=AB.(\text{from hypothesis and (2)})\end{array}$$