Just easy warm-up. With $F$ - midpoint of $AB$ we have $\angle BDF=\angle B=2\angle C=2\angle DEF$. Hence $DE=DF=BF$.

parmenides51 wrote: Let $AD$ and $AE$ be the altitude and median of triangle $ABC$, in with $\angle B = 2\angle C$. Prove that $AB = 2DE$. $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let $T$ be a point in $BC$ such that $\angle ATB=\alpha$ $$\Rightarrow \angle TAB=\alpha$$$$\Rightarrow TB=AB \text{ and } AT=AC...(I)$$$$\Rightarrow TD=DC...(II)$$$$BE=EC=a, ED=b$$$$\Rightarrow BD=a-b$$By $(II):$ $$\Rightarrow TD=a+b$$$$\Rightarrow TB=2b$$By $(I):$ $$\Rightarrow AB=2b=2(DE)_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$