In the obtuse triangle $ABC$, with the obtuse angle $\angle CAB$, angle $\gamma$ is twice the angle $\beta$. There is a point $D\in \overline{BC}$ s.t. $AB\perp AD$. Line parallel to $AD$ passes through the midpoint of $\overline{AB}$ and through the point $E\in\overline{BC}$. Prove $|DE|=|AC|$
Here is my attempt. I was wondering if there are other elegant ways to prove this. Let $M$ be the midpoint of $\overline{AB}$. Then: $$\Delta BEM{\sim}\Delta BDA$$ and $\overline{EM}$ is the midsegment of $\Delta BDA$, so $|BE|=|ED|=\frac{|BD|}{2}$
Also:
$$\frac{|AB|}{\sin(2\beta)}=\frac{|AC|}{\sin\beta}\implies |AB|=2\cos\beta|AC|,$$
furthermore, $$\cos\beta=\frac{|AB|}{|BD|}=\frac{|AB|}{2|DE|}$$
$$\implies|AB|=2\cdot\frac{|AB|}{2|DE|}\cdot|AC|\implies|DE|=|AC|$$
Picture:
I tried to think of a more elementary tool, but I couldn't. Is there any?
Line $ME$ is the perpendicular bisector of $AB$, hence $AM=BM$. Triangle $AEB$ is then isosceles so that $BE=AE$ and $\angle ABE=\angle BAE$.
But $\angle AEC$ is an external angle of $AEB$, hence $$ \angle AEC=\angle ABE+\angle BAE=2\angle ABE=\angle ACB. $$ It follows that triangle $AEC$ is also isosceles and $AC=AE=BE=DE$.