Angle between diagonals of an irregular pentagon 

In pentagon $ABCDE$, $DC$ is parallel to $BE$ and $BC$ is parallel to $AE$. $△DCE$ and $△BCE$ are both isoscles with $$∠CBE= ∠DEC = α,\ ∠EAB= α+\frac{π}{2}.$$ Suppose $AD$ meets $BE$ at $G$. Find $∠BGA$.

I have tried producing $AD$ to meet $BC$ at $F$ and showing that $∠DFC$ must be $α$, but I haven't been able to show definitively a value for $∠BGA$, which I think is $2α$. I have also tried producing lines from $D$ to $BC$ and from $A$ to $BC$ both inclined at an angle $α$ to $BC$, but I haven't been able to show that they are the same line.
Any hints about how to proceed would be much appreciated!
 A: Let $DC=ED=a$.
Hence, $$EC=CB=2a\cos\alpha$$ and
$$EB=2EC\cos\alpha=4a\cos^2\alpha.$$
Thus, by law of sines for $\Delta EAB$ we obtain:
$$\frac{EA}{\sin\left(90^{\circ}-2\alpha\right)}=\frac{4a\cos^2\alpha}{\sin\left(90^{\circ}+\alpha\right)},$$
which gives $$EA=4a\cos\alpha\cos2\alpha.$$
Now, since $\alpha<60^{\circ}$, by law of cosines for $\Delta DEA$ we obtain:
$$DA=\sqrt{a^2+16a^2\cos^2\alpha\cos^22\alpha-8a^2\cos\alpha\cos2\alpha\cos3\alpha}=$$
$$=a\sqrt{1+8\cos\alpha\cos2\alpha(2\cos\alpha\cos2\alpha-\cos3\alpha)}=a\sqrt{1+8\cos^2\alpha\cos2\alpha}=$$
$$=a\sqrt{1+4(1+\cos2\alpha)\cos2\alpha}=a\sqrt{(1+2\cos2\alpha)^2}=a(1+2\cos2\alpha).$$
Thus, by law of cosines again we obtain:
$$\cos\measuredangle DAE=\frac{AE^2+AD^2-DE^2}{2AE\cdot AD}=\frac{16a^2\cos^2\alpha\cos^22\alpha+a^2(1+2\cos2\alpha)^2-a^2}{2\cdot4a\cos\alpha\cos2\alpha\cdot a(1+2\cos2\alpha)}=$$
$$=\frac{16\cos^2\alpha\cos^22\alpha+4\cos^22\alpha+4\cos2\alpha}{8\cos\alpha\cos2\alpha(1+2\cos2\alpha)}=\frac{4\cos^2\alpha\cos2\alpha+\cos2\alpha+1}{2\cos\alpha(1+2\cos2\alpha)}=$$
$$=\frac{4\cos^2\alpha\cos2\alpha+2\cos^2\alpha}{2\cos\alpha(1+2\cos2\alpha)}=\cos\alpha,$$
which says that indeed, $$\measuredangle BGA=2\alpha.$$
A: Solving without trigonometry. Just because.

If $CD||BE$, then $CE$ must be parallel to $BA$:
Since $\angle CBE =\alpha$ and $\angle DCF=\alpha$ then by the Corresponding Angles Postulate, then $CE||BA$.

$\angle DCF = 180^\circ-\angle DCE - \angle ECB $
$\angle DCF=180^\circ-\alpha-(180^\circ-2\alpha)=\alpha$

QED

Since we know that $CE||BA$.
$$
\angle ECB = \angle EAB \qquad{\mathbb{it\ follows\ that:}}$$
$$180-2\alpha=\alpha+\frac{\pi}2$$
And because it is more fun and intuitive to think in degrees instead of radians, then $\alpha = 30^\circ$
$$\angle CBE = \angle CEB$$
$$\angle ECB+\angle CBE+\angle CEB = 180^\circ $$
$$\angle ECB = 180^\circ - 2\alpha$$
$$\angle ECB = 120^\circ$$
Also since we know that $CE||BA$ and $CB||EA$, then:
$$\triangle CEB = \triangle EAB$$
$$\angle EAB = \angle ECB=120^\circ$$
And because $\angle EAB = \alpha +\frac{\pi}2 = 30^\circ+90^\circ$, it necessarily follows that:
$$\angle DAB = 90^\circ$$
and because $\triangle CEB = \triangle EAB$, then $\angle ABE = 30^\circ$, which must mean:
$$\angle BGA = 90^\circ - \angle ABE = 90^\circ-30^\circ$$

$$\therefore \angle BGA = 60^\circ$$
  This is a faster silver bullet for anyone with no expansive knowledge on trigonometry.

