In quadrilateral $ABCD$, $AD$ is parallel to $BC$ and $AB= AC$ . $F$ is a point on $BC$
such that $DF$ is perpendicular to $BC$. $AC$ intersects $DF$ at $E$. If $BE= 2DF$ and
$BE$ bisects $\angle ABC$, find the measure, in degrees, of $∠BAD$ .

I worked out that $\angle ABC = \angle ACB =\angle DAE$
I think the angle bisector theorem could be used but im not sure how to use it in a way that would incorporate the fact that $BE=2DF$
taken from the 2018 BIMC https://chiuchang.org/imc/wp-content/uploads/sites/2/2018/07/BIMC-2018_Keystage-3_Individual.x17381.pdf
Hint. Consider the perpendicular to $BC$ at $B$ and let $G$ be its intersection with line $AC$. Since triangle $ABC$ is isosceles, we have that its height relative to side $BC$ intersects $BC$ in its midpoint. Thus, $BG = 2DF$ and triangle $BGE$ is isosceles.