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 $∠BAD$ . enter image description here

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.

  • $\begingroup$ don't you mean $BGA$ is isosceles, I don't understand how $BGE$ is isosceles? $\endgroup$
    – Tyrone
    Aug 27 '19 at 18:33
  • $\begingroup$ We are using that $BG = 2DF = BE$, where the second equality is a hypothesis. $\endgroup$
    – Daniel
    Aug 27 '19 at 18:36

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