Proving $DF = EF$ without trigonometry

Question

In the isoscales $\triangle ABC$ ($AB=AC$), if $BE = CD$, prove $DF=EF$.

I used Sine Rule for triangles $\triangle BEF$ and $\triangle CDF$ and concluded that $DF = EF$. I'm wondering can you solve this problem without trigonometry?

Answer 1

Take point $K$ on $AB$ such that $BK=BE$.

Since $\triangle ABC$ is an isosceles,

then $KD \parallel BC \parallel BF$

and

$EF = FD$ (line through midpoint $\parallel$ to the second side)

Answer 2

Drop perpendicular $EG$ to the extended side $BC$ and perpendicular $DH$ on $BC$. Then by $ASA$ congruence we have $\triangle EBG\cong\triangle DCH$ and $EG=DH$. But then by $ASA$ again $\triangle FGE\cong\triangle FHD$ and $EF=DF$.