Prove that the triangle is isosceles, based on the intersection of two line segments connecting vertices to sides Given $AB=AC$ and $DE=DF$, prove $DB=DC$

I have no idea how to solve it using elementary way, can somebody help me?
Assume that $DB&gt;DC$, then $\angle BCD&gt;\angle CBD$. So $\alpha&gt;\beta$. From the following picture


we see that $AE&lt;AF$. Thus, in $\triangle AEF$ we have $x&gt;y$. But $x+\alpha=y+\beta$ is a contradiction.
Given that DE=DF, a parallel line can be drawn through EF which is parallel to BC. as AB=AC angleABC=angleACB,THUS following that angleFEA=angleEFA WHICH implies EA=FA .THUS due to all of these angleDBF=angleDCE and thus DB=DC .