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? 
 A: 
Assume that $DB>DC$, then $\angle BCD>\angle CBD$. So $\alpha>\beta$. From the following picture

we see that $AE<AF$. Thus, in $\triangle AEF$ we have $x>y$. But $x+\alpha=y+\beta$ is a contradiction.
A: Assume by contradiction $DB >DC$ (The case $DC>DB$ is identical).
Then in triangle $DBC$ we have $\angle DCB > \angle DBC$. Therefore
$\angle FCA > \angle EBA$. As $\angle EAB =\angle FAC$ we get 
$$\angle AFC < \angle BEA \,.$$
Now, comparing triangles $ABE$ and $AFC$ since $\angle EAB =\angle FAC, AB=AC$ and $\angle AFC < \angle BEA$ you get $BE <FC$ (use for example sin law). But this contradicts $BD> DC$.
A: Draw two perpendicular lines from $CH_1$ to $DB$ and from $BH_2$ to $CD$.
For the sake of contradiction we assume $DB>DC$.
By area of $DBC$ we have $CH_1<BH_2$ hence area of triangle $S_{DEC}<S_{DFB}$ and $S_{AFC}<S_{AEB}$ which implies $AF<AE$ because $AB=AC$ and angle $FAC=EAB$.
Hence we have $BF<CE$ as well. Now ${DF\over sin(DBF)}={BF\over sin(D)} < {EC\over sin(D)} = {DE\over sin(DCE)}$.
This implies $sin(DBF)>sin(DCE)$
(1) When both $DBF$ and $DCE$ are less than $90$ degrees we arrive at a contradiction because angle $DBF>DEC$ implies $DBC>DCB$ and $DB<DC$.
(2) When $DBF$ is larger than $90$ degrees, then $DBC>90$ as well and $DB<DC$ contradiction.
(3) When $DCE$ is larger than $90$ degrees we have $DBF>180-DCE$ and $DBF+DCE>180$ contradiction.
(4) When both are larger than $90$ degrees $DCB+DBC>180$ contradiction.
A: 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 .
