$ABC$ is an isosceles triangle so $AB=BC$, $D$ is the midpoint of $BC$ so $BD=DC$ and $BED$ and $DFC$ are isosceles triangles such that $BE=ED=DF=FC$.
We know that $BED$ and $DFC$ are isosceles triangles but $BD=DC$ then $BED$ and $DFC$ are equilateral triangles where $BE=ED=DF=FC=BD=DC$.
Since $ABC$ is an isosceles triangle let $ABC \sphericalangle=ACB \sphericalangle=a$
$ABE \sphericalangle = ABC \sphericalangle + DBE \sphericalangle =60^o +a $
$ACF \sphericalangle= ACB \sphericalangle +DCF \sphericalangle= 60^o +a$
We can easily show that triagnles $ABE\equiv ACF$ because $AB=BC$, $ABE \sphericalangle=ACF \sphericalangle$ and $AB=AC$ (SAS) we get that triangles $ABE\equiv ACF$.
We also know that $AE=AF$ how to show that $AM=AN$?
Note that the given conditions do not imply the two smaller triangles are equilateral. However, they are congruent by SSS criteria, so that $\angle DBE = \angle DCF$ and you already know $\angle ABC = \angle ACB$.
Thus, $\angle ABE = \angle ACF$, combined with $AB = AC$ and $BE=CF$ gives $\Delta ABE \cong \Delta ACF$, and the desired result follows.