Question 109
The triangle $ABC$ has $\angle BAC = 54^\circ$ and $\angle ACB = 45^\circ$. Let $D$ and $E$ be points on the segments $BC$ and, respectively, $AD$ so that $AD = AB$ and $BE = BD$. Denote $F$ the common point of the lines $BE$ and $AC$, and let $DM$ be the bisector of the angle $\angle ADF$, where $M \in AC$. The perpendicular from $C$ on $AB$ meets $DM$ in $G$. Prove that:
a) the triangle $ABF$ is isosceles;
b) $CG = CM$.
a) the triangle $ABF$ is isosceles;
b) $CG = CM$.