Romanian Mathematical Olympiad

a) From the statement, $\angle ABC = 81^\circ$ and, in the isosceles triangle $ABD$, $\angle ADB = \angle ABD = 81^\circ$, hence $\angle DAB = 18^\circ$ and $\angle CAD = \angle CAB - \angle DAB = 36^\circ$.
The isosceles triangle $BED$ yields $\angle BED = \angle BDE = 81^\circ$, therefore $\angle DBE = 18^\circ$, whence $\angle ABF = \angle ABC - \angle EBD = 63^\circ$.

From the triangle $ABF$, $\angle ABF = 63^\circ$ and $\angle FAB = 54^\circ$, hence $\angle AFB = 63^\circ$, that is triangle $ABF$ is isosceles, with $AF = AB$.

b) The relations $AD = AB$ and $AF = AB$ lead to $AF = AD$, so the triangle $AFD$ is isosceles, with $\angle AFD = \angle ADF = \frac{180^\circ - \angle FAD}{2} = 72^\circ$. Since $DM$ is the bisector of the angle $ADF$, $\angle ADM = \angle FDM = \frac{\angle ADF}{2} = 36^\circ$. The triangle $AMD$ gives $\angle MAD = \angle ADM = 36^\circ$, therefore $\angle CMD = 72^\circ$.
From $CG \perp AB$ follows $\angle ACG = 90^\circ - \angle CAB = 36^\circ$, so the triangle $CMG$ has $\angle ACG = 36^\circ$ and $\angle CMG = 72^\circ$. This gives $\angle CGM = 72^\circ$, that is the triangle $CMG$ is isosceles, with $CG = CM$.