74th 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 $CG=CM$.