75th Romanian Mathematical Olympiad

a) From $△DFG$ isosceles follows $\angle DFG=\angle DGF$, hence $\angle DFB=\angle DGE$. From $△BDE$ isosceles follows $\angle DBF=\angle DEG$.

The above and $BD=DE$ yield $△BDF\equiv △EDG$ (S.A.A.), whence $BF=GE$.

b) Since $△ABC$ is isosceles, $\angle ABC=\angle ACB=\frac{180^{\circ }-\angle BAC}{2}=\frac{150^{\circ }}{2}=75^{\circ }$. Then $BF=GE$ gives $BG=FE=BC$, hence $△BCG$ is isosceles.

So $\angle BCG=\angle BGC=\frac{180^{\circ }-\angle CBG}{2}=\frac{105^{\circ }}{2}=52^{\circ }30^{\prime }.$