58th Ukrainian National Mathematical Olympiad

Since the segment $GE$ is both altitude and median in $\Delta ABG$ (Fig. 2), this triangle is isosceles, therefore $AG=GB$. Analogously, $\Delta CBG$ is isosceles, hence $BG=GC$. Moreover, from the statement of the problem, point $G$ is the middle of the segment $AD$. Hence, we obtain that $DG=AG=BG=CG$, i.e. $\Delta CGD$ is also isosceles, which means $\angle GCD=\angle GDC=70°$.