16th Junior Turkish Mathematical Olympiad

Let $\angle ACB=\alpha$. Since $AB=AC$, we have $\angle ABD=\angle ABC=\alpha$. $DE\perp AC$ and $AD\perp BC$ imply that $\angle EDC=90^{\circ }-\alpha$ and $\angle ADE=\alpha$. Therefore, we get $\angle ABD=\angle ADE=\alpha$ which implies that $DE$ is tangent to the circuncircle of the triangle $ABD$ and hence $G{D}^2=GF\cdot GA$.

On the other hand, $\angle BFA=\angle ADB=90^{\circ }$. Therefore, $EF$ is perpendicular to the hypotenuse of the right triangle $AEG$. Thus, $G{E}^2=GF\cdot GA$ and consequently we get $GD=GE$.