Estonian Math Competitions

Let $BD=DC=x$ and $GP=PE=y$. Then $BG=2\cdot 2y=4y$, $BP=4y+y=5y$ and $BE=4y+2y=6y$. Thus $GP=GD$ if and only if $y^2=B{G}^2-B{D}^2=16y^2-x^2$ or, equivalently, $x^2=15y^2$. The quadrilateral $CEPD$ is cyclic if and only if $BP\cdot BE=BD\cdot BC$, i.e., $5y\cdot 6y=x\cdot 2x$. The latter simplifies to $x^2=15y^2$, too. Hence $GP=GD$ if and only if the quadrilateral $CEPD$ is cyclic, which solves both parts of the problem.