In a triangle $ABC$, let $D$ be a point on the segment $BC$ such that $AB + BD = AC + CD$. Suppose that the points $B,C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB = AC$.
I began this way: Denote the centroid of triangle $ABD$ by $G_1$ and the centroid of triangle $ACD$ by $G_2$. Then by using $Basic$ $Proportionality$ $Theorem$, I got that $BG_1G_2C$ is an isosceles trapezium. After that I am stuck. Please help.