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.
Now, in the standard notation $$c+BD=b+a-BD,$$ which gives $$BD=\frac{a+b-c}{2}$$ and $$CD=\frac{a+c-b}{2}.$$
Thus, $$CG_2=\frac{1}{3}\sqrt{2b^2+2\left(\frac{a+c-b}{2}\right)^2-AD^2}$$ and $$BG_1=\frac{1}{3}\sqrt{2c^2+2\left(\frac{a+b-c}{2}\right)^2-AD^2},$$ which gives $$2b^2+2\left(\frac{a+c-b}{2}\right)^2-AD^2=2c^2+2\left(\frac{a+b-c}{2}\right)^2-AD^2$$ or $$(b-c)(b+c-a)=0$$ or $$b=c.$$