Question

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 $\Delta ABD$ & $\Delta ACD$ lie on a circle. Prove that $AB=AC$.

Answer 1

Let $G_1,G_2$ denote the centroids of triangles ABD and ACD.
Then $G_1,G_2$ lie on the line parallel to BC that passes through the centriod of triangle ABC.
$\therefore B{G}_1{G}_{2}C$ is an isosceles trapezoid.
Therefore it follows that $B{G}_1=C{G}_2$.
$\therefore A{B}^2+B{D}^2=A{C}^2+C{D}^2$.
Hence it follows that $AB\cdot BD=AC\cdot CD$.
Therefore the sets $\{AB,BD\}$ and $\{AC,CD\}$ are the same (since they are both equal to the set of roots of the same polynomial).
Note that if $AB=CD$ then $AC=BD$ and then $AB+AC=BC$, a contradiction.
$\therefore AB=AC$.