Point $ M$ is taken on side $ BC$ of a triangle $ ABC$ such that the centroid $ T_c$ of triangle $ ABM$ lies on the circumcircle of $ \triangle ACM$ and the centroid $ T_b$ of $ \triangle ACM$ lies on the circumcircle of $ \triangle ABM$. Prove that the medians of the triangles $ ABM$ and $ ACM$ from $ M$ are of the same length.
Call $ X,Y$ is the middle point of $ AB,AC$
From the condition we have
$ \angle{AT_{c}X}=\angle {ACB}=\angle{AYX}$
So $ T_c\in (AXY)$
Iin the same method easy to check $ T_{b}\in (AXY)$
So $ MX.MT_c=MY.MT_{b}$
But
$ \frac{MX}{MT_c}=\frac{MY}{MT_b}=\frac{3}{2}$
So $ MX=MY$
April wrote:
Point $ M$ is taken on side $ BC$ of a triangle $ ABC$ such that the centroid $ T_c$ of triangle $ ABM$ lies on the circumcircle of $ \triangle ACM$ and the centroid $ T_b$ of $ \triangle ACM$ lies on the circumcircle of $ \triangle ABM$. Prove that the medians of the triangles $ ABM$ and $ ACM$ from $ M$ are of the same length.
Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200221