# prove that $\angle BGC=90^\circ$

In a $\triangle ABC$ the medians $BE$ and $CF$ meet at the centroid G. Given that $AG = BC$, prove that $\angle BGC=90^\circ$

We have $BC=AG=2GM$, with $M$ is the midpoint of $BC$. Consider triangle $GBC$ which has median $GM$ and $BC=2GM$, then $\angle BGC=90^{\circ}$