
G3 Let $A B C$ be a triangle in which ( $B L$ is the angle bisector of $\widehat{A B C}(L \in A C), A H$ is an altitude of $\triangle A B C(H \in B C)$ and $M$ is the midpoint of the side $[A B]$. It is known that the midpoints of the segments $[B L]$ and $[M H]$ coincides. Determine the internal angles of triangle $\triangle A B C$.


## Solution

Let $N$ be the intersection of the segments $[B L]$ and $[M H]$. Because $N$ is the midpoint of both segments $[B L]$ and $[M H]$, it follows that $B M L H$ is a parallelogram. This implies that $M L \| B C$ and $L H \| A B$ and hence, since $M$ is the midpoint of $[A B]$, the angle bisector [ $B L$ and the altitude $A H$ are also medians of $\triangle A B C$. This shows that $\triangle A B C$ is an equilateral one with all internal angles measuring $60^{\circ}$.
