Equality of areas in a triangle

By definition, the median of $ABC$ from vertex $A$ splits $BC$ into two segments of same length :
$$BM=MC$$Furthermore, the two triangles $ABM$ and $AMC$ have the same height $AH$.

The area of the two triangles are respectively $\frac{BM \times HA}{2}$ and $\frac{MC\times AH}{2}$, which are therefore equal.