Equality of areas in a triangle

Show that the areas of the two triangles $ABM$ and $ACM$ , defined by a median of an arbitrary triangle $ABC$ as in the following figure, are equal.

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Solution by Ancient Tree

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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.