Hello, Cthul!
"Also Sprach" is absolutely correct.
We have $\displaystyle \Delta ABC$ with median $\displaystyle AM$ drawn to side $\displaystyle BC.$
. . Hence: .$\displaystyle BM \,=\,MC.$
Draw altitude $\displaystyle h$ from vertex $\displaystyle A$ to side $\displaystyle BC.$
Code:
A
o
*:* *
* : * *
* : * *
* : * *
* :h * *
* : * *
* : * *
* : * *
* : * *
B o * * * * + * * * * o * * * * * * * * * o C
M
Triangles $\displaystyle ABM$ and $\displaystyle AMC$ have equal bases and the same altitude.
. . Hence, they have equal areas.
A median always divides a triangle into two equal areas.