Here is a triangle, whose area is $\frac{1}{2}(V-V_o)t$, where $V_o$ and $V$ are $y$-coordinates.
$\frac{1}{2}(V-V_o)t$ is also $\frac{1}{2}Vt - \frac{1}{2}V_ot$ i.e., difference between two other triangles.
While this is true, how to geometrically see how this works?
The key is first notice every triangle can produce a parallelogram by joining a copy of itself in inverse direction. Now it suffice to show their parallelograms have equal area. For parallelogram with inner altitude you can see it can be rectangle with equal area by replacing the right triangle that altitude separate it. In case of outer altitude you can check yourself as an exercise.