I'm working on a geometry problem that goes like this:
Points M, N, and P are respectively the midpoints of sides AC, BC, and AB of triangle ABC. Prove that the area of MNP is one fourth of the area of ABC.
I drew out the diagram, and I assumed that the midsegment cut the height in half. I took one midsegment, like NP, and set it to a value of x. This means the base that it is parallel to, AC, is 2x.
The ratio of Area(MNP) and Area(ABC) is xh/2 : 4xh/2, which comes out to 1:4. However, this is assuming that the midsegment divides the height equally. Is this true?
You can say that the two triangles are similar by the intercept theorem with similarity ratio $2$, so it must be true also for the heights ${H\over h}=2$, hence the midsegment divides the height equally.