# Proving that the midpoints of the sides of $\triangle ABC$, and one vertex, determine a parallelogram with area half that of the triangle.

The question says:

The midpoints of the sides of $$\triangle ABC$$ along with any of the vertices as the fourth point make a parallelogram of area equal to what? The answer is, obviously, $$\frac12 \operatorname{area}(\triangle ABC)$$

In the method, I took $$\triangle ABC$$ with $$D$$, $$E$$, and $$F$$ as midpoints of $$AB$$, $$AC$$, and $$BC$$, respectively; and I joined $$DE$$ and $$EF$$ so that I get a parallelogram $$\square DEFB$$.

I know what the answer is because one can easily make that out. Also, those four triangles (four because the parallelogram can still be divided into two triangles and the rest two triangles add up to four) so it's simple that the area of $$\square DEBF$$ will be $$1/4$$ of $$\triangle ABC$$, but how?

Can anyone explain me this?

Help will be appreciated :)

All $$4$$ triangles are congruent using properties of parallel lines and parallelograms and so are all equal in area. All $$4$$ make up triangle $$ABC$$, while $$2$$ make up parallelogram $$DEFB$$. Hence $$DEFB$$ is half the area of $$ABC$$.