The midpoints of the sides of $\mathrm{\u25b3}ABC$ along with any of the vertices as the fourth point make a parallelogram of area equal to what? The answer is, obviously, $\frac{1}{2}\mathrm{area}(\mathrm{\u25b3}ABC)$

In the method, I took $\mathrm{\u25b3}ABC$ 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 $\u25fbDEFB$.

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 $\u25fbDEBF$ will be 1/4 of $\mathrm{\u25b3}ABC$, but how?

Can anyone explain me this?

In the method, I took $\mathrm{\u25b3}ABC$ 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 $\u25fbDEFB$.

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 $\u25fbDEBF$ will be 1/4 of $\mathrm{\u25b3}ABC$, but how?

Can anyone explain me this?