The shaded triangles have heights equal to half the height of \( \triangle ABC\), and their combined bases are equal to \( \overline{DE}\). Because it connects the midpoints of the sides, \(\overline{DE}\) is half as long as \(\overline{AB}\) . If the area of \(\triangle ABC\) is \(\frac{1}{2} bh\) the area of one shaded triangle is \(\frac{1}{2}(\frac{1}{4} b )(\frac{1}{2} h)=\frac{1}{16} bh\) so the shaded area of the two triangles is \(\frac{1}{8} bh
\)

Because \(\frac{\frac{1}{8} bh }{\frac{1}{2 }bh} = \frac{1}{4}\) , one fourth of \(\triangle ABC\) is shaded