The midpoints of the sides create four equal-area triangles

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In a triangle $ABC$ let $D$, $E$, $F$ be the midpoints of sides $BC$, $CA$, $AB$ respectively. Prove that the segments $DE$, $EF$, $FD$ divide the triangle into four equal-area triangles, each with area equal to one quarter of that of $ABC$.

Solution

By the midpoint theorem (the midsegment theorem), each segment joining the midpoints of two sides is parallel to the third side and half as long: $\overline{EF}=\tfrac12\,\overline{BC},\qquad \overline{FD}=\tfrac12\,\overline{CA},\qquad \overline{DE}=\tfrac12\,\overline{AB}.$ Consider the four triangles into which $ABC$ is split: the three “corner” ones $AFE$, $FBD$, $EDC$ and the “central” one $DEF$.

• Triangle $DEF$ has sides $\overline{DE}$, $\overline{EF}$, $\overline{FD}$, i.e. the halves of $AB$, $BC$, $CA$.
• Triangle $AFE$ has $\overline{AF}=\tfrac12\overline{AB}$, $\overline{AE}=\tfrac12\overline{CA}$ and $\overline{FE}=\tfrac12\overline{BC}$: its three sides are the halves of the corresponding sides of $ABC$.

Repeating the argument for $FBD$ and $EDC$, one sees that the four triangles have their three sides respectively equal to $\tfrac12\overline{AB}$, $\tfrac12\overline{BC}$, $\tfrac12\overline{CA}$. By the SSS criterion they are therefore congruent to one another, hence equivalent. Since together they recompose the whole triangle $ABC$, each has area $\mathcal{A}=\tfrac14\,\mathcal{A}(ABC).$ The four triangles are therefore equivalent. $\;\blacksquare$