The Varignon parallelogram is half the quadrilateral

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Given an arbitrary quadrilateral $ABCD$, let $M$, $N$, $P$, $Q$ be the midpoints of sides $AB$, $BC$, $CD$, $DA$. Prove that $MNPQ$ is a parallelogram and that it is equivalent to half of the quadrilateral $ABCD$.

Solution

$MNPQ$ is a parallelogram. Draw the diagonal $AC$. In triangle $ABC$, the segment $MN$ joins the midpoints of $AB$ and $BC$, so $MN\parallel AC$ and $\overline{MN}=\tfrac12\overline{AC}$. In triangle $ACD$, the segment $QP$ joins the midpoints of $DA$ and $CD$, so $QP\parallel AC$ and $\overline{QP}=\tfrac12\overline{AC}$. Sides $MN$ and $QP$ are then parallel and equal: $MNPQ$ is a parallelogram.

Comparing the areas. The parallelogram is obtained by removing from $ABCD$ the four “corner” triangles $MBN$, $NCP$, $PDQ$, $QAM$. Let us find their total area. In triangle $ABC$, triangle $MBN$ has sides $\overline{MB}=\tfrac12\overline{AB}$ and $\overline{BN}=\tfrac12\overline{BC}$ and the same angle at $B$: it is similar to $ABC$ with ratio $\tfrac12$, hence $\mathcal{A}(MBN)=\tfrac14\,\mathcal{A}(ABC).$ Likewise $\mathcal{A}(PDQ)=\tfrac14\,\mathcal{A}(ACD)$. Adding, $\mathcal{A}(MBN)+\mathcal{A}(PDQ)=\tfrac14\big(\mathcal{A}(ABC)+\mathcal{A}(ACD)\big)=\tfrac14\,\mathcal{A}(ABCD).$ Repeating the argument with the diagonal $BD$ for triangles $NCP$ and $QAM$: $\mathcal{A}(NCP)+\mathcal{A}(QAM)=\tfrac14\big(\mathcal{A}(BCD)+\mathcal{A}(ABD)\big)=\tfrac14\,\mathcal{A}(ABCD).$ The total area of the four corner triangles is therefore $\tfrac14\,\mathcal{A}(ABCD)+\tfrac14\,\mathcal{A}(ABCD)=\tfrac12\,\mathcal{A}(ABCD).$ Consequently $\mathcal{A}(MNPQ)=\mathcal{A}(ABCD)-\tfrac12\,\mathcal{A}(ABCD)=\tfrac12\,\mathcal{A}(ABCD).$ The midpoint parallelogram is thus equivalent to half the quadrilateral. $\;\blacksquare$