Question

If $E$, $F$, $G$ and $H$ are respectively the mid-points of the sides of a parallelogram $ABCD$, show that $ar\left( {EFGH} \right) = \dfrac{1}{2}ar\left( {ABCD} \right)$.

Answer 1

Hint:In the solution, we will use the midpoint theorem. From the concept of midpoint theorem, in a triangle, when line segments connect the mid-points of two sides, then that line segments will be parallel to the remaining side.


Complete step-by-step solution
Given: $E$, $F$, $G$ and $H$ are respectively the mid-points of the sides of the parallelogram $ABCD$.

$AD\parallel BC$ and $AD = BC$ (Opposite sides of a parallelogram). Therefore,
$\dfrac{1}{2}AD = \dfrac{1}{2}BC$.
Also,
$AH\parallel BF$ and $DH = CF$. Therefore, $H$ and $F$ are mid points.
Thus, it is clear that $ABFH$ and $HFCD$ are parallelograms.
Now, $\Delta EFH$ and parallelogram $ABFH$ lie on the same base $FH$ and between the same parallel lines $AB$ and $HF$.
Area of $EFH$ is,
${A_1} = \dfrac{1}{2}ar\left( {ABFH} \right)$…..(1)
Also, Area of $GHF$ is,
${A_2} = \dfrac{1}{2}ar\left( {HFCD} \right)$…..(2)
On adding equation (1) and equation (2).
\[\begin{array}{c}
Area{\rm{ }}of\;EFH\; + Area{\rm{ }}of\;GHF = \dfrac{1}{2}ar\left( {ABFH} \right) + \dfrac{1}{2}ar\left( {HFCD} \right)\\
Area(HEFG) = \dfrac{1}{2}\left( {ar\left( {ABFH} \right) + ar\left( {HFCD} \right)} \right)\\
Area(HEFG) = \dfrac{1}{2}Area(ABCD)
\end{array}\]
Hence, it is proved that $are\left( {EFGH} \right) = \dfrac{1}{2}ar\left( {ABCD} \right)$.

Note: Make sure to use the Midpoint theorem when any question is asking about a quadrilateral with midpoints and use Angle Side Angle similar (ASA) triangle properties to compare two triangles.