Question

Prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to the parallel sides are equal to half of their difference.

Answer 1

Let $E$ and $F$ are midpoints of the diagonals $AC$ and $BD$ of trapezium $ABCD$ respectively.

Draw $DE$ and produce it to meet $AB$ at $G$

Consider $△AEG$ and $△CED$

$\Rightarrow$ $\angle AEG=\angle CED$ [ Vertically opposite angles ]

$\Rightarrow$ $AE=EC$ [ $E$ is midpoint of $AC$ ]

$\Rightarrow$ $\angle ECD=\angle EAG$ [ Alternate angles ]

$\Rightarrow$ $△AEG\cong △CED$ [ By $SAA$ congruence rule ]

$\Rightarrow$ $DE=EG$ ---- ( 1 ) [ CPCT ]

$\Rightarrow$ $AG=CD$ ----- ( 2 )

In $△DGB$

$E$ is the midpoint of $DG$ [ From ( 1 ) ]

$F$ is midpoint of $BD$

$\therefore$ $EF\parallel GB$

$\Rightarrow$ $EF\parallel AB$ [ Since $GB$ is part of $AB$ ]

$\Rightarrow$ $EF$ is parallel to $AB$ and $CD.$

Also, $EF=\frac{1}{2}GB$

$\Rightarrow$ $EF=\frac{1}{2}(AB-AG)$

$\Rightarrow$ $EF=\frac{1}{2}(AB-CD)$ [ From ( 2 ) ]