Question

Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel to the parallel sides.

Answer 1

Given: A trapezium $ABCD$ having $BC\parallel AD.$

Also, $E$ and $F$ are the midpoints of the non-parallel sides $AB$ and $CD$ respectively.
To prove: $BC\parallel EF\parallel AD$

Construction: Produced $BF$ to $G$ such that $AG$ is parallel to $BC.$

Proof:

In $\Delta BCF$ and $\Delta GDF,$ we have

$\angle BFC=\angle GFD$ [vertically opposite angles]

$\angle BCF=\angle GDF$ [Alternate interior angles]

Also, $CF=FD$ [Given $F$ is the midpoint of $CD$]

$\Rightarrow \Delta BCF\cong \Delta GDF$ [ by ASA congruency]

$\Rightarrow BF=FG$ [CPCT]

Now, in $\Delta ABG,$ we have

$BE=EA$ [Given, $E$ is the midpoint of $AB$]

and $BF=FG$ [proved above]

$EF\parallel AG$ [If a line joins the midpoints of two sides of a triangle then it is parallel to the third side]
$EF\parallel AD$

$\therefore AD\parallel EF\parallel BC$ [Given $AD\parallel BC$]

[Hence proved]