Question

Question 12
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD, Prove that $EF||ABandEF=\frac{1}{2}(AB+CD)$.

Answer 1

Given ABCD is a trapezium in which AB||CD. Also, E and F are respectively the mid-points of sides AD and BC.

Construction:- Join BE and produce it to meet CD produced at G, also draw BD which intersects EF at O.
To prove EF|| AB and $EF=\frac{1}{2}(AB+CD)$

Proof :- In $\Delta GCB,E$ and F are respectively the mid-points of BG and BC, then by mid-point theorem,

EF||GC

But GC||AB or CD||AB

$\therefore EF||AB$

In $\Delta ADB,AB||EO$ and E is the mid-point of AD.

Therefore by converse of mid-point theorem, O is mid-point of BD.
Also, $EO=\frac{1}{2}AB$

In $\Delta BDC,OF||CD$ and O is the mid-point of BD.
$\therefore$ $OF=\frac{1}{2}CD$ [by converse of mid-point theorem]

On adding Eqs. (i) and (ii), we get

$EO+OF=\frac{1}{2}AB+\frac{1}{2}CD\Rightarrow EF=\frac{1}{2}(AB+CD)$
Hence, proved