Question

Prove that: In a trapezium, the line joining the midpoints of non-parallel sides is
(i) Parallel to the parallel sides and
(ii) Half of the sum of the parallel sides.

Answer 1

Data: In the trapezium $ABCD$,
$AD||BC$, $AX=XB$ and $DY=YC$.

To Prove: (i) $XY||AD$ or $XY||BC$
(ii) $XY=\frac{1}{2}(AD+BC)$

Construction: Extend $BA$ and $CD$ to meet at $Z$. Join $A$ and $C$. Let in cut $XY$ at $P.$

Proof:

$\left(i\right)$ In $△ZBC,$ $AD||BC$
$\frac{ZA}{AB}=\frac{ZD}{DC}$
$\frac{ZA}{2AX}=\frac{ZD}{2DY}$ [$\because$ $X$ and $Y$ are mid points of $AB$ and $DC$]
$\frac{ZA}{AX}=\frac{ZD}{DY}$
$\Rightarrow XY||AD$ [$\because$ Converse of B.P.T.]

(ii) In $△ABC,AX=XB$ [$\because$ Data]
$XP||BC$ [Already Proved]
$AP=PC$ [$\because$ Converse of mid point theorem]
$XP=\frac{1}{2}BC$ [$\because$ Midpoint Theorem]

In $△ADC,$

$PY=\frac{1}{2}AD$.

By adding, we get,

$XP+PY=\frac{1}{2}BC+\frac{1}{2}AD$
$\therefore XY=\frac{1}{2}(BC+AD)$ [Hence Proved]