Question

Prove that "In a trapezium, the line joining the mid points 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 it cut XY at P
Proof: (i) In $△ZBC,AD||BC$ [$\because$ Data]
$\therefore \frac{ZA}{AB}=\frac{ZD}{DC}$ [$\because$ BPT]
$\therefore \frac{ZA}{2AX}=\frac{ZD}{2DY}$ [$\because$ X & Y are mid points of AB and DC]
$\therefore \frac{ZA}{AX}=\frac{ZD}{DY}$
$\Rightarrow XY||AD$ [$\because$ Converse of B.P.T.]
(ii) In $△ABC,AX=XB$ [$\because$ Data]
$XP||BC$ [$\because$ Proved]
$\therefore AP=PC$ [$\because$ Converse of mid point theorem]
$\therefore XP=\frac{1}{2}BC$ [$\because$ Midpoint Theorem]
In $△ADC,PY=\frac{1}{2}AD$
By adding, we get $XP+PY=\frac{1}{2}BC+\frac{1}{2}AD$
$\therefore XY=\frac{1}{2}(BC+AD)$