Question

Prove that the straight line joining the midpoint of the diagonals of a trapezium is parallel to the parallel sides and is equal to half their difference .

Answer 1

R.E.F image

A & B $\Rightarrow$ midpoint of

diagonals

Let PQRS be a trapezium for $\overline{PQ}\parallel \overline{SR}$

To prove : $\overline{PQ}\parallel \overline{AB}$ or $\overline{SR}\parallel \overline{AB}$ and $AB=\frac{SR-PQ}{2}$

Now $\overline{PQ}\parallel \overline{SR}$ and

$\overline{SQ}$ and $\overline{PR}$ arc diagnales,

$\angle QSR=\angle PQS=a$ $\because$ alternate angles

$\angle QPR=\angle PRS=b$ $\because$ alternate angles

also, $SA=AQ$

& $RB=BP$

$\angle PQB=\angle BXR$

$\angle QPB=\angle BRX$ }ASA congruent rule

$△QBP=△XBR$

$\Rightarrow BQ=BX\&PQ=XR$

Again in $△$ QSX,

A & B are mid point of

$QS\&XB$

So, $\overline{AB}\parallel \overline{SX}$

So, $\overline{AB}\parallel \overline{OQ}$

Hence Proved -

Also, $AB=\frac{SX}{2}=\frac{SR-XR}{2}=\frac{SR-PQ}{2}$