Question

PQRS is a trapezium such that PQRS and L, M are the mid points of the diagonals PR and QS, respectively.Then,show that $LM=\frac{1}{2}(PQ-RS)$

Answer 1

PQRS is a trapezium where E and F are the mid - points of PS and RQ respectively.
To prove:
(i) $EF\parallel PQ$
(ii) $EF=\frac{1}{2}(PQ+SR)$
Construction: Join SQ intersecting EF in O.

Proof: Let O' be the mid-point of SQ
In $\Delta SPQ$, E and O' are the mid-points of SP and SQ respectively.
$\therefore E{O}^{{}^{\prime }}\parallel PQ\dots \dots \left(1\right)$ (Mid - point theorem)
Similarly, $O^{{}^{\prime }}F\parallel SR$
Now, $SR\parallel PQ$ (given)
$\therefore O^{{}^{\prime }}F\parallel PQ\dots \dots \left(2\right)$
From (1) and (2), we get
EO'F is a straight line
But EO'F is not a straight line
But EO'F is a straight line only when O and O' coincide
$\therefore$ O is the midpoint of SQ
$\Rightarrow OF\parallel PQ$ (from (2))
$\therefore EF\parallel PQ$
In $\Delta SPQ$, E and O are the mid - points of SP and SQ
$\therefore OE=\frac{1}{2}PQ\dots \dots \left(3\right)$ (mid - point theorem)
In $\Delta SQR$, F and O are the mid - points of QR and SQ
$\therefore OF=\frac{1}{2}SR\dots \dots \left(4\right)$ (mid - point theorem)
Adding (3) and (4), we get
$OE+OF=\frac{1}{2}PQ+\frac{1}{2}SR$
$\Rightarrow EF=\frac{1}{2}(PQ+SR)$