Question

P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that
(i) PQ || AC and PQ = $\frac{1}{2}$AC
(ii) PQ || SR
(iii) PQRS is a parallelogram.

Answer 1

Given: In quadrilateral ABCD, P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA.

To prove:

(i) PQ || AC and PQ = $\frac{1}{2}$AC

(ii) PQ || SR

(iii) PQRS is a parallelogram.

Proof:

(i)
In $∆ABC$,

Since, P and Q are the mid points of sides AB and BC, respectively. (Given)

$\Rightarrow AC\parallel PQ\mathrm{and}PQ=\frac{1}{2}AC$ (Using mid-point theorem.)

(ii)
In $∆ADC$,

Since, S and R are the mid-points of AD and DC, respectively. (Given)

$\Rightarrow SR\parallel AC\mathrm{and}SR=\frac{1}{2}AC$ (Using mid-point theorem.) ...(1)

From (i) and (1), we get

PQ || SR

(iii)
From (i) and (ii), we get

$PQ=SR=\frac{1}{2}AC$

So, PQ and SR are parallel and equal.

Hence, PQRS is a parallelogram.