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 $\parallel ACandPQ=\frac{1}{2}AC$
(ii) PQ $\parallel SR$
(iii) PQRS is a parallelogram.
Proof:
(i)
In$△ABC$
Since, P and Q are the mid points of sides AB and BC, respectively.
$\to AC\parallel PQandPQ=\frac{1}{2}AC$
(Given)
(Using mid-point theorem.)
(ii)
In $triangleADC$
Since, S and R are the mid-points of AD and DC, respectively.
$\to SR\parallel ACandSR=\frac{1}{2}AC$
(Using mid-point theorem.)
From (i) and (1), we get
$PQ\parallel SR$
(iii)
From (i) and (ii), we get
$\to PQ=SR=\frac{1}{2}AC$
So, PQ and SR are parallel and equal.
Hence, PQRS is a parallelogram.