Question

Question 4
P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that $AC\perp BD$. Prove that PQRS is a rectangle.

Answer 1

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

Also, $AC\perp BD$

To prove PQRS is a rectangle.

Proof since, $AC\perp BD$

$\therefore \angle COD=\angle AOD=\angle AOB=\angle COB={90}^{\circ }$

In $\Delta ADC$, S and R are the mid-points of AD and DC respectively, then by mid-point theorem

SR||AC and $SR=\frac{1}{2}AC$ ………………..(i)

In ABC, P and Q are the mid-points of AB and BC respectively, then by mid-point theorem,

$PQ\parallel AC\text{and}PQ\frac{1}{2}=AC$ ……(ii)

From Eqs.(i) and (ii), $PQ\parallel SR\text{and}PQ=SR=\frac{1}{2}AC$ ….(iii)

Similarly SP ∥ RQ and $SP=RQ=\frac{1}{2}BD$ …..(iv)

Now, in quadrilateral EOFR, OE ∥FR, OF ∥ ER

$\therefore \angle EOF=\angle ERF={90}^{\circ }[\because \angle COD={90}^{\circ }\Rightarrow \angle EOF={90}^{\circ }]$ ….(v)

So, PQRS is a rectangle. Hence proved.