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ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.

Given: ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA.

By midpoint theorm, Now, in $$\triangle{ADC}$$, we have,
Thus, SR || AC ...(i)
and SR = ($$\frac{1}{2}$$ ) AC ...(ii)
Similarly, in $$\triangle{ABC}$$, we have,
PQ || AC ...(iii)
and PQ = ($$\frac{1}{2}$$ ) AC ...(iv)
from (i), (ii), (iii) and (iv), we get,
SR = PQ = ($$\frac{1}{2}$$ ) AC and PQ || SR
$$\therefore$$ Quadrilateral PQRS is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at right angles.
$$\therefore$$ $$\angle{EOF}$$ = $$90^\circ$$.

Now, By midpoint theorem, we have,
RQ || BD
Thus, RE || OF
As SR || AC ...(from (i))
Thus, FR || OE

$$\therefore$$ OERF is a parallelogram
So, $$\angle{ERF}$$ = $$\angle{EOF} = 90°$$
(since, Opposite angle of a quadrilateral is equal)

Thus, PQRS is a parallelogram with $$\angle{R}= 90°$$
Hence, it is proved that PQRS is a rectangle.