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P, Q, R and S are respectively the mid-p...

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 `bot` BD. Prove that PQRS is a rectangle.

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Given, ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA.
`therefore` by mid-point theorem,
`In DeltaADC," "SR"||"ACandSR=1/2AC" "...(1)`
`In DeltaABC," "PQ"||"ACandPQ=1/2AC" "...(2)`
From (1) and (2), we get
`PQ"||"SRandPQ=SR=1/2AC`
`thereforePQRS"is a parallelogram."" "(because"one pair of opposite sides is equal and parallel")`
Now, we know that diagonals of a rhombus bisect each other at right angles.
`therefore" "angleEOF=90^(@)`
`Now," "RQ"||"BD" "("by mid-point theorem")`
`implies" "RE"||"FO`
`Also," "SR"||"AC" "["from (1)"]`
`implies" "FR"||"OE`
`thereforeOERF "is parallelogram."" "(because"pairs of opposite sides are parallel")`
`So," "angleERF=angleEOF=90^(@)`
`" "(because"pairs of opposite sides are parallel")`
Thus, PQRS is a parallelogram with `angleR=90^(@).`
Hence, PQRS is a rectangle.
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