Question

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

Answer 1

Given that ABCD is a rectangle and P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.

Now join points AC and BD as shown in the figure.

Now in ΔABC,

since P and Q are the mid points of AB and BC, then from mid point theorem

PQ || AC and PQ = AC ..........1

Now in ΔADC,

since S and R are the mid points of AD and CD, then from mid point theorem

SR || AC and SR = AC ..........2

From equation 1 and 2, we get

PQ || SR and PQ = SR ....3

So quadrilateral ABCD is a parallelogram.

Now

PS || QR and PS = QR ......4 (Opposite side of parallelogram)

Now in ΔBCD,

since Q and R are the mid points of BC and CD, then from mid point theorem

QR || BD and QR = BD ..........5



Again diagonal of a rectangle are equal.

So AC = BD .........6

From equation1 ,2 ,3 4,,5 and 6, we get

PQ = QR = RS = SP

Hense PQRS is a rhombus

Answer 2

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