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Prove that the quadrilateral formed by joining the mid-points of consecutive sides of a rhombus is a rectangle.

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#### Solution

In ΔABC, P and Q are mid points of sides AB and BC respectively.

Therefore, PQ || AC and PQ = `(1)/(2)"AC"`(using mid-point theorem) ...(1)

In ΔADC

R and S are the mid points of CD and AD respectively

Therefore, RS || AC and RS = `(1)/(2)"AC"`(using mid-point theorem) ...(2)

From equations (1) and (2), we have

PQ || RS and PQ = RS

As in quadrilateral PQRS one pair of opposite sides are equal and parallel to each other, so, it is a parallelogram.

Let diagonals of rhombus ABCD intersects each other at point O.

Now in quadrilateral OMQN

MQ || ON ...(PQ || AC)

QN || OM ...(QR || BD)

So, OMQN is parallelogram

∠MQN = ∠NOM

∠PQR = ∠NOM

But, ∠NOM = 90° ...(diagonals of a rhombus are perpendicular to each other)

∠PQR = 90°

Clearly PQRS is a parallelogram having one of its interior angle as 90°.

Hence, PQRS is rectangle.

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(i) ∠DAE

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(iv) ∠AEC

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