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