Mathematics
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.
Rectilinear Figures
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Answer
In Δ ABC,
P and Q are mid points of AB and BC
By mid-point theorem,
PQ || AC and PQ = …..(1)
In Δ ADC,
S and R are mid points of AD and DC.
By mid-point theorem,
⇒ SR || AC and SR = …..(2)
From equations (1) and (2), we get :
⇒ SR = PQ and SR || PQ.
Since, one of the opposite pairs of PQRS are parallel and equal.
∴ PQRS is a parallelogram.
Let the diagonals AC and BD intersect at O.
In Δ BAC,
P and Q are mid points of AB and BC.
By mid-point theorem,
PQ || AC and PQ =
∴ MQ || ON
In Δ BCD,
Q and R are mid points of BC and CD.
By mid-point theorem,
QR || BD and QR =
∴ QN || OM
Since, opposite sides of quadrilateral OMQN are parallel.
∴ OMQN is a parallelogram
⇒ ∠MQN = ∠NOM (Opposite angles of parallelogram are equal)
We know that,
Diagonals of rhombus intersect at 90°
∴ ∠NOM = 90°
⇒ ∠PQR = ∠NOM = 90°
So, ∠PQR = 90°
In PQRS,
One pair of opposite side is parallel and equal and one of its interior angle is 90°.
Hence, proved that PQRS is a rectangle.
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