Q. 24.1( 374 Votes )
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.
Answer :
The figure is given below:
To Prove: PQRS is a rectangle.
For that, we need to prove that, PQ = RS, PS = QR and
also, that angles of PQRS are all right angles.
Proof:
We start by proving the equality of sides of quadrilateral PQRS
In ΔABC, P and Q are the mid-points of sides AB and BC respectively,
The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints
of those sides is parallel to the third side and is half the length of the third side.
∴ From the mid-point theorem,
PQ || AC &
PQ = ½ AC ..(1)
Also, R and S are the mid-points of CD and AD respectively
∴ From the mid-point theorem,
RS || AC
& RS = ½ AC ..(2)
Therefore,
From equations (1) and (2), we obtain
PQ || RS and PQ = RS
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other
So it is a parallelogram.
Let the diagonals of rhombus ABCD intersect each other at point O
In quadrilateral OMQN,
MQ || ON (Because PQ || AC)
QN|| OM (Because QR || BD)
Therefore,
OMQN is a parallelogram
∠MQN = ∠NOM
∠PQR = ∠NOM
However, NOM = 900 (Diagonals of a rhombus are perpendicular to each other)
So, PQR = 900
Clearly, PQRS is a parallelogram having one of its interior angles as 900
Hence, PQRS is a rectangle
Hence, Proved.
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