Q20 of 40 Page 8

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

To Prove: PQRS is a rhombus.

Given: ABCD is a rectangle and P, Q, R, and S are midpoints of AB, BC, CD, and DA.


Concept Used:


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.


Diagram:



Explanation:


PQ || AC and PQ = 1/2 AC (Mid-point theorem) (1)


Similarly, in ΔADC,


SR || AC and SR = 1/2 AC (Mid-point theorem) (2)


Clearly,


PQ || SR and PQ = SR


Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other; it is a parallelogram


PS || QR and PS = QR (Opposite sides of a parallelogram) (3)


In ΔBCD, Q and R are the mid-points of side BC and CD respectively.


QR || BD and QR = 1/2 BD (Mid-point theorem) (4)


However, the diagonals of a rectangle are equal.


AC = BD (5)


By using equation (1), (2), (3), (4), and (5), we obtain


PQ = QR = SR = PS


Now, as all sides of the rhombus are equal.


Hence, PQRS is a rhombus


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