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Question
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
Solution
Given: Let ABCD be a rectangle where P, Q, R, S are the midpoint of AB, BC, CD, DA.
To Prove: PQRS is a rhombus
Construction: Draw two diagonal BD and AC as shown in figure. Where BD = AC
(Since diagonal of the rectangle are equal)
Proof:
From ΔABD and ΔBCD
PS = `1/2` BD = QR and PS || BD || QR
2PS = 2QR = BD and PS || QR ...(1)
Similarly, 2PQ = 2SR = AC and PQ || SR ...(2)
From (1) and (2) we get
PQ = QR = RS = PS
Therefore, PQRS is a rhombus.
Hence, proved.
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