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Question
Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.
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Solution
Given: In a square ABCD, P, Q, R and S are the mid-points of AB, BC, CD and DA, respectively.
To show: PQRS is a square.
Construction: Join AC and BD.
Proof: Since, ABCD is a square.
∴ AB = BC = CD = AD
Also, P, Q, R and S are the mid-points of AB, BC, CD and DA, respectively.
Then, in ΔADC, SR || AC
And SR = `1/2`AC [By mid-point theorem] ...(i)
In ΔABC, PQ || AC
And PQ = `1/2`AC ...(ii)
From equations (i) and (ii),
SR || PQ and SR = PQ = `1/2`AC ...(iii)
Similarly, SP || BD and BD || RQ
∴ SP || RQ and SP = `1/2`BD
And RQ = `1/2`BD
∴ SP = RQ = `1/2`BD
Since, diagonals of a square bisect each other at right angle.
∴ AC = BD
⇒ SP = RQ = `1/2`AC ...(iv)
From equations (iii) and (iv),
SR = PQ = SP = RQ ...[All side are equal]
Now, in quadrilateral OERF,
OE || FR and OF || ER
∴ ∠EOF = ∠ERF = 90°
Hence, PQRS is a square.
Hence proved.
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Remark: Figure is incorrect in Question
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