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Question
The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, ______.
Options
ABCD is a rhombus
diagonals of ABCD are equal
diagonals of ABCD are equal and perpendicular
diagonals of ABCD are perpendicular
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Solution
The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, diagonals of ABCD are equal and perpendicular.
Explanation:
Given, ABCD is a quadrilateral and P, Q, R and S are the mid-points of sides of AB, BC, CD and DA, respectively.
Then, PQRS is a square.
∴ PQ = QR = RS = PS ...(i)
And PR = SQ
But PR = BC and SQ = AB
∴ AB = BC
Thus, all the sides of quadrilateral ABCD are equal.
Hence, quadrilateral ABCD is either a square or a rhombus.
Now, in ΔADB, use mid-point theorem
SP || DB
And SP = `1/2` DB ...(ii)
Similarly in ΔABC ...(By mid-point theorem)
PQ || AC and PQ = `1/2` AC ...(iii)
From equation (i),
PS = PQ
⇒ `1/2` DB = `1/2` AC ...[From equations (ii) and (iii)]
⇒ DB = AC
Thus, diagonals of ABCD are equal and therefore quadrilateral ABCD is a square not rhombus. So, diagonals of quadrilateral are also perpendicular.
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Remark: Figure is incorrect in Question
P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.
P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square.
