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Mathematics

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order is a square only, if :

  1. ABCD is a rhombus

  2. Diagonals of ABCD are equal

  3. Diagonals of ABCD are equal and perpendicular

  4. Diagonals of ABCD are perpendicular

Mid-point Theorem

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Answer

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order is a square only, if. R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Let ABCD be a quadrilateral with P, Q, R and S as mid-points of AB, BC, CD and DA respectively.

Let diagonals be of equal length i.e, AC = BD = x and AC ⊥ BD.

In △BCA,

P and Q are mid-points of AB and BC respectively.

∴ PQ || AC and PQ = 12AC=12x\dfrac{1}{2} AC = \dfrac{1}{2}x [By mid-point theorem] …(1)

Similarly in △ACD,

S and R are mid-points of AD and CD respectively.

∴ SR || AC and SR = 12AC=12x\dfrac{1}{2}AC = \dfrac{1}{2}x [By mid-point theorem] …(2)

In △ABD,

S and P are mid-points of AD and AB respectively.

∴ SP || BD and SP = 12BD=12x\dfrac{1}{2}BD = \dfrac{1}{2}x [By mid-point theorem] …(3)

Similarly in △BCD,

Q and R are mid-points of BC and CD respectively.

∴ QR || BD and QR = 12BD=12x\dfrac{1}{2}BD = \dfrac{1}{2}x [By mid-point theorem] …(4)

From eq.(1), (2), (3) and (4), we have:

PQ = SR = SP = QR

∴ PQRS is a rhombus.

Since, SP || BD and AC ⊥ BD

∴ SP ⊥ AC

⇒ SN ⊥ AC

⇒ ∠SNO = 90°

Since, SR || AC and AC ⊥ BD

∴ SR ⊥ BD

⇒ SM ⊥ BD

⇒ ∠SMO = 90°

From figure,

⇒ ∠MOC + ∠MON = 180° [Linear pair]

⇒ 90° + ∠MON = 180°

⇒ ∠MON = 180° - 90° = 90°.

In quadrilateral sum of angles = 360°

⇒ ∠O + ∠M + ∠N + ∠S = 360°

⇒ 90° + 90° + 90° + ∠S = 360°

⇒ 270° + ∠S = 360°

⇒ ∠S = 360° - 270°

⇒ ∠S = 90°.

Since, in rhombus adjacent angles sum = 180°

Thus, in rhombus PQRS.

⇒ ∠S + ∠R = 180°

⇒ 90° + ∠R = 180°

⇒ ∠R = 180° - 90°

⇒ ∠R = 90°.

⇒ ∠Q + ∠R = 180°

⇒ 90° + ∠Q = 180°

⇒ ∠Q = 180° - 90°

⇒ ∠Q = 90°.

⇒ ∠S + ∠P = 180°

⇒ 90° + ∠P = 180°

⇒ ∠P = 180° - 90°

⇒ ∠P = 90°.

Since, PQ = QR = RS = SP and ∠P = ∠Q = ∠R = ∠S = 90°.

∴ PQRS is a square.

Thus, we can say that :

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.

Hence, option 3 is the correct option.

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