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Question
ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the
mid-points of the sides, in order, is a rectangle.
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Solution
Given,
A kite ABCD having AB = AD and BC = CD. P,Q,R, S are the midpoint of sides
AB,BC,CD andDArespectively PQ,QR,RS and spare joined
To prove:
PQRS is a rectangle
Proof:
In ΔABC, P and Q are the midpoints of AB and BC respectively.
∴PQ || AC and PQ = `1/2` AC ....(i )
In Δ ADC, R and S are the midpoint of CD and AD respectively.
∴ RS || AC and RS = `1/2` AC .....(ii )
From (i) and (ii), we have
PQ || RS and PQ = RS
Thus, in quadrilateral PQRS, a pair of opposite sides are equal and parallel. So PQRS is a
parallelogram. Now, we shall prove that one angle of parallelogram PQRS it is a right angle
Since AB = AD
⇒ `1/2` AB = AD `(1/2)`
⇒ AP = AS ...(iii) [∵ P and S are the midpoints of B and AD respectively]
⇒ `∠`1 = `∠`2 ....(iv)
Now, in ΔPBQ and ΔSDR, we have
PB = SD [ ∴AD = AB ⇒ `1/2` AD = `1/2` AB ]
BQ = DR ∴ PB = SD
And PQ = SR [ ∴ PQRS is a parallelogram]
So by SSS criterion of congruence, we have
Δ PBQ ≅ Δ SOR
⇒ `∠`3 = `∠`4 [CPCT ]
Now, `∠`3 + `∠`SPQ + `∠`2 = 180°
And `∠`1+ `∠`PSR + `∠`4 =180°
∴ `∠`3 + `∠`SPQ + `∠`2 = `∠`1+ `∠`PSR + `∠`4
⇒ `∠`SPQ = `∠`PSR (`∠`1 = `∠`2 and `∠`3 = `∠`4)
Now, transversal PS cuts parallel lines SR and PQ at S and P respectively.
∴ `∠`SPQ + `∠`PSR = 180°
⇒ 2`∠`SPQ = 180° = `∠`SPQ = 90° [ ∵ `∠`PSR = `∠`SPQ]
Thus, PQRS is a parallelogram such that `∠`SPQ = 90°
Hence, PQRS is a parallelogram.
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