Option : (B)
To prove :
That the quadrilateral formed by joining the mid points of sides of a rhombus is a rectangle.
ABCD is a rhombus P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.
Construction :
Join AC
Proof :
In ΔABC,
P and Q are the mid points of AB and BC respectively
Therefore,
PQ || AC and PQ = AC ..(i) (Mid-point theorem)
Similarly,
RS || AC and RS = AC ...(ii) (Mid-point theorem)
From (i) and (ii), we get
PQ ‖ RS and PQ = RS
Thus,
PQRS is a parallelogram
(A quadrilateral is a parallelogram, if one pair of opposite sides is parallel and equal)
AB = BC (Given)
Therefore,
\(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC
PB = BQ
(P and Q are mid points of AB and BC respectively)
In ΔPBQ,
PB = BQ
Therefore,
∠BQP = ∠BPQ ...(iii)
(Equal sides have equal angles opposite to them)
In ΔAPS and ΔCQR,
AP = CQ (AB = BC = \(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC = AP = CQ)
AS = CR (AD = CD = \(\frac{1}{2}\)AD = \(\frac{1}{2}\)CD = AS = CR)
PS = RQ
(Opposite sides of parallelogram are equal)
Therefore,
ΔAPS ≅ ΔCQR
(By SSS congruence rule)
∠APS = ∠CQR ...(iv) (By c.p.c.t)
Now,
∠BPQ + ∠SPQ + ∠APS = 180°
∠BQP + ∠PQR + ∠CQR = 180°
Therefore,
∠BPQ + ∠SPQ + ∠APS = ∠BQP + ∠PQR + ∠CQR
∠SPQ = ∠PQR ...(v)
[From (iii) and (iv)]
PS || QR and PQ is the transversal,
Therefore,
∠SPQ + ∠PQR = 180°
(Sum of adjacent interior an angles is 180°)
∠SPQ + ∠SPQ = 180° [From (v)]
2∠SPQ = 180°
∠SPQ = 90°
Thus,
PQRS is a parallelogram such that
∠SPQ = 90°
Hence,
PQRS is a rectangle.
