Ex 8.2, 2
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Given: ABCD is rhombus where
P, Q, R and S are the mid-points of the
sides AB, BC, CD and DA respectively
To prove: PQRS is a rectangle
Construction: Join A & C
Proof: A rectangle is a parallelogram with one angle 90°
First we will prove PQRS is a parallelogram,
and prove one angle 90 °
From (1) & (2)
PQ ∥ RS and PQ = RS
In PQRS,
one pair of opposite side is parallel and equal.
Hence, PQRS is a parallelogram.
Now we prove have to prove PQRS is a rectangle
Since AB = BC
1/2AB = 1/2BC
So, PB = BQ
Now, in Δ BPQ
PB = BQ
∴ ∠ 2 = ∠ 1
In ∆ APS & ∆ CQR
AP = CQ
AS = CR
PS = QR
∴ ∆ APS ≅ ∆ CQR
∠ 3 = ∠ 4
Now,
AB is a line
So, ∠ 3 + ∠ SPQ + ∠ 1 = 180°
Similarly, for line BC
∠ 2 + ∠ PQR + ∠ 4 = 180°
∠ 1 + ∠ PQR + ∠ 3 = 180°
From (5) & (6)
∠ 1 + ∠ SPQ + ∠ 3 = ∠ 1 + ∠ PQR + ∠ 3
∴ ∠ SPQ = ∠ PQR
Now,
PS ∥ QR
& PQ is a transversal
So, ∠ SPQ + ∠ PQR = 180°
∠ SPQ + ∠ SPQ = 180°
2∠ SPQ = 180°
∠ SPQ = 180"°" /2 = 90°
So, PQRS is a parallelogram with one angle 90°
∴ PQRS is a rectangle
Hence proved

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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