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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.