Question 17
In given figure ∠1 = ∠2 and ∆NSQ ≅ ∆MTR , then prove that ∆PTS ~ ∆ PRQ .
Given: ∠ 1 = ∠ 2
& ΔNSQ ≅ ΔMTR
To Prove: ΔPTS ∼ PRQ
Proof:
Given ΔNSQ ≅ ΔMTR
∴ ∠ NQS = ∠ MRT
i.e. ∠ PQR = ∠ PRQ
Now,
In Δ PST
By angle sum property
∠ P + ∠ 1 + ∠ 2 = 180°
Since ∠ 1 = ∠ 2 given
∠ P + ∠ 1 + ∠ 1 = 180°
∠ P + 2 ∠ 1 = 180°
In Δ PQR
By angle sum property
∠ P + ∠ PQR + ∠ PRQ = 180°
From (1), ∠ PQR = ∠ PRQ
∠ P + ∠ PQR + ∠ PQR = 180°
∠ P + 2 ∠ PQR = 180°
Right side of (2) and (3) are same
So, from (2) and (3)
∠ P + 2 ∠ 1 = ∠ P + 2 ∠ PQR
2 ∠ 1 = 2 ∠ PQR
∠ 1 = ∠ PQR
In ΔPTS & Δ PRQ
∠P = ∠P (Common angle)
∠ PST = ∠ PQR (From (4))
∴ ΔPTS ∼ PRQ (AA similarity)
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.