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 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, Social Science, Physics, Chemistry, Computer Science at Teachoo.
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