In given figure ∠1 = ∠2 and ∆NSQ ≅ ∆MTR , then prove that ∆PTS ~ ∆ PRQ .

OR

In an equilateral triangle ABC, D is a point on the side BC such that BD = 1/3 BC. Prove that 9AD 2 = 7AB 2

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Question 17

In an equilateral triangle ABC, D is a point on the side BC such that BD = 1/3 BC. Prove that 9AD 2 = 7AB 2

 

This is exactly same as Ex 6.5, 15.

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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 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° Now, 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 ∠ PST = ∠ PQR ∴ ΔPTS ∼ PRQ Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.