### 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 a question of CBSE Sample Paper - Class 10 - 2017/18.   ### 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.

1. Class 10
2. Solutions of Sample Papers for Class 10 Boards
3. CBSE Class 10 Sample Paper for 2018 Boards

Transcript

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

CBSE Class 10 Sample Paper for 2018 Boards

Class 10
Solutions of Sample Papers for Class 10 Boards 