find the solution from pic
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Heya !!
Given :- An equilateral ∆ ABC in which D is a point on BC such that BD = 1/3 BC
To prove :- 9AD² = 7AB²
Construction :- Draw AE perpendicular to BC. Join AD.
Proof :- In ∆ADE,
AD² = AE² + DE²
=> AD² = AE² + (BE–BD)²
=> AD² = AE² + BE² + BD² – 2(BE)(BC)
=> AD² = AB² + (BC/3)² – 2 (BC/2) (BC/3)
=> AD² = AB² + (AB²/9) – (BC²/3)
=> AD² = ( 9AB² + AB² – 3AB²) / 9
=> AD² = 7AB²/9
=> 9AD² = 7AB²
Hence, proved.
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Hope my ans.'s satisfactory.☺
Given :- An equilateral ∆ ABC in which D is a point on BC such that BD = 1/3 BC
To prove :- 9AD² = 7AB²
Construction :- Draw AE perpendicular to BC. Join AD.
Proof :- In ∆ADE,
AD² = AE² + DE²
=> AD² = AE² + (BE–BD)²
=> AD² = AE² + BE² + BD² – 2(BE)(BC)
=> AD² = AB² + (BC/3)² – 2 (BC/2) (BC/3)
=> AD² = AB² + (AB²/9) – (BC²/3)
=> AD² = ( 9AB² + AB² – 3AB²) / 9
=> AD² = 7AB²/9
=> 9AD² = 7AB²
Hence, proved.
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Hope my ans.'s satisfactory.☺
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