Question

I an equilateral triangle abc, d is a point on the side bc such that BD=1/3.prove that 9AD2=7AB2

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Here we go

ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC .

Now, ∆ABE and ∆AEC
∠AEB = ∠ACE = 90°
AE is common side of both triangles ,
AB = AC [ all sides of equilateral triangle are equal ]

From R - H - S congruence rule ,

∆ABE ≡ ∆ACE
∴ BE = EC = BC/2

Now, from Pythagoras theorem ,
∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)
∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2)

From equation (1) and (2)
AB² - AD² = BE² - DE²
= (BC/2)² - (BE - BD)²
= BC²/4 - {(BC/2) - (BC/3)}²
= BC²/4 - (BC/6)²
= BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9
∵AB = BC = CA
So, AB² = AD² + 2AB²/9 j
9AB² - 2AB² = 9AD²
Hence, 9AD² = 7AB²

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