Question

in triangle ABC drawn AD as BD=1/3BC prove that 9BD=7AD​

Answer 1

Answer:

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Step-by-step explanation:

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

9AB² - 2AB² = 9AD²

Hence, 9AD² = 7AB²