Question

in an equilateral triangle abc d is a point on the side bc such that bd=1/3bc.
✓prove 9ad^2=7ab^2

Answer 1

Answer:

9ad² = 7ab²

Step-by-step explanation:

Let us assume that each side of triangle is 'a' and AE be the altitude.

Then, BE = EC = BC/2 = a/2.

∴ AE = (a√3)/2.

Given that BD = (1/3)BC.

∴ BD = a/3

DE = BE - BD

     = a/2 - a/3

     = a/6.

Apply Pythagoras theorem in ΔADE, we get

⇒ AD² = AE² + DE²

⇒ AD² = (a√3/2)² + (a/6)²

⇒ AD² = 3a²/4 + a²/36

⇒ AD² = (28/36)a²

⇒ AD² = (7/9)a²

⇒ AD² = (7/9)AB²

⇒ 9AD² = 7AB².

Hope it helps you!

Answer 2

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