Math, asked by shanaya9866, 1 year ago

in an equilateral triagleABC , D is a point on BC such that BD =1/3BC .prove that 9AD²=7AB²

Answers

Answered by Anonymous
1

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

And, AE = a√3/2

Given that, BD = 1/3BC

∴ BD = a/3

DE = BE - BD = a/2 - a/3 = a/6

Applying Pythagoras theorem in ΔADE, we get

AD2 = AE2 + DE2 


⇒ 9 AD2 = 7 AB2





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Answered by Anonymous
2

Step-by-step explanation:

➡ Given :-

→ A ∆ABC in which AB = BC = CA and D is a point on BC such that BD = ⅓BC.

➡ To prove :-

→ 9AD² = 7AB² .

➡ Construction :-

→ Draw AL ⊥ BC .

➡ Proof :-

In right triangles ALB and ALC, we have

AB = AC ( given ) and AL = AL ( common )

∴ ∆ALB ≅ ∆ ALC [ By RHS axiom ] .

So, BL = CL .

Thus, BD = ⅓BC and BL = ½BC .

In ∆ALB, ∠ALB = 90° .

∴ AB² = AL² + BL² .......(1) [ by Pythagoras' theorem ] .

In ∆ALD , ∠ALD = 90° .

∴ AD² = AL² + DL² . [ by Pythagoras' theorem ] .

⇒ AD² = AL² + ( BL - BD )² .

⇒ AD² = AL² + BL² + BD² - 2BL.BD .

⇒ AD² = ( AL² + BL² ) + BD² - 2BL.BD .

⇒ AD² = AB² + BD² - 2BL.BD. [ using (1) ]

⇒ AD² = BC² + ( ⅓BC )² - 2( ½BC ). ⅓BC .

[ ∵ AB = BC, BD = ⅓BC and BL = ½BC ] .

⇒ AD² = BC² + 1/9BC² - ⅓BC² .

⇒ AD² = 7/9BC² .

⇒ AD² = 7/9AB² [ ∵ BC = AB ] .

 \huge \green{ \boxed{ \sf \therefore 9AD^{2} = 7AB^{2}. }}

Hence, it is proved.

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