Question

In an equilateral triangle ABC,D is the point on the side BC such that BD =1/3BC prove that 9ADsquare =7ABsquare

Answer 1

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

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Answer 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 ] .

⇒ 9 AD² = 7 AB²

Hence proved