Question

in an equilateral ∆ABC,Dis a point on side BC such that BD=⅓BC.prove that9(AD)²=7(AB)² ..,......plzz answer fast ....​

Answer 1

Step-by-step explanation:

Heya! mates here's Ur ans..widout..spam..

plz..guys follow..thnx.. Mark as brainlist..

hope it's helpful..

Answer 2

Answer:

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..

HOPE THIS HELPS YOU

PLZZ MARK AS BRAINLIEST