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