Question

BD=BC/3 angleE=90' ABC IS TRIANGLE PROVE THAT 9AD^=7AB^

Answer 1

tr aeb congruent to tr aec by rhs rule
be=ec=1/2 bc
de=be-bd
=1/2bc-1/3 bc
=1/6bc
in tr aed.= ae2+de2=ad2

in tr aeb=ae2 +be2=ab2
=de2-be2 =ad2 -ab2
1/36bc2 -1/4bc2 =ad2 -ab2
bc2-9bc2 /36=ad2-ab2
-8bc2/36=ad2 -ab2
-2/9 bc2=ad2 -ab2
-2 bc2=9ad2-9ab2
-2bc2 +9ab2=9ad2
7ab2=9ad2
hence proved
hope helps you
this is right answer☝

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 ] . ....

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

Hence, it is proved.