Question

in an equilateral triangle ABC, D is a point on side BC such that BD=BC/3. prive that 9AD^ = 7AB^

Answer 1

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In a ΔABC and ΔACE
AB = AC ( Given)
AE = AE ( common)
∠AEB = ∠AEC = 90°
∴ ΔABC ≅ ΔACE ( For RHS criterion)
BE = EC (By C.P.C.T)
BE = EC = BC / 2
In a right angled triangle ADE
AD2 = AE2 + DE2 ---------(1)
In a right angled triangle ABE
AB2 = AE2 + BE2 ---------(2)
From equ (1) and (2) we obtain
⇒ AD2  - AB2 =  DE2 - BE2 .
⇒ AD2  - AB2 = (BE – BD)2 - BE2 .
⇒ AD2  - AB2 = (BC / 2 – BC/3)2 – (BC/2)2 
⇒ AD2  - AB2 = ((3BC – 2BC)/6)2 – (BC/2)2 
⇒ AD2  - AB2 = BC2 / 36 – BC2 / 4 ( In a equilateral triangle ΔABC, AB = BC = CA)
⇒ AD2 = AB2 + AB2 / 36 – AB2 / 4
⇒ AD2 = (36AB2 + AB2– 9AB2) / 36
⇒ AD2 = (28AB2) / 36
⇒ AD2 = (7AB2) / 9
9AD2 = 7AB2 .

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.