in an equilateral triagleABC , D is a point on BC such that BD =1/3BC .prove that 9AD²=7AB²
Answers
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
And, AE = a√3/2
Given that, BD = 1/3BC
∴ BD = a/3
DE = BE - BD = a/2 - a/3 = a/6
Applying Pythagoras theorem in ΔADE, we get
AD2 = AE2 + DE2
⇒ 9 AD2 = 7 AB2
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 ] .
Hence, it is proved.