Question

In an equilateral triangle ABC ,D is the mid point on side BC such that BD =1/3 BC .Prove that 9(AD)^2 = 7 (AB)^2

Answer 1

Hey...............

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 .

hope it helps:)

Answer 2

Here is your solution

Given that :-

In  Δ ABC is an equilateral triangle.

D is point on BC such that BD =BC.

To prove:-

 9 AD² = 7 AB²

Construction: Draw AE ⊥ BC.

Proof :-


In a ΔABC and ΔACE

AB = AC ( given)

AE = AE (common)

∠AEB = ∠AEC = (Right angle)

Hence ΔABC ≅ ΔACE (By RHS Creation)

Again,

BE = EC (By C.P.C.T)

BE = EC = BC²

In a right angled ΔADE

AD²= AE2 + DE² ---(1)

In a right angled ΔABE

AB² = AE² + BE² ---(2)

From equation (1) and (2) ;

 =) 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 )

( we know that In a equilateral triangle, All sides are equal to each other)

AB = BC = AC

 = ) AD²= AB² + AB²/ 36 – AB² / 4

 = )AD² = (36AB² + AB²– 9AB²) / 36

 = ) AD² = (28AB²) / 36

=) AD² = (7AB²) / 9

= ) 9AD² = 7AB² ‎‎‎‎‎‎‎‎‎‎‎‎‎ ‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎proved

Hope it helps you