Question

IN AN EQUILATERAL TRIANGLE ABC, D IS A POINT ON THE SIDE BC SUCH THAT BD=1÷3 PROVE THAT 9AD^2=7AB^2

Answer 1

Construction:  Draw AE ⊥ BC.
Proof :

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.

Hence, proved.

Hope this helps you !!
# Dhruvsh

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