in an equilateral triangle ABC, D …

Sol:

Given:   In an equilateral triangle ΔABC. The side BC is trisected at D such that BD = (1/3) BC.

To prove:  9AD²  = 7AB² 

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 .