Solution:

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

To prove: 9AD^{2} = 7AB^{2}

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

AD^{2} = AE^{2} + DE^{2} ---------(1)

In a right angled triangle ABE

AB^{2} = AE^{2} + BE^{2} ---------(2)

From equ (1) and (2) we obtain

⇒ AD^{2} - AB^{2} = DE^{2} - BE^{2} .

⇒ AD^{2} - AB^{2} = (BE – BD)^{2} - BE^{2} .

⇒ AD^{2} - AB^{2} = (BC / 2 – BC/3)^{2} – (BC/2)^{2}

⇒ AD^{2} - AB^{2} = ((3BC – 2BC)/6)^{2} – (BC/2)^{2 }

⇒ AD^{2} - AB^{2} = BC^{2} / 36 – BC^{2} / 4 ( In a equilateral triangle ΔABC, AB = BC = CA)

⇒ AD^{2} = AB^{2} + AB^{2} / 36 – AB^{2} / 4

⇒ AD^{2} = (36AB^{2} + AB^{2}– 9AB^{2}) / 36

⇒ AD^{2} = (28AB^{2}) / 36

⇒ AD^{2} = (7AB^{2}) / 9

**9AD**^{2} = 7AB^{2}