In an equilateral triangle ABC, D is a point on the side BC such that BD = 1/3 BC. Prove that 9AD² = 7AB².

Given: AB = BC = CA = x (say)

BD = 1/3 BC

To prove: 9 AD² = 7 AB²

Proof: Construct AE perpendicular to BC.

As BC = x,

BD = x/3

In ∆ABE and ∆ACE,

AB = AC [∵ AB = AC = x]

AE = AE [common sides in both triangles]

∠AEB = ∠AEC [∵ ∠AEB = ∠AEC = 90°]

Thus, ∆ABE ≅ ∆ACE by RHS congruency, i.e., Right angle-Hypotenuse-Side congruency.

If ∆ABE ≅ ∆ACE,

BE = CE [∵ corresponding parts of congruent triangles are congruent]

So, BE = CE = 1/2 BC

⇒ BE = CE = x/2

We have BE = x/2, BD = x/3 and clearly

BD + DE = BE

⇒

⇒

In ∆ABE using Pythagoras theorem,

(hypotenuse)² = (perpendicular)² + (base)²

⇒ x² = AE² +

⇒ AE² = x² –

⇒ AE² = …(i)

Similarly, using pythagoras theorem in right ∆AED,

AD² = AE² + ED²

⇒ AD² = + [∵ ED = x/6]

⇒ AD² =

⇒ AD² =

⇒ AD² =

⇒ 9 AD² = 7 x²

⇒ 9 AD² = 7 AB² [∵ AB = x ⇒ AB² = x²]

Hence, proved.