ABC is an equilateral triangle. P is a point on BC such that BP : PC = 2 : 1. Prove that : 9AP² = 7AB²

Given: In an equilateral triangle ABC, P is a point on side BC such that the ratio BP:PC = 2:1.

To Prove: 9AP² = 7AB²

Construction: Draw equilateral triangle ABC. Join AP and draw AM perpendicular to BC.

Proof: Let each side of equilateral triangle ABC be 3x.

AB = BC = CA = 3x

BP = 2x and PC = x

BC = BP + PC = 2x + x = 3x

Since AM ⊥ BC and the perpendicular from a vertex in an equilateral triangle bisects opposite side, so, M is the mid-point of BC.

BM = MC = $\dfrac{1}{2} \times$ BC = $\dfrac{3x}{2}$

MC = MP + PC

⇒ $\dfrac{3x}{2}$ = MP + x

⇒ MP = $\dfrac{3x}{2}$ - x

⇒ MP = $\dfrac{3x - 2x}{2}$

⇒ MP = $\dfrac{x}{2}$

In Δ AMP,

AP² = PM² + AM²

⇒ AM² = AP² - PM² …………….(1)

From Δ ABM,

AB² = BM² + AM²

= $\Big(\dfrac{3x}{2}\Big)^2$ + AM²

Using equation (1), we get

⇒ AB² = $\dfrac{9x^2}{4}$ + (AP² - MP²)

= $\dfrac{9x^2}{4}$ + AP² - $\Big(\dfrac{x}{2}\Big)^2$

= $\dfrac{9x^2}{4}$ + AP² - $\dfrac{x^2}{4}$

= $\dfrac{9x^2 - x^2}{4}$ + AP²

= $\dfrac{8x^2}{4}$ + AP²

= 2x² + AP²

⇒ AB² = 2x² + AP²

⇒ AB² - 2x² = AP²

Multiplying both side with 9, we get

⇒ 9AB² - 9(2x²) = 9AP²

⇒ 9AB² - 2(9x²) = 9AP²

⇒ 9AB² - 2(3x)² = 9AP²

⇒ 9AB² - 2(AB)² = 9AP²

⇒ 9AB² - 2AB² = 9AP²

⇒ 7AB² = 9AP²

Hence, 9AP² = 7AB².