Question

ABC is an equilateral triangle. Side BC is trisected at D. Prove that 9AD² = 7AB².

[Hint: Draw AP ⊥ BC. Let BD = 2x.
∴ DC = 4x, BC = 6x = AB and DP = x.]

Answer 1

Given:

△ABC is equilateral, and side BC is trisected at D.

AP ⊥ BC

To prove: 9AD² = 7AB²

Proof:

Let BD = 2x

Then, DC = 4x and BC = 6x = AB = AC

In right △ABP:

AB² = AP² + BP²

(6x)² = AP² + (3x)²

36x² = AP² + 9x²

AP² = 27x²

In right △ADP:

AD² = AP² + PD²

= 27x² + x²

= 28x²

Now,

AB² = 36x²

⇒ 9AD² = 9(28x²)

= 252x²

⇒ 7AB² = 7(36x²)

= 252x²

Hence,

9AD² = 7AB²