Question

ΔABC is right-angled at B and D is the mid-point of BC.
Prove that: $A{C}^2=(4A{D}^2-3A{B}^2).$

Answer 1

Given: In Δ ABC, $\angle$B = 90° and D is the mid-point of BC.
To prove: AC² = (4AD² - 3AB²)
Proof: In Δ ABC , $\angle$B = 90°

∴ AC² = AB² + BC² ( By Pythagoras' theorem)
= AB² + (2BD)² [ Since BC = 2BD]
= AB² + 4BD²
= AB² + 4(AD² - AB²) [ Since AB² + BD² = AD²]
= 4AD² - 3AB²
Hence, AC² = 4AD² - 3AB²