Question

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB² + AC² = 2(AD² + CD²)

Answer 1

We have ∠AED = 90°
∴ ∠ADE < 90° and ∠ADC > 90°
i.e. ∠ADE is acute and ∠ADC is obtuse.

From (iii), we have

AB² + AC² = `2"AD"^2 + (1)/(2)"BC"^2`

⇒ AB² + AC² = `2"AD"^2 + (1)/(2)(2 xx "CD")^2`

⇒ AB² + AC² = `2"AD"^2 + (1)/(2) xx 4"CD"^2`

⇒ AB² + AC² = 2AD² + 2CD²
⇒ AB² + AC² = 2(AD² + CD²).