Question

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB² + AC² = 2AD² + `(1)/(2)"BC"^2`

Answer 1

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

Adding (i) and (ii), we have
AC² + AB² = AD² + BC x DE + `(1)/(4)"BC"^2  + "AD"^2 - "BC" xx "DE" + (1)/(4)"BC"^2`

⇒ AB² + AC² = `2"AD"^2 + (1)/(2)"BC"^2`.    ....(iii)