Question

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC² - AB² = 2BC x ED

Answer 1

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

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

⇒ AC² - AB² = 2BC x DE.