Question

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB² = AD² - BC x CE + `(1)/(4)"BC"^2`

Answer 1

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

In ΔABD, ∠ADE is an acute angle.
∴ AB² = AD² + BD² - 2 x BD x DE

⇒ AB² = AD² + `(1/2"BC")^2 - 2 xx (1)/(2)"BC" xx "DE"`

⇒ AB² = AD² + `(1)/(4)"BC"^2 - "BC" xx "DE"`

⇒ AB² = AD² + BC x DE - `(1)/(4)"BC"^2`.   ....(ii)