Question

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC² = AD² + BC x DE + `(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 ΔADC, ∠ADC is an obtuse angle.
∴ AC² = AD² + DC² + 2 x DC x DE

⇒ AC² = AD² + `(1/2"BC")^2 + 2 xx (1)/(2)"BC" xx "DE"`

⇒ AC² = AD² + `(1)/(4)"BC"^2 + "BC" xx "DE"`

⇒ AC² = AD² + BC x DE + `(1)/(4)"BC"^2` .       ....(i)