Question

In triangle ABC, ∠B = 90^o and D is the mid-point of BC.

Prove that: AC² = AD² + 3CD².

Answer 1

In ΔABC

AB² + BC² = AC²

as [BC = 2CD]

AB² + 4CD² = AC²           ....(1)

In ΔABD,

AD² = AB² + BD² = AB² + CD²                    ....(2)

Subtracting (1) and (2)

AC² - AD² = AB² + 4CD² - (AB² + CD²)

AC² - AD² = 3CD²

AC² = AD² + 3CD²

Hence proved.