In triangle ABC, ∠B = 90° and D is the mid-point of side BC. Prove that : AC² = AD² + 3CD².

Given: ABC is a triangle such that ∠ABC = 90°. D is the mid-point of BC.

To prove: AC² = AD² + 3CD²

Proof: In Δ ABD, using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AD² = AB² + BD² ………….(1)

Similarly, in Δ ABC,

⇒ AC² = AB² + BC²

⇒ AC² = AB² + (BD + DC)²

⇒ AC² = AB² + BD² + DC² + 2 x BD x DC

As D is the midpoint of BC, BD = DC.

⇒ AC² = AB² + BD² + CD² + 2 x CD x CD

⇒ AC² = AB² + BD² + CD² + 2CD²

⇒ AC² = AB² + BD² + 3CD²

Using equation (1), we get

⇒ AC² = AD² + 3CD²

Hence, AC² = AD² + 3CD².