Question

triangle ABC is right angled at B and D is the mid point of BC. prove that AC^2 = 4AD^2 - 3AB^2

Answer 1

In △ABC, ∠B=90° By the Pythagorean theorem, AC²=AB²+BC² …Eq.1,

In △ABD,  in the same manner, AD²=AB²+BD² …Eq.2   

Here, BD=½BC   Thus Eq.2 is rewritten as follows: AD²=AB²+(½BC)²

⇔ 4AD²=4AB²+BC² ⇔ BC²=4AD²－4AB² …Eq.3    Plug Eq.3 into Eq.1

⇒ AC²=AB²+(4AD²－4AB²) ⇔AC²=4AD²－3AB²