Given: A ΔABC right-angled at B, and D is the mid-point of BC, i.e. BD = CD

To Prove: AC² = (4AD² - 3AB²)

Proof:

In ΔABD,

By Pythagoras theorem, [i.e. Hypotenuse² = Base²+ Perpendicular²]

AD² = AB² + BD²

[ as D is mid-point of BC, therefore,

⇒ 4AD² = 4AB² + BC²

⇒ BC² = 4AD² - 4AB² [1]

Now, In ΔABC, again By Pythagoras theorem

AC² = AB² + BC²

AC² = AB² + 4AD² - 4AB² [From 1]

AC² = 4AD² - 3AB²

Hence Proved !