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 Q.E.D.