In Fig. 6.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that

Applying Pythagoras theorem in ΔADB, we obtain

AD² + DB² = AB²

AD² = AB²– DB²     (i)

Applying Pythagoras theorem in ΔADC, we obtain

AD² + DC² = AC²

AB²– BD² + DC² = AC² [Using equation (i)]

AB²– BD² + (BC - BD)² = AC²

AC² = AB²– BD² + BC² + BD² -2BC x BD

= AB² + BC² - 2BC x BD