Question

In ∆ ABC, AD ⊥ BC.
Prove that  AC² = AB² +BC² − 2BC x BD

In ∆ ABC (Figure 3), AD ⊥ BC.
Prove that  AC² = AB² +BC² − 2BC x BD

Answer 1

Applying Pythagoras theorem in ΔADB, we obtain

AD² + DB² = AB²

⇒ AD² = AB² − DB²                                               .....(1)

Applying Pythagoras theorem in ΔADC, we obtain

AD² + DC² = AC²

AB² − BD² + DC² = AC²    ...[Using equation (1)]

AB² − BD² + (BC − BD)² = AC²

AC² = AB² − BD² + BC² + BD² −2BC x BD

AC² = AB² + BC² − 2BC x BD