In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² − 2BC.BD.

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Solution

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 × BD

= AB² + BC² − 2BC × BD

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