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