Question 3:
In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC.BD.
Answer:
Applying Pythagoras theorem in ΔADB, we obtain
AB2 = AD2 + DB2 … (1)
Applying Pythagoras theorem in ΔACD, we obtain
AC2 = AD2 + DC2
AC2 = AD2 + (DB + BC)2
AC2 = AD2 + DB2 + BC2 + 2DB × BC
AC2 = AB2 + BC2 + 2DB × BC [Using equation (1)]
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