In Fig. 6.58, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that
Using Pythagoras theorem in ΔADB, we get:
AB2 = AD2 + DB2 (i)
Applying Pythagoras theorem in ΔACD, we obtain
AC2 = AD2 + DC2
AC2 = AD2 + (DB + BC)2
AC2 = AD2 + DB2 + BC2 + 2DB x BC
AC2 = AB2 + BC2+ 2DB x BC [Using equation (i)]