In Fig. 6.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that
Applying Pythagoras theorem in ΔADB, we obtain
AD2 + DB2 = AB2
AD2 = AB2– DB2 (i)
Applying Pythagoras theorem in ΔADC, we obtain
AD2 + DC2 = AC2
AB2– BD2 + DC2 = AC2 [Using equation (i)]
AB2– BD2 + (BC - BD)2 = AC2
AC2 = AB2– BD2 + BC2 + BD2 -2BC x BD
= AB2 + BC2 - 2BC x BD