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

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