Ex 6.6, 4 (Optional)

Transcript

Ex 6.6, 4 In Fig. 6.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2 BC . BD. Given: ABC is a triangle where ∠ABC < 90° and AD ⊥ BC To Prove: AC2 = AB2 + BC2 − 2BC.BD Proof: In right ∆ APB By Pythagoras Theorem, AB2 = AD2 + BD2 In right ∆ ADC By Pythagoras Theorem, AC2 = AD2 + CD2 From (2) AC2 = AD2 + CD2 AC2 = AD2 + (BC − BD)2 AC2 = AD2 + BD2 + BC2 − 2BC.BD Putting AD2 + BD2 = AB2 from equation (1) AC2 = AB2 + BC2 − 2BC . BD Hence Proved.