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Slide12.JPG

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Ex 6.6, 3 In Fig. 6.58, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2 BC . BD. Given : ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB extended D. To Prove: AC2 = AB2 + BC2 + 2BC.BD Proof: In right ∆ ADB Applying Pythagoras theorem, AB2 = AD2 + BD2 In right ∆ ADC Applying Pythagoras theorem, AC2 = AD2 + CD2 From (2) AC2 = AD2 + CD2 AC2 = AD2 + (BD + BC)2 AC2 = AD2 + BD2 + BC2 + 2BD.BC Put AD2 + BD2 = AB2 from equation (1) AC2 = AB2 + BC2 + 2BC.BD. Hence Proved.

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