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  • In Fig. 6.59, ABC is a triangle in which angle ABC < 90° and AD perpendicular BC. Prove that AC square equals AB square plus BC square minus 2 BC . BD.

Q4   In Fig. 6.59, ABC is a triangle in which \angle ABC < 90° and AD \perp BC. Prove that
      AC^2 = AB^2 + BC^2 - 2 BC . BD.

        

Answers (1)

best_answer

In \triangleADB, by Pythagoras theorem

AB^2=AD^2+DB^2

AD^2=AB^2-DB^2...........................1

In \triangleACD, by Pythagoras theorem

AC^2=AD^2+DC^2

AC^2=AB^2-BD^2+DC^2         (From 1)

\Rightarrow AC^2=AB^2-BD^2+(BC-BD)^2

\Rightarrow AC^2=AB^2-BD^2+(BC)^2+(BD)^2-2.BD.BC

AC^2 = AB^2 + BC^2 - 2 BC . BD.         

 

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seema garhwal

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