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Q3   In Fig. 6.58, ABC is a triangle in which \angle  ABC > 90° and AD \perp CB produced. Prove that
       AC^2 = AB^2 + BC^2 + 2 BC . BD.

        

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In \triangleADB, by Pythagoras theorem

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

In \triangleACD, by Pythagoras theorem

AC^2=AD^2+DC^2.......................2

AC^2=AD^2+(BD+BC)^2

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

AC^2 = AB^2 + BC^2 + 2 BC . BD.            (From 1)

 

 

 

 

 

 

 

 

Posted by

seema garhwal

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