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# In Fig. 6.58, ABC is a triangle in which Ð ABC > 90° and AD perpendicular CB produced. Prove that AC square equals AB2 plus BC square plus 2 BC . BD.

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 $\triangle$ADB, by Pythagoras theorem

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

In $\triangle$ACD, 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)

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