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

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In $\triangle$ADB, by Pythagoras theorem

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

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

In $\triangle$ACD, 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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