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# In Fig. 6.60, AD is a median of a triangle ABC and AM perpendicular BC. Prove that : (question 2 )

Q5 (2)   In Fig. 6.60, AD is a median of a triangle ABC and  AM $\perp$ BC. Prove that :

$AB ^2 = AD ^2 - BC .DM + \left ( \frac{BC}{2} \right ) ^2$

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In $\triangle$ABM, by Pythagoras theorem

$AB^2=AM^2+MB^2$

$AB^2=(AD^2-DM^2)+MB^2$

$\Rightarrow AB^2=(AD^2-DM^2)+(BD-MD)^2$

$\Rightarrow AB^2=AD^2-DM^2+(BD)^2+(MD)^2-2.BD.MD$

$\Rightarrow AB^2=AD^2+(BD)^2-2.BD.MD$

$\Rightarrow AB^2 = AD^2 + (\frac{BC}{2})^2 -2(\frac{BC}{2}). MD.=AC^2$           (BC=2 BD)

$\Rightarrow AD^2 + (\frac{BC}{2})^2 -BC. MD.=AC^2$

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