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# In Fig. 6.60, AD is a median of a triangle ABC and AM perpendicular BC. Prove that : (question 1 )

Q5 (1)   In Fig. 6.60, AD is a median of a triangle ABC and  AM $\perp$ BC. Prove that :

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

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Given: AD is a median of a triangle ABC and  AM $\perp$ BC.

In $\triangle$AMD, by Pythagoras theorem

$AD^2=AM^2+MD^2.......................1$

In $\triangle$AMC, by Pythagoras theorem

$AC^2=AM^2+MC^2$

$AC^2=AM^2+(MD+DC)^2$

$\Rightarrow AC^2=AM^2+(MD)^2+(DC)^2+2.MD.DC$

$AC^2 = AD^2 + DC^2 + 2 DC . MD.$            (From 1)

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

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

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