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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           

Answers (1)
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Given: AD is a median of a triangle ABC and  AM \perp BC.

In \triangleAMD, by Pythagoras theorem

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

In \triangleAMC, 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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