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  • In Fig. 6.60, AD is a median of a triangle ABC and AM perpendicular BC. Prove that : (question 3 )

Q5 In Fig. 6.60, AD is a median of a triangle ABC and  AM \perp BC. Prove that :    AC ^2 + AB ^2 = 2 AD^2 + \frac{1}{2} BC ^2           

Answers (1)

best_answer

In \triangleABM, by Pythagoras theorem

AB^2=AM^2+MB^2.......................1

In \triangleAMC, by Pythagoras theorem

AC^2=AM^2+MC^2..................................2

Adding equation 1 and 2,

AB^2+AC^2=2AM^2+MB^2+MC^2

\Rightarrow AB^2+AC^2=2AM^2+(BD-DM)^2+(MD+DC)^2

\Rightarrow AB^2+AC^2=2AM^2+(BD)^2+(DM)^2-2.BD.DM+(MD)^2+(DC)^2+2.MD.DC

\Rightarrow AB^2+AC^2=2AM^2+2.(DM)^2+BD^2+(DC)^2+2.MD.(DC-BD)\Rightarrow AB^2+AC^2=2(AM^2+(DM)^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2+2.MD.(\frac{BC}{2}-\frac{BC}{2})\Rightarrow AB^2+AC^2=2(AM^2+(DM)^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2

AC ^2 + AB ^2 = 2 AD^2 + \frac{1}{2} BC ^2

 

 

 

 

Posted by

seema garhwal

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