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0
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In triangle ABC, the medians AD, BE, and CF concur at the centroid G.  

 (a) Prove that AD < AB + AC/2.  (b) Let P=AB+AC+BC be the perimeter of $\triangle ABC.$ Prove that 3P/4 < AD + BE + CF < P.

AdminMod2  Aug 15, 2017
 #1
avatar+87635 
+1

 

 

Since we can orient the triangle any way we want ...let  A  =(0,0) , B  = (0, b)  and C  = (c, 0)

 

 

Let  D  =   ( c/2 , b/2)    ...so  AD  =  sqrt [ (c/2 - 0)^2   +  (b /2 - 0) ^2  ]  =  sqrt [ ( b^2 + c^2 ) / 4 ]   = 

sqrt [ b^2 + c^2 ] / 2

 

AB  = b

 

And 

 

AC  = c

 

So

 

AD  <  [ AB  + AC] / 2

 

sqrt [ b^2 + c^2  ] / 2  <  [ b + c ] / 2

 

sqrt [ b^2 + c^2 ] < b + c        square both sides

 

b^2 + c^2  <  b^2  + 2bc + c^2

 

0 <  2bc       which is true  since  b, c   are both > 0 

 

 

 

cool cool cool

CPhill  Aug 15, 2017

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