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

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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
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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

CPhill  Aug 15, 2017

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