+167 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.
+128407
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
