Question 448086
Whoops, mixed up AM with BM. It occasionally happens when I try to solve a question without pen or paper. Anyway, I have come up with a new proof, which is similar to that of the other tutor's, but uses a different method of constructing it. Note that neither proof is necessarily more "correct" than the other.


Start with the original diagram (I have used the other tutor's as a reference):

{{{drawing(400,1000/3,-6,6,-5,5, triangle(1,4,-5,0,5,0), 
green(line(1,4,0,0)), locate(-5,0,B), locate(5,0,C), locate(1,4.5,A),
locate(0,0,M)
 )}}}

We can extend CA past A, and draw a line through B parallel to AM, and call their intersection E:

{{{drawing(400,1000/3,-6,6,-5,5, triangle(1,0,-5,-4,5,-4), 
green(line(1,0,0,-4)), locate(-5,-4,B), locate(5,-4,C), locate(1,0.5,A),
locate(0,-4,M),


line(1, 0, -3, 4), line(-5, -4, -3, 4), locate(-3, 4.5, E)
 )}}}


By AAA similarity, we can show that triangles EBC and AMC are similar, with a ratio of 2:1 (since BC = CM). Hence, we can establish the following ratios:


EB = 2AM
AC = AE

From the triangle inequality, EB < AB + AE. Since EB = 2AM and AC = AE, we obtain 2AM < AB + AC --> AM < (AB + AC)/2.