Question 448086
<pre><font face = "consolas" color = "indigo" size = 3><b>
<s>The other tutor did not prove the theorem because
AM is not half of BC, as he stated.  Here is the correct proof:</s>

The other tutor came back, edited it, and gave an alternate correct proof. 

{{{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)
 )}}}

Extend AM to twice its length, to D, so that AM = DM

{{{drawing(400,1000/3,-6,6,-5,5, triangle(1,4,-5,0,5,0), 
green(line(1,4,-1,-4)), locate(-5,0,B), locate(5,0,C), locate(1,4.5,A),
locate(0,0,M), locate(-1,-4,D), arc(0,0,2,-2,75,180),arc(0,0,2,-2,255,360)

 )}}}

Draw DC


{{{drawing(400,1000/3,-6,6,-5,5, triangle(1,4,-5,0,5,0), 
green(line(1,4,-1,-4)), locate(-5,0,B), locate(5,0,C), locate(1,4.5,A),
locate(0,0,M), locate(-1,-4,D), arc(0,0,2,-2,75,180),arc(0,0,2,-2,255,360),
red(line(-1,-4,5,0))

 )}}}

       AM = DM       by construction

     &#8736;AMB = &#8736;DMC    because they are are vertical angles.

       BM = CM       because a median bisects the side it's drawn to

     &#11616;ABM &#8773; &#11616;DCM    by Side-angle-side

        AB = DC      corresponding parts of congruent triangles.

   AC + CD > AD      by triangle inequality on &#11616;ACD

   AC + AB > AD      substituting equals for equals.

        AD = 2AM     by construction

   AC + AB > 2AM     substituting equals for equals.

½(AC + AB) > AM      multiplying both sides by ½ .

Edwin</pre>