Question 1021728
If M is the midpoint of AC then it would have coordinates (b/2, c/2) by a direct application of the midpoint formula of analytic geometry.  In the same way, if N is the midpoint of BC then it would have the coordinates ({{{(a+b)/2}}}, c/2).

==> d(MN) = {{{sqrt(( (a+b)/2 - b/2)^2+(c/2-c/2)^2) = sqrt(a^2/4+0) = abs(a)/2}}}

Also, 
d(AB) = {{{sqrt( (a-0)^2+ (0-0)^2) = abs(a)}}}

==> {{{d(MN) = d(AB)/2}}}