Let AD be the midline on the BC side of △ABC, where cosB=(a? +c？ -c？ ) /2ac, in △ABC, BD=a/2, which is derived from cosine theorem, AD? =AB？ +BD？ -2AB BDcosB=c？ +a？ /4-ac(a？ +c？ -B? )/2ac=(b？ +c？ )/2-a？ /4,∴AD=m(a)={√[2(b？ +c？ )-a？ ]}/2. It can also be proved that m (b) = {√ [2 (a? +c？ )-b？ ]}/2，m(c)={√[2(a？ +b？ )-c？ ]}/2,

Note: (a)(b)(b) M is a subscript. Draw this picture yourself. I can't get it in.