a. Xet tgAMB va tgAMC co:
AM chung
AB=AC (gt)
MB=MC(gt)
=> tgAMB = tgAMC (c.c.c)
b. ta co gAMB + gAMC = 180 ( ke bu)
Ma gAMB = gAMC ( do cau a)
=> gAMB = gAMC =180/2 = 90 do
=> tgAMB vuong tai M
=> BM<AB=AC
Do do BM<AC (dpcm)
c. Xet tg ABC co 
MD//AC
M la trung diem BC
=> D la trung diem AB
=> MD la duong trung binh cua tgABC
=> MD=1/2 AC = 1/2 AB
<=> AB=2MD (dpcm)