vectors

Given MA=DM and BN=NC

MN=MA+AB+BN . . . . . (1)

MN=MD+DC+CN . . . . . (2)

(1)+(2):
2 MN=MA+AB+BN+MD+DC+CN
=(MA+MD)+(BN+CN)+AB+DC
=(MA-DM)+(BN-NC)+AB+DC
=AB+DC

∴ MN=(1/2)(AB+DC)


Given AB//DC
Let DC=kAB where k is a constant

MN=(1/2)(AB+DC)
=(1/2)(AB+kAB)
=(1/2)(1+k)AB
=cAB where c=(1/2)(1+k) is a constant

∴ MN//AB