Consider the quadrilateral #ABCD#, and let #E, N, F, M# be the midpoints of the edges #AB, BC, CD, DA# respectively. How do you prove that #vec(EF)=1/2(vec(AD)+vec(BC))# and #vec(AC)=vec(MN)+vec(EF)#?

Question

Consider the quadrilateral #ABCD#, and let #E, N, F, M# be the midpoints of the edges #AB, BC, CD, DA# respectively. Prove that
#vec(EF)=1/2(vec(AD)+vec(BC))# and #vec(AC)=vec(MN)+vec(EF)#
(From Izu Vaisman's book)

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Answer 1 · 2018-05-28T12:44:59

Please see the proof below.

Apply Chasles' relation

#vec(EF)=vec(EB)+vec(BC)+vec(CF)#..............#(1)#

#vec(EF)=vec(EA)+vec(AD)+vec(DF)#...............#(2)#

Therefore,

Adding #(1)# and #(2)#

#2vec(EF)=vec(EB)+vec(EA)+vec(BC)+vec(AD)+vec(CF)+vec(DF)#

But,

#vec(EB)+vec(EA)=0#

#vec(CF)+vec(DF)=0#

#2vec(EF)=vec(BC)+vec(AD)#

#vec(EF)=1/2(vec(BC)+vec(AD))#

Similarly, using the mid point theorem,

#vec(AC)=2vec(EN)=2(vec(EF)+vec(FN))#

#vec(AC)=2vec(MF)=2(vec(MN)+vec(NF))#

#2vec(AC)=2vec(EF)+2vec(FN)+2vec(MN)+2vec(NF)#

But #vec(FN)+vec(NF)=0#

#vec(AC)=vec(EF)+vec(MN)#