Given #ABC# a triangle where #bar(AD)# is the median and let the segment line #bar(BE)# which meets #bar(AD)# at #F# and #bar(AC)# at #E#. If we assume that #bar(AE)=bar(EF)#, show that #bar(AC)=bar(BF)#?.

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Answer 1 · 2016-12-28T17:53:41

See below.

We will apply the theorem of Menelaus of Alexandria to the sub-triangle #Delta ECB# and the line #[AD]#

According to Menelaus,

#abs(AC)/abs(AE) xx abs(EF)/abs(FB) xx abs(BD)/abs(DC)=1#

but #abs(AE)=abs(EF)# and #abs(BD)=abs(DC)# so we have

#abs(AC)/abs(FB)=1#

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