Can you prove it?

Question

ABC is a triangle with AB = AC. D is any point on the side BC. E and F are point on the side AB and AC, respectively; such that DE is parallel to AC, and DE is parallel to AB prove that

DF + FA + AE + ED = AB + AC

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Answer 1 · 2017-04-06T10:04:09

see explanation

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I assume that #DF# is parallel to #AB#

#AB=AE+EB#
#AC=AF+FC#
#=> AB+AC=AE+EB+AF+FC#

Note that #DeltaABC, DeltaEBD and DeltaFDC# are similar triangles,
and given #AB=AC#,
#=> (AB)/(AC)=(EB)/(ED)=(FD)/(FC)=1#
#=> EB=ED, and FD=FC#

#=> AB+AC=AE+ED+AF+FD#
#=> AB+AC=DF+FA+AE+ED# (proved)