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)