Answer:
The conclusion supported by the diagram is;
(A) [tex]\dfrac{AB}{BC} = \dfrac{FE}{ED}[/tex]
Step-by-step explanation:
The question is examines the theorem that two transversals cut by three or more parallel lines are proportionally divided by the transversals
The given parallel lines are;
Line CD, line BE, and line line AB
The given transversals are;
Line AC and line DF
Therefore, by the above three or more parallel lines cutting two transversal theorem, we have;
BC/AB = ED/FE, BC/DF = AB/FE, CA/AB = DF/FE
From BC/AB = ED/FE, we find the inverse of both sides to get;
[tex]\dfrac{1}{\dfrac{BC}{AB} } = \dfrac{1}{\dfrac{ED}{FE} }[/tex]
[tex]\dfrac{1}{\dfrac{BC}{AB} } = \dfrac{AB}{BC} \ and \ \dfrac{1}{\dfrac{ED}{FE} } = \dfrac{FE}{ED}[/tex]
[tex]\therefore \ \dfrac{AB}{BC} = \dfrac{FE}{ED}[/tex]
The correct option is [tex]\dfrac{AB}{BC} = \dfrac{FE}{ED}[/tex]
