The triangle can be of ANY FORM that satisfies the given constraints.
Redraw ABC as a 6-8-10 triangle such that AC=6, BC=8 and AB=10:
∆ABC:
Since D is the midpoint of BC, BD=DC=4.
∆ACD:
When a line is drawn from the midpoint of the hypotenuse to a leg, the leg is BISECTED.
Thus, EH bisects CD, and EG bisects AC.
Result:
DH=2, implying that BH=6.
CG=3, implying that EH=3.
∆BEH and ∆BFC:
Since EH is parallel to CF, ∆BEH and ∆BFC have the same combination of angles and thus are SIMILAR.
Corresponding sides of similar triangles are in the SAME RATIO.
Side BH in ∆BEH corresponds to side BC in ∆BFC.
Side EH in ∆BEH corresponds to side FC in ∆BFC.
Thus:
BH/BC = EH/FC
6/8 = 3/FC
6(FC) = 24
FC= 4.
Side AC:
Since AC=6 and FC=4, AF=2.
Thus:
AF:FC = 2:4 = 1:2.
The correct answer is
B.
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