Consider two triangles AEB and CEF. In these triangles:
- ∠AEB≅∠CEF (as vertical angles);
- ∠EAB≅∠ECF (lines AB and CD are parallel and AC is transversal, then these angles are congruent as alternate interior angles);
- ∠ABE≅∠CFE (lines AB and CD are parallel, BF is transversal, then these angles congruent as alternate interior angles).
Thus, by AAA theorem, [tex]\triangle AEB\sim \triangle CEF.[/tex]
Corresponding sides of similar triangles are proportional, then
[tex]\dfrac{AB}{CF}=\dfrac{AE}{EC},\\ \\\dfrac{AB}{CF}=\dfrac{2}{1},\\ \\AB=2CF.[/tex]
ABCD is a parallelogram, then AB=CD.
Therefore,
CD=2CF and CD=DF+CF. Equate these two expressions:
DF+CF=2CF,
DF=CF.
This gives you that DF:CF=1:1.
Answer: 1:1
