Question

Point E belongs to diagonal AC of parallelogram ABCD (labeled counterclockwise) so that AE:EC=2:1. Line BE intersects CD at point F. Find the ratio of DF:FC.

Answer 1

Consider two triangles AEB and CEF. In these triangles:

1. ∠AEB≅∠CEF (as vertical angles);
2. ∠EAB≅∠ECF (lines AB and CD are parallel and AC is transversal, then these angles are congruent as alternate interior angles);
3. ∠ABE≅∠CFE (lines AB and CD are parallel, BF is transversal, then these angles congruent as alternate interior angles).

Thus, by AAA theorem, $\triangle AEB\sim \triangle CEF.$

Corresponding sides of similar triangles are proportional, then

$\dfrac{AB}{CF}=\dfrac{AE}{EC},\\ \\\dfrac{AB}{CF}=\dfrac{2}{1},\\ \\AB=2CF.$

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