Question

the figure below shows segments ac and ef which intersect at point b. segment af is parallel to segment ec: which of these facts is used to prove that triangle abf is similar to triangle cbe? angle fab is equal to angle ceb because corresponding angles are congruent. angle abf is congruent to angle ceb because vertically opposite angles are congruent. angle afb is congruent to angle ceb because alternate interior angles are congruent. angle afb is congruent to angle ceb because supplementary angles are congruent. i chose b, but i'm not positive it is the answer. can anyone give some clarification?

Answer 1

Answer:  Angle AFB is congruent to angle CEB because alternate interior angles are congruent.

Step-by-step explanation:

Given: $fa\parallel ec$,

And, ac and ef are intersecting each other at point b.

Prove: Triangle abf is similar to triangle cbe

Since, $\angle abf \cong \angle cbe$   (Reflexive)

$fa\parallel ec$,

⇒ ef is the common transversal of parallel lines fa and ec.

$\angle afb \cong \angle ceb$         (Because Alternative interior angles are congruent)

Thus, By AA similarity postulate,

$\triangle abf\sim \triangle cbe$

Answer 2

angle A = angle C angle F = angle E because of some line through parallel lines postulate