Final answer:
By utilizing the properties of angle bisectors, parallel lines, and isosceles triangles, it can be demonstrated that OE is indeed parallel to AC in △ABC.
Step-by-step explanation:
To prove that OE is parallel to AC, we rely on the given conditions that AD bisects ∣BAC, BO is parallel to AD, and BE equals EC. Those conditions imply certain relationships and properties within △ABC that can be used to establish the desired parallelism.
Since AD is an angle bisector, we have two triangles, △ABD and △ACD, where ∣BAD = ∣CAD. By the Alternate Interior Angles Theorem, since BO is parallel to AD, the angles ∣OBD and ∣BAC are congruent, which further implies that ∣OBC is congruent to ∣ABC. Now, if BE equals EC, then △BEC is isosceles with ∣BEC congruent to ∣BCE. Since OE bisects ∣BEC, it must also bisect ∣OBC if it is to be parallel to AC; this is due to the Converse of the Alternate Interior Angles Theorem. Given that OB is parallel to AD and therefore, OE and AC are on the same directional path separated by transversal BO, OE is necessarily parallel to AC based on these bisection properties.