Final answer:
To prove that segments BE and AC are parallel, we utilize properties of isosceles triangles and the alternate interior angles theorem or vector operations, demonstrating the congruence of angles and parallelism of vectors.
Step-by-step explanation:
The question requires us to prove that two line segments, BE and AC, are parallel given that ∆A ≅ ∆EBC and AB ≅ BC. To approach this proof, we can use the concept of similar triangles, properties of parallel lines, and Euclidean postulates like the alternate interior angles theorem and the transversal postulate.
First, we can consider that ∆ABC is isosceles since AB ≅ BC which implies that angle BAC is congruent to angle BCA. From the given information that angle A is congruent to angle EBC, we can apply the alternate interior angles theorem. If the alternate interior angles formed by a transversal cutting through two lines are congruent, then the lines are parallel. In this case, BE acts as a transversal for line segments AC and a hypothetical line passing through points E and C. Since angles A and EBC are congruent and located in the alternate interior positions relative to this setup, we can conclude that the line segments BE and AC are parallel.
Another approach could be using vector operations. If vectors А and Б represent the line segments AB and BC respectively, and they are congruent, it signifies that these vectors not only have the same magnitude but also the same direction. As segments AB and BC have the same vector representation and since the vector addition of AB and BC would mathematically give us the vector representation for AC, this sets a transitivity property establishing BE and AC as parallel vectors and the corresponding line segments BE and AC as parallel lines.