One of the properties of a parallelogram is that its opposite sides are parallel and congruent.
Segment AB and CD are opposite sides of the parallelogram and is therefore, congruent.
Therefore, the reason for CD≅ AB is: "Opposite sides of a parallelogram/rhombus/rectangle/square are congruent."
For the next statement, since CD≅AB and AB≅CE, then by Transitive Property, CD≅CE.
Since CD and CE are sides of a triangle and are congruent as stated in Statement 3, then ∠E ≅ ∠CDE because in a triangle, angles opposite of congruent sides are congruent.
In addition, we can say that ∠A ≅ ∠CDE because parallel lines (AB and CD) cut by a transversal (AE) form congruent corresponding angles.
Lastly, since ∠A ≅ ∠CDE and ∠CDE ≅ ∠E, we can say that ∠A ≅ ∠E by Transitive Property.