Short Answer 
A parallelogram has congruent opposite sides and angles, and its diagonals intersect to form congruent vertical angles. By applying the Angle-Angle-Side (AAS) criterion, we demonstrate that triangles AEB and CED are congruent, leading to the equalities of segments AE = CE and BE = DE.
Step 1: Understand the Properties of a Parallelogram
A parallelogram has specific properties that make it unique, particularly that its opposite sides are congruent and that opposite angles are equal. In parallelogram ABCD, the sides AB and CD are parallel, as are the sides BC and AD. This relationship implies that the angles formed by these parallel lines are congruent: ‚a†A = ‚a†C and ‚a†B = ‚a†D, which is key to our proof.
Step 2: Analyze the Intersecting Diagonals
The diagonals of the parallelogram, AC and BD, intersect at point E, forming vertical angles. Vertical angles are always congruent, meaning ‚a†AEB = ‚a†CED and ‚a†BEA = ‚a†CDE. This congruence of angles will help us demonstrate that the triangles AEB and CED hold certain properties that will lead to the equality of specific segments.
Step 3: Apply the Angle-Angle-Side (AAS) Criterion
Using the congruence from step 1 and step 2, we now apply the Angle-Angle-Side (AAS) criterion, which states that if two angles and a non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent. Thus, triangles AEB and CED are congruent, leading us to conclude that AE = CE and BE = DE, confirming the segment equalities we set out to prove.