Question

prove: $\angle a \cong \angle e$.
step | statement | reason
1 | $abcd$ is a parallelogram; $\overline{ab} \cong \overline{ce}$ | given
2 | $\overline{cd} \cong \overline{ab}$ | opposite sides of a parallelogram / rhombus / rectangle / square are congruent
3 | $\overline{cd} \cong \overline{ce}$ | transitive property
4 | $\angle e \cong \angle cde$ | in a triangle, angles opposite of congruent sides are congruent
5 | $\overline{ab} \parallel \overline{cd}$ | opposite sides of a parallelogram / rhombus / rectangle / square are parallel
6 | $\angle a \cong \angle cde$ | parallel lines cut by a transversal form congruent corresponding angles
7 | $\angle a \cong \angle e$ | select a reason...
(image of a parallelogram $abcd$ and triangle $cde$ with markings for congruent sides and angles)

Explanation:

Step1: Identify known congruences

From Step 4, we have \( \angle E \cong \angle CDE \), and from Step 6, we have \( \angle A \cong \angle CDE \).

Step2: Apply Transitive Property

The Transitive Property of Congruence states that if \( \angle A \cong \angle CDE \) and \( \angle E \cong \angle CDE \), then \( \angle A \cong \angle E \).

Answer:

The reason for \( \angle A \cong \angle E \) is the Transitive Property of Congruence (if \( \angle A \cong \angle CDE \) and \( \angle E \cong \angle CDE \), then \( \angle A \cong \angle E \)).