Question

read the proof.
given: \\( \overline { a b } \parallel \overline { d e } \\)
prove: \\( \triangle a c b \sim \triangle d c e \\)
we are given \\( \overline { a b } \parallel \overline { d e } \\). because the lines are parallel
and segment cb crosses both lines, we can consider
segment cb a transversal of the parallel lines. angles
ced and cba are corresponding angles of transversal
\\( \overline { c b } \\) and are therefore congruent, so \\( \angle c e d \cong \angle c b a \\).
we can state \\( \angle c \cong \angle c \\) using the reflexive property.
therefore, \\( \triangle a c b \sim \triangle d c e \\) by the
\\( \bigcirc \\) aa similarity theorem.
\\( \bigcirc \\) sss similarity theorem.
\\( \bigcirc \\) aas similarity theorem.
\\( \bigcirc \\) asa similarity theorem.

Explanation:

Step1: Identify congruent angles

We have $\angle CED\cong\angle CBA$ (corresponding angles) and $\angle C\cong\angle C$ (reflexive property).

Step2: Apply similarity theorem

The AA (Angle - Angle) similarity theorem states that if two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. Since we have two pairs of congruent angles in $\triangle ACB$ and $\triangle DCE$, we use the AA similarity theorem.

Answer:

AA similarity theorem.