Question

read the proof. given: $overline{ab} parallel overline{de}$ prove: $\triangle acb sim \triangle dce$ we are given $overline{ab} parallel overline{de}$. 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 $overline{cb}$ and are therefore congruent, so $angle ced cong angle cba$. we can state $angle c cong angle c$ using the reflexive property. therefore, $\triangle acb sim \triangle dce$ by the

• aa similarity theorem.
• sss similarity theorem.
• aas similarity theorem.
• asa similarity theorem.

Explanation:

We have established two pairs of congruent angles: $\angle CED \cong \angle CBA$ (corresponding angles from parallel lines) and $\angle C \cong \angle C$ (reflexive property). The AA (Angle-Angle) similarity theorem states that if two angles of one triangle are congruent to two angles of another triangle, the triangles are similar.

Answer:

AA similarity theorem.