Question

refer to the diagram shown. if $overline{ab}paralleloverline{cd}$, the choose... theorem can be used to show that $\triangle acdcong\triangle cab$.

Explanation:

Step1: Identify common side

In \(\triangle ACD\) and \(\triangle CAB\), \(AC = AC\) (common side).

Step2: Use parallel - side property

Since \(\overline{AB}\parallel\overline{CD}\), \(\angle BAC=\angle DCA\) (alternate - interior angles). Also, assume \(AB = CD\) (if not given explicitly, we can consider the property of a parallelogram - like figure where opposite sides are equal in a parallelogram. Here we can use the SSS (Side - Side - Side) or SAS (Side - Angle - Side) congruence criteria). If we consider the fact that in a parallelogram \(AB = CD\), \(AC\) is common and \(\angle BAC=\angle DCA\), we can use the SAS (Side - Angle - Side) congruence theorem.

Answer:

SAS (Side - Angle - Side)