Question

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

Explanation:

Step1: Identify equal - side conditions

Since \(AB\parallel CD\), in quadrilateral \(ABCD\), we know that \(AB = CD\) (opposite sides of a parallelogram if \(AB\parallel CD\) and assume it's a parallelogram - not given explicitly but from parallel sides we can infer for congruence). Also, \(AC\) is common to both \(\triangle ACD\) and \(\triangle CAB\), so \(AC=AC\). And \(AD = CB\) (opposite sides of a parallelogram).

Step2: Apply congruence theorem

The SSS (Side - Side - Side) congruence theorem states that if three sides of one triangle are equal to three corresponding sides of another triangle, the two triangles are congruent. Here, in \(\triangle ACD\) and \(\triangle CAB\), we have \(AB = CD\), \(AC = AC\), and \(AD=CB\). So, by SSS, \(\triangle ACD\cong\triangle CAB\).

Answer:

The SSS theorem is correct as it can be used to show \(\triangle ACD\cong\triangle CAB\) based on the equal - side relationships in the figure.