Question

given: $\angle b \cong \angle d$ and $\overline{ad} \parallel \overline{bc}$. prove: $\triangle abc \cong \triangle cda$. note: quadrilateral properties are not permitted in this proof. step 1 statement $\angle b \cong \angle d$ $\overline{ad} \parallel \overline{bc}$ reason given try type of statement

Explanation:

Step1: Find alternate interior angles

Since \(\overline{AD}\parallel\overline{BC}\), by the Alternate Interior Angles Theorem, \(\angle DAC\cong\angle BCA\).

Step2: Consider the common side

\(\overline{AC}\cong\overline{CA}\) (Reflexive Property of Congruence).

Step3: Use AAS congruence

We have \(\angle B\cong\angle D\) (given), \(\angle BCA\cong\angle DAC\) (from step 1), and \(\overline{AC}\cong\overline{CA}\) (step 2). By the Angle - Angle - Side (AAS) Congruence Theorem, \(\triangle ABC\cong\triangle CDA\).

Answer:

\(\triangle ABC\cong\triangle CDA\) by AAS (Angle - Angle - Side) congruence.