Question

the proof $\triangle abc \cong \triangle dcb$ that is shown. given: $\angle a \cong \angle d$; $\overline{cd}\parallel\overline{ab}$ prove: $\triangle abc \cong \triangle dcb$ diagram: points c, a, d, b with triangles, c connected to a and d, b connected to a and d, angles marked at a and d, cb as a common side what is the missing reason in the proof? \

$$\begin{tabular}{|c|c|} \\hline statement & reason \\\\ \\hline 1. $\\angle a \\cong \\angle d$ & 1. given \\\\ \\hline 2. $\\overline{cd}\\parallel\\overline{ab}$ & 2. given \\\\ \\hline 3. $\\overline{cb} \\cong \\overline{bc}$ & 3. refl. prop. \\\\ \\hline 4. $\\angle abc \\cong \\angle dcb$ & 4. alt. int. $\\angle$s are $\\cong$ \\\\ \\hline 5. $\\triangle abc \\cong \\triangle dcb$ & 5.? \\\\ \\hline \\end{tabular}$$

options: alt. ext. $\angle$s are $\cong$, asa, aas, corr. int. $\angle$s are $\cong$

Explanation:

Step1: Identify known congruences

We know $\angle A \cong \angle D$ (given), $\angle ABC \cong \angle DCB$ (alternate interior angles), and $\overline{CB} \cong \overline{BC}$ (reflexive property).

Step2: Match with triangle congruence criteria

In $\triangle ABC$ and $\triangle DCB$:

• $\angle A \cong \angle D$ (angle)
• $\overline{CB} \cong \overline{BC}$ (side)
• $\angle ABC \cong \angle DCB$ (angle)

This matches the AAS (Angle - Angle - Side) congruence criterion, where two angles and a non - included side of one triangle are congruent to the corresponding two angles and non - included side of another triangle.

Answer:

C. AAS