Question

the proof that \\( \triangle abc \cong \triangle dcb \\) is shown. given: \\( \angle a \cong \angle d ; c d \parallel a b \\) prove: \\( \triangle a b c \cong \triangle d c b \\) what is the missing reason in the proof? corr. int. \\( \angle s \\) are \\( \cong \\) alt. ext. \\( \angle s \\) are \\( \cong \\) asa aas

Explanation:

• In the proof, we have two angles ($\angle A\cong\angle D$ and $\angle ABC\cong\angle DCB$) and a non - included side ($CB\cong BC$).
• The AAS (Angle - Angle - Side) congruence criterion states that if two angles and a non - included side of one triangle are congruent to two angles and the corresponding non - included side of another triangle, then the two triangles are congruent.

Answer:

AAS