Question

given: \\( \angle abc \\) and \\( \angle fgh \\) are right angles; \\( \overline{ba} \parallel \overline{gf} \\); \\( \overline{bc} \cong \overline{gh} \\) prove: \\( \triangle abc \cong \triangle fgh \\) step 1: we know that \\( \angle abc \cong \angle fgh \\) because all right angles are congruent. step 2: we know that \\( \angle bac \cong \angle gfh \\) because corresponding angles of parallel lines are congruent. step 3: we know that \\( \overline{bc} \cong \overline{gh} \\) because it is given. step 4: \\( \triangle abc \cong \triangle fgh \\) because of the asa congruence theorem. aas congruence theorem. third angle theorem. reflexive property.

Explanation:

Step1: Recall congruence theorems

ASA (Angle - Side - Angle) theorem states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. AAS (Angle - Angle - Side) theorem 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 triangles are congruent.

Step2: Analyze the given congruent parts

We have \(\angle ABC\cong\angle FGH\) (right angles), \(\angle BAC\cong\angle GFH\) (corresponding angles of parallel lines), and \(\overline{BC}\cong\overline{GH}\) (given). The side \(\overline{BC}\) is not included between \(\angle ABC\) and \(\angle BAC\) in \(\triangle ABC\), and \(\overline{GH}\) is not included between \(\angle FGH\) and \(\angle GFH\) in \(\triangle FGH\)

Answer:

AAS congruence theorem.