Question

given: \\( \angle g \cong \angle i \\); \\( \overline{fh} \\) bisects \\( \angle gfi \\) prove: \\( \triangle gfh \cong \triangle ifh \\)

Explanation:

Step1: Use the angle - bisector definition

Since \(\overline{FH}\) bisects \(\angle GFI\), by the definition of an angle bisector, \(\angle GFH=\angle IFH\).

Step2: Identify the common side

In \(\triangle GFH\) and \(\triangle IFH\), \(\overline{FH}\) is a common side. So \(FH = FH\) (reflexive property of equality).

Step3: Apply the AAS (Angle - Angle - Side) congruence criterion

We are given that \(\angle G=\angle I\), we have \(\angle GFH=\angle IFH\) (from step 1), and \(FH = FH\) (from step 2). By the AAS (Angle - Angle - Side) congruence criterion, \(\triangle GFH\cong\triangle IFH\).

Answer:

\(\triangle GFH\cong\triangle IFH\) by the AAS (Angle - Angle - Side) congruence criterion.