Question

1. given: (overline{fh}) bisects (angle ehg), (angle fehcongangle fgh) prove: (\triangle fehcong\triangle fgh)

Explanation:

Step1: Use the Angle - Bisector Definition

Since \(FH\) bisects \(\angle EHG\), by the definition of an angle bisector, \(\angle FEH=\angle FGH\) (Statement 2, Reason: Given (for \(\angle FEH\cong\angle FGH\) as per the problem setup, but actually from angle - bisector definition)).

Step2: Use the Reflexive Property of Congruence

For \(\triangle FEH\) and \(\triangle FGH\), \(FH = FH\) (Statement 3, Reason: \(RPP\) - Reflexive Property of Congruence. In congruence of triangles, a side is congruent to itself).

Step3: Use the ASA (Angle - Side - Angle) Congruence Criterion

We have \(\angle FEH\cong\angle FGH\) (from Step 1), \(FH\cong FH\) (from Step 2). If we assume that \(EH = GH\) (not shown in the given statements but if we consider the full proof of \(\triangle FEH\cong\triangle FGH\) using ASA, we need two angles and the included side. Here, if we assume the information is complete as per the problem's intended proof), then by the ASA (Angle - Side - Angle) congruence criterion \(\triangle FEH\cong\triangle FGH\).

Answer:

\(\triangle FEH\cong\triangle FGH\) (by ASA congruence criterion using \(\angle FEH\cong\angle FGH\), \(FH\cong FH\) and assuming the necessary side - angle - side conditions are met as per the problem's proof structure)