Question

proof complete the flow proof by dragging the reasons. given: given (overline{ge}) bisects (angle hef), (angle hge) and (angle fge) are right angles. prove: (\triangle egh cong \triangle egf) (overline{ge}) bisects (angle hef) given (angle heg cong angle feg) (overline{eg} cong overline{eg}) (angle hge cong angle fge) (angle hge) and (angle hge) are right angles. (\triangle egh cong \triangle egf) asa def of angle bisector reflexive property right angles are (cong) given

Explanation:

Step1: Analyze the first missing reason

We know that \( \overline{GE} \) bisects \( \angle HEF \). By the definition of an angle bisector, this means \( \angle HEG \cong \angle FEG \). So the reason for \( \angle HEG \cong \angle FEG \) is "Def of angle bisector".

Step2: Analyze the second missing reason

We are given that \( \angle HGE \) and \( \angle FGE \) are right angles. By the property that right angles are congruent, we can say \( \angle HGE \cong \angle FGE \). So the reason for \( \angle HGE \cong \angle FGE \) is "Right angles are \( \cong \)".

Step3: Analyze the third missing reason (for triangle congruence)

We have \( \angle HEG \cong \angle FEG \) (from angle bisector), \( \overline{EG} \cong \overline{EG} \) (reflexive property), and \( \angle HGE \cong \angle FGE \) (right angles congruent). This is the ASA (Angle - Side - Angle) congruence criterion. So the reason for \( \triangle EGH \cong \triangle EGF \) is "ASA".

Answer:

1. For \( \angle HEG \cong \angle FEG \): Def of angle bisector
2. For \( \angle HGE \cong \angle FGE \): Right angles are \( \cong \)
3. For \( \triangle EGH \cong \triangle EGF \): ASA