Question

given: \\( \overline { f g } \cong \overline { f h } \\), \\( \overline { f k } \\) bisects \\( \angle h f g \\). prove: \\( \triangle g j k \cong \triangle h j k \\) 8. \\( \angle g j k \cong \angle h j k \\) 9. 10. \\( \triangle g j k \cong \triangle h j k \\) 8. transitive property 9. 10.

Explanation:

Step1: Use the definition of angle bisector

Since \( \overline{FK} \) bisects \( \angle HFG \), then \( \angle GFK=\angle HFK \).

Step2: Use the Side - Angle - Side (SAS) congruence criterion

We know that \( \overline{FG}\cong\overline{FH} \) (given), \( \angle GFK = \angle HFK \) (from step 1), and \( \overline{FK}\cong\overline{FK} \) (reflexive property). So, \( \triangle FGK\cong\triangle FHK \) by SAS. Then \( \overline{GK}\cong\overline{HK} \).

Step3: Use the reflexive property for another side

\( \overline{JK}\cong\overline{JK} \) (reflexive property).

Step4: Use the Side - Angle - Side (SAS) congruence criterion for \( \triangle GJK \) and \( \triangle HJK \)

We have \( \angle GJK\cong\angle HJK \) (given), \( \overline{JK}\cong\overline{JK} \) (reflexive), and \( \overline{GK}\cong\overline{HK} \) (from step 2). So, \( \triangle GJK\cong\triangle HJK \) by SAS.

Answer:

\( \triangle GJK\cong\triangle HJK \) by the Side - Angle - Side (SAS) congruence criterion.