given: $ \overline { fg } \cong \overline { fh } $,$ \overline { fk } $ bisects $ \angle hfg $.
prove: $ \triangle gjk \cong \triangle hjk $
1. $ \overline { fg } \cong \overline { fh } $, $ \overline { fk } $ bisects $ \angle hfg $.
1. given
2. definition of an angle
bisector
3. reflexive property
5. corresponding parts

Question

Accepted Answer

# Explanation:
## Step1: Use the definition of an angle bisector
Since $ \overline{FK} $ bisects $ \angle HFG $, we have $ \angle GFK=\angle HFK $.
## Step2: Use the reflexive property
We know that $ \overline{FK}=\overline{FK} $ (reflexive property of congruence).
## Step3: Use the SAS (Side - Angle - Side) congruence criterion
We are given $ \overline{FG}\cong\overline{FH} $, $ \angle GFK = \angle HFK $ (from step 1), and $ \overline{FK}\cong\overline{FK} $ (from step 2). By the SAS congruence criterion, $ \triangle FGK\cong\triangle FHK $. Then, since $ \triangle FGK\cong\triangle FHK $, we have $ \overline{GK}=\overline{HK} $ and $ \angle FGK=\angle FHK $. Also, $ \overline{FJ}$ is common in some sub - triangles (but focusing on the main triangles for the final proof). Another way:
We know that $ \triangle FGK\cong\triangle FHK$ (SAS: $ FG = FH$, $ \angle GFK=\angle HFK$, $ FK = FK$). Then, for $ \triangle GJK$ and $ \triangle HJK$:
We have $ \overline{GK}=\overline{HK}$ (from $ \triangle FGK\cong\triangle FHK$), $ \overline{JK}=\overline{JK}$ (reflexive property), and $ \angle GKJ=\angle HKJ$ (since $ \triangle FGK\cong\triangle FHK$ implies $ \angle FGK=\angle FHK$ and using angle - addition or subtraction depending on the figure's exact structure, but more directly from the fact that $ \triangle FGK\cong\triangle FHK$ gives us the necessary angle and side equalities for $ \triangle GJK$ and $ \triangle HJK$). By SSS (if we consider $ GK = HK$, $ JK=JK$, and $ GJ = HJ$ (from $ \triangle FGK\cong\triangle FHK$ and properties of the figure) or SAS (if we use $ GK = HK$, $ \angle GKJ=\angle HKJ$, $ JK = JK$).

# Answer:
2. $ \angle GFK=\angle HFK $; 3. $ \overline{FK}=\overline{FK} $; 4. $ \triangle FGK\cong\triangle FHK$ (by SAS: $ FG = FH$, $ \angle GFK=\angle HFK$, $ FK = FK$); 5. $ \triangle GJK\cong\triangle HJK$ (by SSS or SAS depending on the full set of equal sides and angles derived from $ \triangle FGK\cong\triangle FHK$)

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Question

Explanation:

Step1: Use the definition of an angle bisector

Step2: Use the reflexive property

Step3: Use the SAS (Side - Angle - Side) congruence criterion

Answer: