Question

△fij≅△ghj. complete the proof that △fgi≅△gfh.
1 △fij≅△ghj
2 (overline{fi}congoverline{gh})
3 (overline{fj}congoverline{gj})
4 (overline{hj}congoverline{ij})
5 (∠fijcong∠ghj)
6 (fh = fj+hj)
7 (gi = gj+ij)
8 (fh = gj+ij)
9 (fh = gi)
10 △fgi≅△gfh

Explanation:

Step1: Given

$\triangle FIJ\cong\triangle GHJ$ (This is the given information for the proof)

Step2: Corresponding parts of congruent triangles are congruent (CPCTC)

Since $\triangle FIJ\cong\triangle GHJ$, then $\overline{FI}\cong\overline{GH}$ (CPCTC)

Step3: CPCTC

Since $\triangle FIJ\cong\triangle GHJ$, then $\overline{FJ}\cong\overline{GJ}$ (CPCTC)

Step4: CPCTC

Since $\triangle FIJ\cong\triangle GHJ$, then $\overline{HJ}\cong\overline{IJ}$ (CPCTC)

Step5: CPCTC

Since $\triangle FIJ\cong\triangle GHJ$, then $\angle FIJ\cong\angle GHJ$ (CPCTC)

Step6: Segment addition postulate

$FH = FJ+HJ$ (By the segment addition postulate which states that if $J$ is between $F$ and $H$, then $FH=FJ + HJ$)

Step7: Segment addition postulate

$GI=GJ + IJ$ (By the segment addition postulate which states that if $J$ is between $G$ and $I$, then $GI = GJ+IJ$)

Step8: Substitution

Substitute $\overline{FJ}\cong\overline{GJ}$ and $\overline{HJ}\cong\overline{IJ}$ into $FH = FJ+HJ$. So $FH=GJ + IJ$ (Substitution property)

Step9: Substitution

Since $GI=GJ + IJ$ and $FH=GJ + IJ$, then $FH = GI$ (Substitution property)

Step10: Side - Angle - Side (SAS) congruence criterion

In $\triangle FGI$ and $\triangle GFH$, we have $\overline{FI}\cong\overline{GH}$, $\angle FIJ\cong\angle GHJ$ (which is $\angle FGI\cong\angle GFH$ as they are the angles between the sides), and $FH = GI$. So $\triangle FGI\cong\triangle GFH$ (SAS)

Answer:

1. Given; 2. CPCTC; 3. CPCTC; 4. CPCTC; 5. CPCTC; 6. Segment Addition Postulate; 7. Segment Addition Postulate; 8. Substitution; 9. Substitution; 10. SAS