QUESTION IMAGE
Question
complete the proof that △efj≅△hgi.
statement reason
1 ij≅fg given
2 ej≅hi given
3 gh≅ef given
4 fj = ij + fi additive property of length
5 gi = fg + fi substitution
6 fj = fg + fi transitive property of equality
7 gi = fj
8 △efj≅△hgi sss
Step1: Recall given equal - segments
We are given $\overline{IJ}\cong\overline{FG}$, $\overline{EJ}\cong\overline{HI}$, $\overline{GH}\cong\overline{EF}$.
Step2: Apply additive property of length
We know that by the additive property of length, $FJ = IJ+FI$ and $GI = FG + FI$.
Step3: Use substitution
Since $\overline{IJ}\cong\overline{FG}$, we can substitute $IJ$ with $FG$ in the equation $FJ = IJ+FI$ to get $FJ=FG + FI$.
Step4: Use transitive property
We have $GI = FG+FI$ and $FJ = FG + FI$. By the transitive property of equality, if $a=b$ and $c = b$, then $a = c$. So $GI=FJ$.
Step5: Prove congruence
We have $\overline{GH}\cong\overline{EF}$, $\overline{EJ}\cong\overline{HI}$, and now $\overline{GI}\cong\overline{FJ}$. By the Side - Side - Side (SSS) congruence criterion, $\triangle EFJ\cong\triangle HGI$.
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- Reason: Transitive Property of Equality
- Reason: SSS (Side - Side - Side congruence criterion)