QUESTION IMAGE
Question
complete the proof that $\triangle fgkcong\triangle ijh$.
| statement | reason | |
|---|---|---|
| 2 | $overline{hi}congoverline{fk}$ | given |
| 3 | $overline{jk}congoverline{gh}$ | given |
| 4 | $gk = jk+gj$ | |
| 5 | $hj = gh + gj$ | |
| 6 | $gk = gh+gj$ | substitution |
| 7 | $hj = gk$ | |
| 8 | $\triangle fgkcong\triangle ijh$ | sss |
Step1: Segment - addition postulate
By the segment - addition postulate, if a point $J$ lies on line segment $\overline{GK}$, then $GK = JK+GJ$.
Step2: Segment - addition postulate
By the segment - addition postulate, if a point $G$ lies on line segment $\overline{HJ}$, then $HJ = GH + GJ$.
Step3: Substitution
Since $JK\cong GH$ (given), we can substitute $JK$ with $GH$ in the equation $GK = JK+GJ$ to get $GK=GH + GJ$.
Step4: Transitive property
Since $HJ = GH + GJ$ and $GK=GH + GJ$, by the transitive property of equality, $HJ = GK$.
Step5: SSS congruence
We have $\overline{FG}\cong\overline{IJ}$ (given), $\overline{HI}\cong\overline{FK}$ (given), and $\overline{GK}\cong\overline{HJ}$ (proven above). So, by the Side - Side - Side (SSS) congruence criterion, $\triangle FGK\cong\triangle IJH$.
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