(7) Given: $\overline{H I} \cong \overline{J G}, \overline{H I} \| \overline{G J}$
Prove: $\triangle G J H \cong \Delta I H J$
\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{ Statements } & \multicolumn{1}{r}{ Reasons } \\
\hline 1. $\overline{H I} \cong \overline{J G}$ & 1. \\
\hline 2. $\overline{H I} \| \overline{G J}$ & 2. \\
\hline 3. $\angle G J H \cong \angle H J$ & 3. \\
\hline 4. $\overline{H J} \cong \overline{J H}$ & 4. \\
\hline 5. $\Delta G J H \cong \Delta H J$ & 5. \\
\hline
\end{tabular}
(c) Gina Wilson (All Things Algebra

Step 1 ：Given that line segment HI is congruent to line segment JG (\(\overline{H I} \cong \overline{J G}\))

Step 2 ：Given that line segment HI is parallel to line segment GJ (\(\overline{H I} \parallel \overline{G J}\))

Step 3 ：Since HI is parallel to GJ, the alternate interior angles are congruent. Therefore, angle GJH is congruent to angle HJI (\(\angle G J H \cong \angle H J I\))

Step 4 ：Line segment HJ is common to both triangles, so it is congruent to itself (\(\overline{H J} \cong \overline{H J}\))

Step 5 ：With these three pairs of congruent parts, we can conclude that the two triangles are congruent by the ASA (Angle-Side-Angle) postulate (\(\triangle G J H \cong \Delta I H J\))

Step 6 ：\(\boxed{\triangle G J H \cong \Delta I H J}\)