1. **State the problem:** We have two lines $f$ and $g$ crossed by a transversal $n$. We want to find which angle measure among $a$, $b$, $c$, and $d$ should equal $105^\circ$ to prove that $f \parallel g$. 2. **Recall the rule for parallel lines and transversals:** When two lines are parallel, corresponding angles, alternate interior angles, and consecutive interior angles have specific relationships. - Corresponding angles are equal. - Alternate interior angles are equal. - Consecutive interior angles are supplementary (sum to $180^\circ$). 3. **Given:** One angle is $75^\circ$ at the right intersection on the lower-right side. 4. **Find the angle that should be $105^\circ$:** Since $75^\circ + 105^\circ = 180^\circ$, the angle supplementary to $75^\circ$ is $105^\circ$. 5. **Identify which angle is supplementary to $75^\circ$:** The angle $a$ is above the transversal at the right intersection, adjacent to the $75^\circ$ angle, so $a$ and $75^\circ$ are consecutive interior angles. 6. **Therefore, to prove $f \parallel g$, angle $a$ must be $105^\circ$ because consecutive interior angles must sum to $180^\circ$:** $$a + 75^\circ = 180^\circ$$ $$a = 180^\circ - 75^\circ = 105^\circ$$ **Final answer:** Angle $a$ should equal $105^\circ$ to prove that $f \parallel g$.