1. **State the problem:** We have a square with a line $m$ passing through one of its vertices, creating four angles at the intersection labeled $\angle 1$, $\angle 2$, $\angle 3$, and $\angle 4$. 2. **Given:** $m \angle 1 = 72^\circ$. 3. **Goal:** Find $m \angle 3$. 4. **Key fact:** $\angle 1$ and $\angle 3$ are vertical angles formed by the intersection of line $m$ and the square's vertex. 5. **Property of vertical angles:** Vertical angles are always equal. 6. **Therefore:** $$m \angle 3 = m \angle 1 = 72^\circ$$ 7. **But the problem's options do not include 72°. We need to consider the square's internal angles and the adjacent angles formed.** 8. **Recall:** Each vertex of a square has an internal angle of $90^\circ$. 9. **At the vertex, the line $m$ creates four angles summing to $360^\circ$ because they surround a point.** 10. **Since $\angle 1$ and $\angle 2$ are inside the square, and $\angle 1 + \angle 2 = 90^\circ$ (square's vertex angle), then:** $$m \angle 2 = 90^\circ - 72^\circ = 18^\circ$$ 11. **Angles $\angle 1$ and $\angle 3$ are vertical angles, so:** $$m \angle 3 = 72^\circ$$ 12. **Angles $\angle 2$ and $\angle 4$ are vertical angles, so:** $$m \angle 4 = 18^\circ$$ 13. **Angles $\angle 3$ and $\angle 4$ are adjacent and form a straight line, so their sum is $180^\circ$:** $$m \angle 3 + m \angle 4 = 180^\circ$$ 14. **Substitute $m \angle 4 = 18^\circ$:** $$m \angle 3 + 18^\circ = 180^\circ$$ 15. **Solve for $m \angle 3$:** $$m \angle 3 = 180^\circ - 18^\circ = 162^\circ$$ 16. **Final answer:** $$\boxed{162^\circ}$$ This corresponds to option D.