1. **State the problem:** Given that $m\angle 1 = m\angle 3$, prove that $\angle AEC \cong \angle DEB$. 2. **Understand the setup:** At point $E$, rays $EA$, $EB$, $EC$, and $ED$ form angles $\angle 1$ between $EA$ and $EB$, $\angle 2$ between $EB$ and $EC$, and $\angle 3$ between $EC$ and $ED$. 3. **Recall angle addition postulate:** The sum of adjacent angles around a point on a straight line equals the larger angle formed by the outer rays. 4. **Express $\angle AEC$ and $\angle DEB$ in terms of $\angle 1$, $\angle 2$, and $\angle 3$:** - $\angle AEC$ is the angle between rays $EA$ and $EC$, so $$m\angle AEC = m\angle 1 + m\angle 2$$ - $\angle DEB$ is the angle between rays $ED$ and $EB$, so $$m\angle DEB = m\angle 3 + m\angle 2$$ 5. **Given that $m\angle 1 = m\angle 3$, substitute into the expressions:** $$m\angle AEC = m\angle 1 + m\angle 2 = m\angle 3 + m\angle 2 = m\angle DEB$$ 6. **Conclude:** Since $m\angle AEC = m\angle DEB$, by definition of congruent angles, $$\angle AEC \cong \angle DEB$$ This completes the proof.