1. **State the problem:** We need to prove that triangle ABC is a right triangle given that line segment BE is perpendicular to DE and that the measure of angle ABE equals the measure of angle DEB. 2. **Given information:** - BE \perp DE means $\angle B E D = 90^\circ$. - $m\angle ABE = m\angle DEB$. 3. **Goal:** Prove that triangle ABC has a right angle, i.e., one of its angles is $90^\circ$. 4. **Analyze the given angles:** Since $BE \perp DE$, $\angle B E D = 90^\circ$. Given $m\angle ABE = m\angle DEB$, and these two angles share vertex B and rays BA and BE, and DE and BE respectively. 5. **Use the fact that angles around point B on line BE sum to $180^\circ$:** The angles $\angle ABE$ and $\angle DEB$ are adjacent and together with $\angle B E D$ form a straight line or sum to $180^\circ$. 6. **Since $\angle B E D = 90^\circ$ and $m\angle ABE = m\angle DEB$, let each equal $x$:** Then $x + x + 90^\circ = 180^\circ$. 7. **Solve for $x$:** $$2x + 90 = 180$$ $$2x = 90$$ $$x = 45^\circ$$ 8. **Therefore, $m\angle ABE = m\angle DEB = 45^\circ$.** 9. **Consider triangle ABE:** It has angles $45^\circ$, $45^\circ$, and $90^\circ$. 10. **Since triangle ABE is right-angled at B, and ABC shares this angle at B, triangle ABC is right-angled at B.** **Final answer:** Triangle ABC is a right triangle with a right angle at vertex B.