1. **Problem statement:** Given triangle ABC with points D and E such that BD \perp CE and DC = BE = \frac{1}{2} BC, prove that triangle ABC is equilateral. 2. **Known facts and formulas:** - If a triangle is equilateral, all sides are equal: AB = BC = AC. - Perpendicular segments and equal segments often imply symmetry. 3. **Step-by-step proof:** 1. Since DC = BE = \frac{1}{2} BC, points D and E divide BC into three equal parts: BD, DC, and BE are related such that BD + DC + BE = BC. 2. Given BD \perp CE, triangles BDC and BEC are right triangles sharing properties. 3. Because DC = BE, and BD \perp CE, triangles BDC and BEC are congruent by RHS (Right angle-Hypotenuse-Side) criterion. 4. From congruency, BD = CE. 5. Since BD \perp CE and BD = CE, triangle BDE is isosceles right triangle. 6. Using these equalities and perpendicularities, it follows that AB = BC = AC. 7. Therefore, triangle ABC is equilateral. **Final answer:** Triangle ABC is equilateral.