1. **State the problem:** We are given triangle ABC with a right angle at E inside it. AE = BE, AC = 12, and DC = 5. We need to find the area of triangle BED. 2. **Identify known lengths and relationships:** - AC = 12 - DC = 5 - AE = BE (given) - E is the right angle in triangle BED 3. **Use the given information:** Since AE = BE and AC = 12, point E divides AC into AE and EC such that AE + EC = 12. 4. **Find AE and BE:** Since AE = BE, and E lies on AC, AE = BE = x (unknown length). 5. **Find EC:** EC = AC - AE = 12 - x. 6. **Use right triangle BED:** Triangle BED is right angled at E, with legs BE and DC. 7. **Calculate area of triangle BED:** Area = \frac{1}{2} \times BE \times DC = \frac{1}{2} \times x \times 5 = \frac{5x}{2}. 8. **Find x:** Since AE = BE = x, and DC is perpendicular to AB at E, we can use the Pythagorean theorem in triangle ADC. 9. **Triangle ADC:** AC = 12, DC = 5, AD is unknown. 10. **Calculate AD:** Using Pythagoras in triangle ADC, $$AD = \sqrt{AC^2 - DC^2} = \sqrt{12^2 - 5^2} = \sqrt{144 - 25} = \sqrt{119}.$$ 11. **Since AE = BE = x and E lies on AC, and DC is perpendicular to AB at E, the length BE = AE = x is the projection of AD on AC. Given the problem's constraints, the length x equals AD. 12. **Therefore, x = \sqrt{119}**. 13. **Calculate area:** $$\text{Area} = \frac{5x}{2} = \frac{5 \times \sqrt{119}}{2}.$$ 14. **Final answer:** $$\boxed{\frac{5 \sqrt{119}}{2}}.$$