In a triangle $$ABC$$, we have $$\angle ACB = \angle BAC + 90^\circ$$. The point $$D$$ is taken on the extended $$BC$$ such that $$AC = AD$$. A point $$E$$ is taken such that points $$A$$ and $$E$$ are on the opposite sides of side $$BC$$ and $$\angle EBC = \angle BAC$$, $$\angle EDC = \frac{1}{2}\angle BAC$$. Prove that $$\angle CED = \angle ABC$$.

Let $$\angle BAC = x$$, so $$\angle ACB = 90^\circ + x$$, $$\angle ABC = 90^\circ - 2x$$, $$\angle EBC = x$$ and $$\angle EDC = \frac{x}{2}$$. Since $$AC = AD$$, in $$\triangle ADC$$ we have $$\angle ADC = \angle ACD = 90^\circ - x$$ and $$\angle DAC = 2x$$. Let $$O$$ be the circumcentre of $$\triangle BED$$; then $$\angle DOE = 2\angle DBE = 2x$$ and $$OD = OE$$ give $$\angle ODE = \angle OED = 90^\circ - x$$, so $$\triangle DAC \sim \triangle DOE$$ and $$\frac{DA}{DO} = \frac{DC}{DE}$$. Also $$\angle ADO = \angle ADC - \angle ODC = (90^\circ - x) - \left(90^\circ - x - \frac{x}{2}\right) = \frac{x}{2} = \angle CDE$$, so $$\triangle DAO \sim \triangle DCE$$ by SAS and $$\angle CED = \angle AOD$$. In $$\triangle BED$$, $$\angle BED = 180^\circ - \frac{3x}{2}$$, so $$\angle BOD = 2(180^\circ - \angle BED) = 3x = \angle BAC + \angle CAD = \angle BAD$$, hence $$B, O, A, D$$ are concyclic. Therefore $$\angle AOD = \angle ABD = \angle ABC$$, and so $$\angle CED = \angle ABC$$.