3. A right triangle ABC is inscribed in a circle with hypotenuse AB. On the larger leg BC, a point D is taken such that AC = BD, and point E is the midpoint of the arc AB containing point C. Find the angle DEC.
3. Point $\mathrm{E}$ is the midpoint of arc $\mathrm{AB}$, so $\mathrm{AE}=\mathrm{BE}$. Moreover, the inscribed angles $\mathrm{CAE}$ and $\mathrm{EBC}$, which subtend the same arc, are equal. Also, by the given condition, $\mathrm{AC}=\mathrm{BD}$. Therefore, triangles $\mathrm{ACE}$ and $\mathrm{BDE}$ are congruent, from which it follows that $\angle \mathrm{CEA}=\angle \mathrm{BED}$. But then $\angle \mathrm{DEC}=\angle \mathrm{BEA}=90^{\circ}$, since $\angle \mathrm{BEA}=\angle \mathrm{BCA}$.

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