Angle bisector - Geometry problem

Let <ECA = x. ΔODC is right triangle. <DAB=45 – x. So <AEC =135

Draw radius $OD$ , and let $F$ be the intersection of $OD$ and $EC$ , and let $x = \angle DCE$ . Then $\angle DCO = 2x$ .

Since $CD$ is a tangent, $\angle ODC = 90°$ , and since the angle sum of $\triangle DOC$ is $180°$ and $\angle DCO = 2x$ , $\angle COD = 180° - 90° - 2x = 90° - 2x$ .

Since the angle sum of $\triangle OFC$ is $180°$ and $\angle FCO = x$ and $\angle COD = 90° - 2x$ , $\angle CFO = 180° - x - (90° - 2x) = 90° + x$ .

Since $\angle CFO$ and $\angle EFO$ form a straight line equaling $180°$ and $\angle CFO = 90° + x$ , $\angle EFO = 180° - (90° + x) = 90° - x$ , and since $\angle COD$ and $\angle AOD$ form a straight line equaling $180°$ and $\angle COD = 90° - 2x$ , $\angle AOD = 180° - (90° - 2x) = 90° + 2x$ .

Since $\triangle AOD$ is an isosceles formed by two radii sides, $\angle OAD = \angle ODA$ , and since the angle sum of $\triangle AOD$ is $180°$ and $\angle AOD = 90° + 2x$ , $\angle OAD = \frac{180° - (90° + 2x)}{2} = 45° - x$ .

Finally, since the angle sum of quadrilateral $AEFO$ is $360°$ , and $\angle OAD = 45° - x$ , $\angle AOD = 90° + 2x$ , and $\angle EFO = 90° - x$ , $\angle AEC = 360° - (45° - x) - (90° + 2x) - (90° - x) = \boxed{135°}$ .