Determine the angle $\angle DEC$ in a triangle (Euclidean Geometry)

Question

Any ideas how to find the angle $\angle DEC$ in the following situation shown in the image:

In the above figure we have that $\angle BAC = 90, \angle ABD = \alpha, \angle DBC = 2\alpha$, and $\angle ACE = 30+\alpha$ and we need to find $\angle DEC$.

Answer 1

Let $F$ be the point symmetric to $C$ with respect to the bisector of angle $CBD$. Then $F$ lies on $BD$ and $$\angle DFC = 90^\circ - \frac 12 \angle CBF = 90^\circ - \alpha = \angle ADB = \angle CDF,$$ hence $CD=CF$ and $\angle FCD = 2\alpha$. Build an equilateral triangle $CFG$ so that $B$ and $G$ lie on different sides of the line $CF$. Note that $D$, $F$, $G$ lie on the circle with center $C$ and radius $CD=CF=CG$. Thus $$\angle GDF = \frac 12 \angle GCF = 30^\circ.$$ Let $H$ be the point symmetric to $G$ with respect to $BD$. Note that $H$ lies on the line $AB$ as $\angle GBD = \alpha = \angle DBA$. We have $\angle GDH = 2\angle FDG = 2\cdot 30^\circ = 60^\circ$ and of course $DG=DH$, hence $DGH$ is equilateral. Thus $CDHG$ is a kite and in particular $\angle HCD = \frac 12 \angle GCD = 30^\circ + \alpha$ and $\angle DHC = \frac 12 \angle DHG = 30^\circ$. Since $\angle HCA = 30^\circ + \alpha = \angle ACE$ and $CA \perp AB$, it follows that $E$ is symmetric to $H$ with respect to $CD$. Thus $$\angle CED = \angle DHC = 30^\circ.$$