Since $BD$ is the bisector of the angle $\angle CBA$, it follows that $|CD| : |AD| = |BC| : |AB|$, and because of that $\frac{|CD|}{|CE|} = \frac{|CD|}{|AD|} = \frac{|BC|}{|AB|} = \frac{|CB|}{|CA|}$.

Triangles $ABC$ and $EDC$ have a common angle in vertex $C$ and equal ratios of corresponding sides so they are similar. Hence $\angle CED = \angle CAB$.

Let $\angle ABC = 2x$. Then $\angle CBD = \angle DBA = x$ and $\angle ACB = 2x$. Furthermore, $\angle CED = \angle BAC = 180^\circ - 4x$.

Since the triangle $BED$ is isosceles, we have $\angle DEB = 90^\circ - \frac{x}{2}$. As $\angle DEB + \angle CED = 180^\circ$, we get
$$(90^\circ - \frac{x}{2}) + (180^\circ - 4x) = 180^\circ,$$
whence follows $x = 20^\circ$. The angles of the observed triangle are $\angle ABC = \angle ACB = 40^\circ$ and $\angle BAC = 100^\circ$.