## Task B-4.5.

In triangle $A B C$, $\varangle C A B=50^{\circ}$ and $\varangle A B C=60^{\circ}$. On side $\overline{A B}$ there is a point $D$, and on side $\overline{B C}$ there is a point $E$ such that $\varangle C A E=\varangle A C D=30^{\circ}$. Calculate the measure of angle $\varangle C D E$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_d18d2bbdee068dba95dfg-21.jpg?height=983&width=973&top_left_y=1593&top_left_x=450)

Let $S = AE \cap CD$.

Triangle $ASC$ has two congruent angles, which means it is isosceles, i.e., $|AS| = |SC|$ and

$$
\angle ASC = 180^\circ - 2 \cdot 30^\circ = 120^\circ
$$

Since $\angle ABC = 60^\circ$, by Thales' theorem on central and inscribed angles, point $S$ is the center of the circumcircle of triangle $ABC$. Its radius is equal to $|SA| = |SC| = |SB|$.

Then triangles $ASC$ and $BSC$ are also isosceles, so

$$
\begin{aligned}
& \angle SAB = \angle SBA = 50^\circ - 30^\circ = 20^\circ, \\
& \angle SBC = \angle SCB = 70^\circ - 30^\circ = 40^\circ.
\end{aligned}
$$

Consider the angles of quadrilateral $DBES$.

$$
\angle ESD = \angle ASC = 120^\circ
$$

so

$$
\angle ESD + \angle DBE = 120^\circ + 60^\circ = 180^\circ
$$

The sum of opposite angles is $180^\circ$, so quadrilateral $DBES$ is cyclic, meaning a circle can be circumscribed around it. Angles $\angle SDE$ and $\angle SBE$ are smaller inscribed angles subtending the same chord $\overline{SE}$, which means they are congruent. Therefore,

$$
\angle CDE = \angle SDE = \angle SBE = \angle SBC = 40^\circ
$$