## Task A-3.3.

Given is a triangle $A B C$. Let point $D$ be the foot of the altitude from vertex $A$, and point $E$ be the intersection of the angle bisector of $\varangle C B A$ with the opposite side. If $\varangle B E A=45^{\circ}$, determine $\varangle E D C$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_76afda38443ee3ddf2f1g-13.jpg?height=497&width=1430&top_left_y=1716&top_left_x=313)

Let $\alpha, \beta$, and $\gamma$ be the measures of the angles of triangle $ABC$ at vertices $A, B$, and $C$ respectively, and let $\varphi$ be the measure of angle $\angle EDC$. Since $\angle BEA$ is an exterior angle of triangle $BEC$, we have $\angle EBC + \angle BCE = \angle BEA$, or $\frac{\beta}{2} + \gamma = 45^\circ$. Therefore,

\[
\alpha = 180^\circ - \beta - \gamma = 180^\circ - \beta - 2\gamma + \gamma = 180^\circ - 2\left(\frac{\beta}{2} + \gamma\right) + \gamma = 180^\circ - 2 \cdot 45^\circ + \gamma = 90^\circ + \gamma.
\]

For the angles of triangle $DAE$, we have

\[
\begin{gathered}
\angle EAD = 180^\circ - \angle ADC - \angle DCA = 90^\circ - \gamma \\
\angle EDA = \angle CDA - \angle CDE = 90^\circ - \varphi.
\end{gathered}
\]

Applying the Law of Sines to triangles $DEC$ and $DAE$ respectively, we get

\[
\frac{|EC|}{|DE|} = \frac{\sin \varphi}{\sin \gamma}, \quad \frac{|DE|}{|AE|} = \frac{\sin (90^\circ - \gamma)}{\sin (90^\circ - \varphi)} = \frac{\cos \gamma}{\cos \varphi}
\]

Multiplying these two equalities, we obtain

\[
\frac{|EC|}{|AE|} = \frac{\sin \varphi \cos \gamma}{\cos \varphi \sin \gamma}
\]

On the other hand, since $BE$ is the angle bisector of $\angle CBA$ and the Law of Sines applied to triangle $ABC$ gives

\[
\frac{|EC|}{|AE|} = \frac{|BC|}{|AB|} = \frac{\sin \alpha}{\sin \gamma} = \frac{\sin (90^\circ + \gamma)}{\sin \gamma} = \frac{\cos \gamma}{\sin \gamma}
\]

Thus, we have

\[
\frac{\sin \varphi \cos \gamma}{\cos \varphi \sin \gamma} = \frac{|EC|}{|AE|} = \frac{\cos \gamma}{\sin \gamma}
\]

which implies $\tan \varphi = 1$. Since angle $\angle EDC$ is acute, we conclude that the measure of the desired angle $\varphi = 45^\circ$.