3. Let $D$ be the foot of the altitude from $A$ to side $BC$ of triangle $ABC$. The angle bisector of the internal angle at $C$ intersects the opposite side at point $E$. What is the measure of angle $\Varangle EDB$, if $\Varangle CEA=\frac{\pi}{4}$?
IV /3. Let's denote the angles of the triangle as $\alpha, \beta$, and $\gamma$ in the usual way. In triangle $AEC$, we have $\alpha + \frac{\pi}{4} + \frac{\gamma}{2} = \pi$. From this, considering $\gamma = \pi - \alpha - \beta$, we derive $\alpha = \beta + \frac{\pi}{2}$.

Since $AD \perp DB$, we have $\frac{|AD|}{|DB|} = \tan \beta$. Since $CE$ is the angle bisector, we have $|AE| : |EB| = |AC| : |BC|$. By the Law of Sines, $\frac{|AC|}{|BC|} = \frac{\sin \beta}{\sin \alpha} = \frac{\sin \beta}{\sin \left(\beta + \frac{\pi}{2}\right)} = \frac{\sin \beta}{\cos \beta} = \tan \beta$. Therefore, $|AE| : |EB| = |AD| : |DB|$, and $ED$ is the angle bisector of the right angle

![](https://cdn.mathpix.com/cropped/2024_06_07_44798e3c82dc47fe22e9g-8.jpg?height=372&width=605&top_left_y=2087&top_left_x=1268)
$\angle ADB$, so $\angle EDB = \frac{\pi}{4}$.