Let $ABC$ be a triangle in which $\measuredangle{A}=135^{\circ}$. The perpendicular to the line $AB$ erected at $A$ intersects the side $BC$ at $D$, and the angle bisector of $\angle B$ intersects the side $AC$ at $E$.
Find the measure of $\measuredangle{BED}$.
1. Given that $\angle A = 135^\circ$ in $\triangle ABC$, we can determine the other angles in the triangle. Let $\angle ABC = \beta$. Then, $\angle ACB = 180^\circ - 135^\circ - \beta = 45^\circ - \beta$.
2. The perpendicular to $AB$ at $A$ intersects $BC$ at $D$. This means $\angle DAB = 90^\circ$.
3. The angle bisector of $\angle B$ intersects $AC$ at $E$. By the Angle Bisector Theorem, $\frac{AE}{EC} = \frac{AB}{BC}$.
4. We need to find $\angle BED$. To do this, we first analyze the angles in $\triangle ADC$:
   - $\angle ADC = 90^\circ + \beta$ because $\angle DAB = 90^\circ$ and $\angle BAC = 135^\circ$.
5. Using the Law of Sines in $\triangle ADC$:
   \[
   \frac{\sin(45^\circ - \beta)}{AD} = \frac{\sin 45^\circ}{CD} \implies \frac{AD}{CD} = \frac{\sin(45^\circ - \beta)}{\sin 45^\circ} = \cos \beta - \sin \beta
   \]
6. Since $\frac{AD}{BD} = \sin \beta$, we have:
   \[
   \frac{BD}{CD} = \frac{\cos \beta - \sin \beta}{\sin \beta} \implies \frac{BD}{CD} + 1 = \cot \beta \implies \frac{CD}{BC} = \tan \beta
   \]
7. By the Angle Bisector Theorem in $\triangle ABC$:
   \[
   \frac{AB}{BC} = \frac{AE}{EC} \implies \frac{AD}{CD} = \frac{AE}{EC}
   \]
8. Since $\overline{DE}$ bisects $\angle ADC$, we have:
   \[
   \angle BED = 180^\circ - \angle DBE - \angle BDE
   \]
9. Substituting the known angles:
   \[
   \angle DBE = \frac{\beta}{2}, \quad \angle BDE = 90^\circ - \beta + 45^\circ + \frac{\beta}{2}
   \]
10. Therefore:
    \[
    \angle BED = 180^\circ - \frac{\beta}{2} - \left(90^\circ - \beta + 45^\circ + \frac{\beta}{2}\right) = 180^\circ - \frac{\beta}{2} - 135^\circ + \beta - \frac{\beta}{2} = 45^\circ
    \]

The final answer is $\boxed{45^\circ}$.