Let's construct a triangle, given the side $a$ and the opposite angle $\alpha$, and we know that the angle bisector of $\alpha$ forms a $45^{\circ}$ angle with side $a$.
We denote the vertex opposite to the side of length $a$ as $A$, the endpoint of the shorter side starting from $A$ as $B$, and the third vertex as $C$. We measure on the extensions of the lines $AB$ and $AC$ beyond $A$ the segments $AD = AC$ and $AE = AB$, and connect the resulting points $D$ and $E$ to $C$ and $B$, respectively. This way, we obtain isosceles triangles whose bases form angles of $\alpha / 2$ with the sides $AC$ and $AB$, respectively, and are therefore parallel to the angle bisector of $\alpha$. In our case, $\angle BCD = 45^\circ$ and $\angle CBE = 135^\circ$.

![](https://cdn.mathpix.com/cropped/2024_05_02_08955fb5561d2a7d3dcag-1.jpg?height=697&width=718&top_left_y=389&top_left_x=690)

Based on this, the construction proceeds as follows. Draw the segment $BC$ of length $a$, and measure angles of $135^\circ$ at $B$ and $45^\circ$ at $C$. Rotate the rays thus obtained towards each other by $\alpha / 2$ around the points $B$ and $C$, respectively, and obtain the lines of the missing two sides of the desired triangle, whose intersection point is $A$. In the thus constructed triangle $ABC$, the angle at $A$ is indeed $\alpha$, the opposite side has length $a$, and the angle bisector of $\alpha$ forms a $45^\circ$ angle with $a$.

Since $\angle ACD$ is part of $\angle BCD$, the problem is only solvable if $\alpha / 2 < 45^\circ$, or in other words, if $\alpha$ is an acute angle.