Problem 8.5. In a convex quadrilateral $A B C D$, the bisectors of angles $A$ and $C$ are parallel, and the bisectors of angles $B$ and $D$ intersect at an angle of $46^{\circ}$, as shown in the figure. How many degrees does the acute angle between the bisectors of angles $A$ and $B$ measure?

![](https://cdn.mathpix.com/cropped/2024_05_06_191a5994d6b917032c66g-03.jpg?height=490&width=616&top_left_y=94&top_left_x=419)
Answer: 67.

Solution. Let's mark the points of intersection of the angle bisectors $K, L, M, N$ (see Fig. 2). Additionally, let $\angle A=2 \alpha, \angle B=2 \beta, \angle C=2 \gamma, \angle D=2 \delta$. Since the sum of the angles of quadrilateral

![](https://cdn.mathpix.com/cropped/2024_05_06_191a5994d6b917032c66g-03.jpg?height=663&width=713&top_left_y=807&top_left_x=368)

Fig. 2: to the solution of problem 8.5

$ABCD$ is $360^{\circ}$, we have:

$$
\begin{gathered}
2 \alpha+2 \beta+2 \gamma+2 \delta=360^{\circ} \\
\alpha+\beta+\gamma+\delta=180^{\circ}
\end{gathered}
$$

Consider triangle $KMN$. In it:

- $\angle MKN=46^{\circ}$,
- $\angle KMN=\angle ALM=\alpha+\beta$, since $\angle ALM$ is the exterior angle of triangle $ABL$ (this is the angle we need to find in the problem),
- $\angle KNM=\angle CND=180^{\circ}-\gamma-\delta=\alpha+\beta$.

Therefore, triangle $KMN$ is isosceles, and $\angle KMN=\frac{180^{\circ}-46^{\circ}}{2}=67^{\circ}$.

Another solution. Fix angle $A$ and translate angle $C$ parallel so that vertex $C$ lies on the angle bisector of angle $A$. (Or more formally, mark a point $C'$ on the angle bisector of angle $A$ and draw rays from it in the same direction as rays $CB$ and $CD$; the intersections of these rays with rays $AB$ and $AD$ are denoted by $B'$ and $D'$, respectively, as shown in Fig. 3.)

![](https://cdn.mathpix.com/cropped/2024_05_06_191a5994d6b917032c66g-04.jpg?height=662&width=703&top_left_y=547&top_left_x=369)

Fig. 3: to the solution of problem 8.5

The sides of the new quadrilateral $AB'C'D'$ are parallel to the sides of the original quadrilateral; thus, the angles between these sides are the same. Consequently, the angle bisectors of the new quadrilateral are parallel to the corresponding angle bisectors of the original, and the angles between them are also preserved.

But this means that the angle bisectors of angles $A$ and $C'$ coincide, so the entire new picture is symmetric with respect to the line $AC'$ (from the congruence of triangles $AB'C'$ and $AD'C'$ by the common side and the adjacent angles).

From the symmetry, it follows that the other two angle bisectors intersect on the line $AC'$ and form equal angles with it. Then the required angle, after doubling, will complement $46^{\circ}$ to a straight angle. Therefore, it is equal to $\frac{1}{2}\left(180^{\circ}-46^{\circ}\right)=67^{\circ}$.