3. Let $S$ be the intersection of the diagonals of the convex quadrilateral $A B C D$. If

$$
\measuredangle S A B=\measuredangle S B C=30^{\circ} \text { and } \measuredangle S C D=\measuredangle S D A=45^{\circ}
$$

determine the angle between the diagonals of the quadrilateral.
Solution. For definiteness, let us assume that $\measuredangle A S D=\varphi$ is acute, as shown in the right diagram. The case when this angle is not acute is considered analogously. This angle is external to triangles $S C D$ and $S A B$, so we easily obtain

$$
\varphi=\measuredangle C D A=\measuredangle A B C
$$

Thus, points $B$ and $D$ lie on the arcs of circles $k_{1}$ and $k_{2}$ with equal radii (say $r$) which are the geometric loci of points from which the segment $A C$ is viewed at an angle $\varphi$. Let the centers of these circles be denoted by $O_{1}$ and $O_{2}$, and the intersections of the segment

![](https://cdn.mathpix.com/cropped/2024_06_05_8091cd2ad09499b0556cg-05.jpg?height=497&width=387&top_left_y=612&top_left_x=1095)
$B D$ (different from $B$ and $D$) with circles $k_{1}$ and $k_{2}$ be $P$ and $Q$ respectively (see diagram).

Since $\measuredangle A Q D=\measuredangle A C D=45^{\circ}$ and $\measuredangle Q D A=45^{\circ}$, we conclude that triangle $A Q D$ is isosceles right-angled and the center $O_{2}$ of the circumscribed circle $k_{2}$ around it is the midpoint of the hypotenuse $D Q$. Furthermore, $\measuredangle B P C=\measuredangle B A C=30^{\circ}$ and $\measuredangle P B C=30^{\circ}$, so triangle $B C P$ is isosceles and $\measuredangle B C P=120^{\circ}$, which means $B C=C P=r$. Now, since $C P=C O_{2}$ and $O_{2} \in B D$, we have $P=O_{2}$. Furthermore, $\measuredangle C O_{2} A=120^{\circ}$ and $\measuredangle C D A=60^{\circ}$, so $\varphi=60^{\circ}$.