7.2. Given a quadrilateral $A B C D$. Its diagonals $A C$ and $B D$ intersect inside the quadrilateral. These diagonals divide the angles of the quadrilateral into two smaller parts, i.e., 8 angles are formed, two at each vertex of the quadrilateral. Can 3 of these 8 angles be obtuse?
Solution. Let the diagonals of the quadrilateral intersect at point $O$. At point $O$, two pairs of vertical angles meet. Two of these angles are not less than $90^{\circ}$, so even if the triangles they form are obtuse, the obtuse angle is not at the vertex of the quadrilateral. In the other two triangles, there are no more than two obtuse angles in total, so 3 out of 8 angles cannot be obtuse.

Answer. They cannot.