Draw a convex nondegenerate quadrilateral $ABCD$ and draw in the diagonals $AC,BD$. Write $O$ for the meeting point of the diagonals. There are twelve angles you could label in the picture using $A,B,C,D,O$ that aren't identically $\pi$.
Four relationships such as $\angle OAB +\angle ABO +\angle BOA = \pi $ hold by setting $A,B$ to be any two adjacent vertices. Two relationships of the form $\angle AOB = \angle COD$ hold by cyclically permuting $A,B,C,D$. The relationship $\angle AOB + \angle BOC = \pi$ also holds.
This gives 5 degrees of freedom among choosing these angles when there should be 4 (the space of all convex quadrilaterals up to similarity is 4-dimensional, and certainly all of these angles together determine a quadrilateral up to similarity). Where's the missing relationship? Is it easy to see what it ought to be? Is it a linear relationship, like the others?
Let $\,A_2 = \angle OAD\,$, $\,A_1 = \angle OAB\,$, $\,B_2= \angle OBA\,$, $\,B_1= \angle OBC\,$ etc, then by the law of sines:
$$ \quad\quad\dfrac{OC}{OD} = \dfrac{OC}{OB} \cdot \dfrac{OB}{OA} \cdot \dfrac{OA}{OD} \\[15px] \iff\quad \dfrac{\sin D_2}{\sin C_1} = \dfrac{\sin B_1}{\sin C_2} \cdot \dfrac{\sin A_1}{\sin B_2} \cdot \dfrac{\sin D_1}{\sin A_2} \\[20px] \iff\quad \sin A_1 \cdot \sin B_1 \cdot \sin C_1 \cdot \sin D_1 = \sin A_2 \cdot \sin B_2 \cdot \sin C_2 \cdot \sin D_2 $$