23. In a parallelogram with area $S$, the bisectors of its internal angles are drawn. The area of the quadrilateral formed by their intersection is $Q$. Find the ratio of the lengths of the sides of the parallelogram.
23. Show that the bisectors of a parallelogram intersect to form a rectangle, the diagonals of which are parallel to the sides of the parallelogram and equal to the difference of the sides of the parallelogram. Consequently, if $a$ and $b$ are the sides of the parallelogram, $\alpha$ is the angle between them, then $S=a b \sin \alpha, Q=\frac{1}{2}(a-b)^{2} \sin \alpha, \frac{S}{Q}=$ $=\frac{2 a b}{(a-b)^{2}}$.

Answer: $\frac{S+Q+\sqrt{Q^{2}+2 Q S}}{S}$.