The $I$-altitudes of triangles $A I B$ and $C I D$ are both equal to the radius of $\omega$, hence have equal length. Therefore $\frac{[A I B]}{[C I D]}=\frac{A B}{C D}$. Also note that $[A I B]=I A \cdot I B \cdot \sin A I B$ and $[C I D]=I C \cdot I D \cdot \sin C I D$, but since lines $I A, I B, I C, I D$ bisect angles $\angle D A B, \angle A B C, \angle B C D, \angle C D A$ respectively we have that $\angle A I B+\angle C I D=\left(180^{\circ}-\angle I A B-\angle I B A\right)+\left(180^{\circ}-\angle I C D-\angle I D C\right)=180^{\circ}$. So, $\sin A I B=\sin C I D$. Therefore $\frac{[A I B]}{[C I D]}=\frac{I A \cdot I B}{I C \cdot I D}$. Hence $$\frac{A B}{C D}=\frac{I A \cdot I B}{I C \cdot I D}=\frac{35}{36}$$