, 2015

Solution:

The $I$-altitudes of triangles $AIB$ and $CID$ are both equal to the radius of $\omega$, hence have equal length. Therefore $\frac{[AIB]}{[CID]} = \frac{AB}{CD}$. Also note that $[AIB] = IA \cdot IB \cdot \sin AIB$ and $[CID] = IC \cdot ID \cdot \sin CID$, but since lines $IA, IB, IC, ID$ bisect angles $\angle DAB, \angle ABC, \angle BCD, \angle CDA$ respectively we have that $\angle AIB + \angle CID = (180^\circ - \angle IAB - \angle IBA) + (180^\circ - \angle ICD - \angle IDC) = 180^\circ$. So, $\sin AIB = \sin CID$. Therefore $\frac{[AIB]}{[CID]} = \frac{IA \cdot IB}{IC \cdot ID}$. Hence
$$\frac{AB}{CD} = \frac{IA \cdot IB}{IC \cdot ID} = \frac{5 \cdot 7}{4 \cdot 9} = \frac{35}{36}$$