Hint: Find the angles of ABCD first then it's straight forward.

The radius of the inscribed circle must be r = 4.38034593.

i have another approach that avoids rigorous calculation. let the inverse point of $A,B,C$ and $D$ w.r.t center $O$ of the circle, be $A',B',C'$ and $D"$ respectively. then by simple angle chasing , we can see that $A'B'C'D'$ forms a parallelogram . since diagonals of a parallelogram bisect each other $\Longrightarrow$ midpoints of $A'C'$ and $B'D'$ concur at one point. (say at X) now by the inversion formula we can see by similarity and say that $OX=\frac{r^2}{OA.OC}.OM=\frac{r^2}{OB.OD}.ON$ thus the ratio turns out to be equal to $\frac{OM}{ON}=\frac{OA.OC}{OB.OD}=\frac{35}{48}$ i propose a tougher question than this , find the radius of the circle?

The radius of the circle was up there.

what is the generalized answer?

palashahuja wrote: l $OX=\frac{r^2}{OA.OC}.OM=\frac{r^2}{OB.OD}.ON$ thus the ratio turns out to be equal to $\frac{OM}{ON}=\frac{OA.OC}{OB.OD}=\frac{35}{48}$ How did you get this?

It's all about Newton's theorem.

palashahuja wrote: what is the generalized answer? General formula: see http://forumgeom.fau.edu/FG2010volume10/FG201005.pdf $OA=a, OB=b, OC=c, OD=d$ from the quadrilateral$ABCD$ and an inscribed circle, centre $O$. Calculate $ M = \frac{abc+bcd+cda+dab}{2}$; then \[r= 2 \cdot \sqrt{\frac{(M-abc)(M-bcd)(M-cda)(M-dab)}{abcd (ab+cd)(ac+bd)(ad+bc)}}\]

Oh... Van Straelen, Thank you for posting the formula. However, I did not use that formula to find the exact value for r. I developed my own formula which is very interesting. Once I get some free time, I will post it here. The way I solve the problem, I did not need to find the angles but rather I found their sine values, and from there I applied the Stewart's theorem to get the ratio.

This problem is almost the same as Russian 2005 MO http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=35309&p=3182086#p3182086

Vo Duc Dien, can you post your solution from last year? It sounds like an interesting approach.