Opposite angles of a cyclic quadrilateral are supplementary, because they are based on the same chord but on different sides. Because of this, their cosines are opposite.

Denote the angle BAD as a and apply the cosine law to the triangles BAD and BCD:

`AB^2 + AD^2 - 2 AB AD cos a = BD^2 , ` `BC^2 + CD^2 + 2 BC CD cos a = BD^2 .`

It is a linear system for `cos a ` and `BD^2. ` Let's eliminate `cos a ` by multiplying the first equation by `BC* CD , ` multiplying the second by `AB* AD ` and adding the results:

`BC* CD ( AB^2 + AD^2 ) + AB* AD ( BC^2 + CD^2 ) = ( BC* CD + AB* AD ) BD^2 .`

Find the second diagonal AC the same way:

`AD *CD ( AB^2 + BC^2 ) + AB* BC ( AD^2 + CD^2 ) = ( AD* CD + AB *BC ) AC^2 .`

Now we know

`BD^2 = ( 110 * 221 + 70 * 221 ) / ( 110 + 70 ) = 221 ,`

`AC^2 = ( 154 * 125 + 50 * 317 ) / ( 154 + 50 ) = 35100 / 204 = 5850 / 34 = 2925 / 17.`

So, `AC + BD = sqrt(221) + sqrt ( 25*117/17) = sqrt(13*17)+9 sqrt(13/17)=`

`=1/sqrt(13*17) (13*17+15*13) = 416 / sqrt(13*17).`

Finally, n = 416, p = 13, q = 17 and n + p + q = **446**.

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