The value of (x-y) of quadrilateral inscribed in circle A quadrilateral that has consecutive sides of length $70,90,110,130$ is inscribed in a circle and also has a circle inscribed in it . The point of tangency of the inscribed circle to the side of length $130$ divides that side into segments of $x$ and $y$ . If $y\geq x$ then what is the value of $(y-x)$ 

Call the quadrilateral $ABCD$, with sides $AB=70, BC=90, CD=130, DA=110$, i.e. in the order 
$$\begin{array}{ccc}
A&\leftarrow70\rightarrow&B\\
\begin{array}{c}\uparrow\\110\\\downarrow\end{array}&O&\begin{array}{c}\uparrow\\90\\\downarrow\end{array}\\
D&E&C\\
&\leftarrow130\rightarrow&\\
\end{array}$$
For $ABCD$ to be cyclic, $\angle B + \angle D = 180^\circ$, so $\cos\angle B = -\cos\angle D$. Consider square of length of $AC$:
$$\begin{align*}
AC^2 = 130^2+110^2 -2\times130\times110\cos\angle D =& 70^2+90^2-2\times70\times90\cos\angle B\\
\cos\angle D =& \frac{16000}{41200}\\
=& \frac{40}{103}
\end{align*}$$
Similarly,
$$\begin{align*}
BD^2 = 110^2+70^2 -2\times110\times70\cos\angle A =& 130^2+90^2-2\times130\times90\cos\angle C\\
\cos\angle C =& \frac{8000}{38800}\\
=& \frac{20}{97}
\end{align*}$$
Let the centre of the inscribed circle be $O$, and the tangent point of the inscribed circle on $CD$ be $E$. The line $OC$ bisects $\angle BCD$, and the line $OD$ bisects $\angle CDA$. $OE$ is a line perpendicular with $CD$. Consider $\triangle OCD$,
$$OD\sin\angle ODE = OC\sin\angle OCE\\
OD\cos\angle ODE + OC\cos\angle OCE = 130$$
With half-angle formulae,
$$\sin\angle ODE = \sqrt{\frac{1-\cos\angle CDA}{2}}\\
\cos\angle ODE = \sqrt{\frac{1+\cos\angle CDA}{2}}\\
\sin\angle OCE = \sqrt{\frac{1-\cos\angle BCD}{2}}\\
\cos\angle OCE = \sqrt{\frac{1+\cos\angle BCD}{2}}$$
These should be enough to find the ratio $OD:OC$ and hence find out both of $DE$ and $CE$.
