20.89 The four sides of a quadrilateral are 70, 90, 130, and 110, respectively. It is inscribed in a circle and has an incircle. The points of tangency of the incircle divide the side of length 130 into segments of lengths \(x\) and \(y\). The value of \(|x-y|\) is
(A) 12.
(B) 13.
(C) 14.
(D) 15.
(E) 16.
(41st American High School Mathematics Examination, 1990)
[Solution 1] Given that $ABCD$ is inscribed in a circle, hence
$$
\angle A + \angle C = \angle B + \angle D = \pi.
$$

Thus, by the cosine rule, we have
$$
\begin{array}{c}
110^{2} + 130^{2} - 2 \cdot 110 \cdot 130 \cdot \cos A \\
= BD^{2} = 70^{2} + 90^{2} + 2 \cdot 70 \cdot 90 \cdot \cos A,
\end{array}
$$

and
$$
\begin{aligned}
& 90^{2} + 130^{2} - 2 \cdot 90 \cdot 130 \cdot \cos B \\
= & AC^{2} = 70^{2} + 110^{2} + 2 \cdot 70 \cdot 110 \cdot \cos B,
\end{aligned}
$$

yielding $\cos A = \frac{40}{103}, \quad \cos B = \frac{20}{97}$.
Since $ABCD$ has an inscribed circle, its center $O$ is the intersection of the angle bisectors of $\angle A$ and $\angle B$. If the circle $O$ touches $AB$ at $H$, then $OH \perp AB$.
Thus, if we set $AH = x, HB = y$, then
$$
x: y = \operatorname{ctg} \frac{A}{2}: \operatorname{ctg} \frac{B}{2} = \frac{1 + \cos A}{\sin A}: \frac{1 + \cos B}{\sin B},
$$

and
$$
\begin{aligned}
\frac{1 + \cos A}{\sin A}: \frac{1 + \cos B}{\sin B} & = \frac{1 + \cos A}{\sqrt{1 - \cos^2 A}}: \frac{1 + \cos B}{\sqrt{1 - \cos^2 B}} \\
& = 143: 117, \\
x: y & = 143: 117,
\end{aligned}
$$

thus $\quad|x - y| = 130 \cdot \frac{143 - 117}{143 + 117} = 13$.
Therefore, the answer is $(B)$.
[Solution 2] Let the center of the inscribed circle of the quadrilateral be $O$, and the radius be $r$. In quadrilateral $OHAE$ (as shown in the figure),
$S_{\text{OHAE}} = \frac{1}{2}\left(x^{2} + r^{2}\right) \sin A = x r$,
so $\quad \sin A = \frac{2 x r}{x^{2} + r^{2}}$.
Similarly, $\sin C = \frac{2(x-40) r}{(x-40)^{2} + r^{2}}$,
and since $\sin A = \sin C$,
we have $\frac{2 x r}{x^{2} + r^{2}} = \frac{2(x-40) r}{(x-40)^{2} + r^{2}}$,
or $r^{2} = x(x-40)$,
Similarly, $\cos B = \frac{2 y r}{y^{2} + r^{2}}, \quad \cos D = \frac{2(y-20) r}{(y-20)^{2} + r^{2}}$,

thus
$$
r^{2} = y(y-20).
$$

From (1) and (2), we have $x(x-40) = y(y-20)$, so
$$
\frac{x}{y} = \frac{y-20}{x-40} = \frac{110-x}{90-y} = \frac{x+(110-x)}{y+(90-y)} = \frac{11}{9}.
$$

Also, $\left\{\begin{array}{l}x + y = 130, \\ \frac{x}{y} = \frac{11}{9},\end{array}\right.$ solving gives $\left\{\begin{array}{l}x = 71.5, \\ y = 58.5,\end{array}\right.$
$$
\therefore \quad|x - y| = 13.
$$

Therefore, the answer is $(B)$.