$20 \cdot 68$ In a circle, the sides of an inscribed triangle are 20, 21, and 29. The circle is divided into four parts by the triangle. Let $A, B, C$ represent the three parts that are not the triangle, with $C$ being the largest. Which of the following relationships is true?
(A) $A+B=C$.
(B) $A+B+210=C$.
(C) $A^{2}+B^{2}=C^{2}$.
(D) $20 A+21 B=29 C$.
(E) $\frac{1}{A^{2}}+\frac{1}{B^{2}}=\frac{1}{C^{2}}$.
(50th American High School Mathematics Examination, 1999)
[Solution] Notice that $29^{2}=20^{2}+21^{2}$, so the triangle is a right triangle, and its hypotenuse is the diameter of the circumcircle.
Obviously, $A+B+S_{\triangle}=C$,
which means $A+B+\frac{1}{2} \times 20 \times 21=C$.
Simplifying, we get $A+B+210=C$.
Therefore, the answer is $(B)$.