18.10 The lengths of the sides $AB, BC, CD, DA$ of quadrilateral $ABCD$ are $1, 9, 8, 6$ respectively (as shown in the figure). For the following statements:
(1) Quadrilateral $ABCD$ is circumscribed about a circle;
(2) Quadrilateral $ABCD$ is not inscribed in a circle;
(3) The diagonals are not perpendicular;
(4) $\angle ADC \geqslant 90^{\circ}$;
(5) $\triangle BCD$ is an isosceles triangle.

The correct understanding is
(A) (1) true, (2) false, (4) true.
(B) (3) true, (4) false, (5) true.
(C) (3) true, (4) false, (3) false.
(D) (2) false, (3) false, (4) true.
(China Junior High School Mathematics League, 1986)
[Solution] If $\angle A D C=90^{\circ}$, then $A C=10$;
and if $\angle A D C>90^{\circ}$, then $A C>10$, thus $A C \geqslant 10$.
However, in $\triangle A B C$, $A C<A B+B C=10$, which is a contradiction. $\therefore$ (4) is false.
Also, since $1^{2}+8^{2} \neq 6^{2}+9^{2}$, we know that $A C$ and $B D$ are not perpendicular, $\therefore$ (3) is true.
At the same time, since $B D<1+6$, $\triangle B C D$ is not an isosceles triangle, i.e., (5) is false. Therefore, the answer is $(C)$.