5. The sides $A B, B C, C D, D A$ of quadrilateral $A B C D$ are 1, 9, 8, 6 respectively (as shown in the figure). For the following statements:
(1) Quadrilateral $A B C D$ is circumscribed about a circle;
(2) Quadrilateral $A B C D$ is not inscribed in a circle;
(3) The diagonals are not perpendicular to each other;
(4) $\angle A D C \geqslant 90^{\circ}$;
(5) $\triangle B C D$ 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, (5) false;
(D) (2) false, (3) false, (4) true.
5. (C)

Solution: Given $\angle A D C=90^{\circ}$, we have $A C=10$.
Also, $\angle A D C>90^{\circ}$, so $A C>10$.
From the above, we get $A C \geqslant 10$.
However, in $\triangle A B C$, $A C<A B+B C=10$, which is a contradiction!
$\therefore$ (4) is false.
$\because 1^{2}+8^{2} \neq 6^{2}+9^{2}$, hence $A C, B D$ are not perpendicular,
$\therefore$ (3) is true.
$\because B D<1+6$,
$\therefore \triangle B C D$ is not an isosceles triangle, so (5) is false.
Since only one of (A), (B), (C), (D) is correct,
$\therefore$ the correct choice is (C).