3. As shown in the figure, $AB$ is the diameter of a semicircle, $O$ is the center of the circle, $CD \perp AB$, $DE \perp OC$. If the lengths of $AD$, $BD$, and $CD$ are all rational numbers, then the statements:
甲. The length of $OE$ is a rational number.
乙. The length of $DE$ is a rational number.
丙. The lengths of all line segments marked with letters in the figure are rational numbers.
(A) Only 甲 is correct;
(B) Only 乙 is correct;
(C) Only 甲 and 乙 are correct:
(D) 甲, 乙, and 丙 are all correct.

Answer $(\quad)$
3. (D)

Solution Since the lengths of $A D$, $B D$, and $C D$ are all rational numbers,
$O A=O B=O C=\frac{A D+B D}{2}$ is a rational number.
Thus, $O D=O A-A D$ is a rational number.
Therefore, $O E=\frac{O D^{2}}{O C}$ and $D E=\frac{D C \cdot D O}{O C}$ are both rational numbers,
$$
C E=O C-O E, A B=A D+D B
$$

are also rational numbers. Therefore, statements 甲, 乙, and 丙 are all correct.
Solution Let $x$ be the negative root of the equation, then $-x=a x+1$, hence $x=$ $\frac{-1}{a+1}0, a>-1$, clearly when $a>-1$, the equation has a negative root.

Assume the equation has a positive root $x$, then $x=a x+1$, hence $\quad(1-a) x=$ $1, x=\frac{1}{1-a}>0$, so $a-1$ holds.
Moreover, $a<1$ does not hold,
thus $a \geqslant 1$.
This problem can also be solved using function graphs, by plotting the functions $y=|x|$ and $y=a x+1$ on the same Cartesian coordinate system.
$\because$ The equation $|x|=a x+1$ has one negative root and no positive roots.
$\therefore$ The line $y=a x+1$ intersects the broken line $y=|x|$ only at $x<0$, from the graph we know $a \geqslant 1$