21. In the mechanism shown in the figure, $\theta$ is measured in radians, $C$ is the center of the circle, $BCD$ and $ACE$ are straight line segments, and $AB$ is the tangent to the circle at point $A$. Given $0<\theta<\frac{\pi}{2}$, the necessary condition for the areas of the two shaded regions to be equal is
(A) $\operatorname{tg} \theta=\theta$;
(B) $\operatorname{tg} \theta=2 \theta$,
(C) $\operatorname{tg} \theta=4 \theta$;
(D) $\operatorname{tg} 2 \theta=\theta$;
(E) $\operatorname{tg} \frac{\theta}{2}=\theta$
$\frac{21}{\mathrm{~B}}$

21. The area of the shaded sector is $\frac{\theta}{2}(A C)^{2}$, which must equal half the area of $\triangle A B C$, so it is true if and only if
$\frac{\theta}{2}(A C)^{2}=\frac{1}{4}(A C) \quad(A B)$, that is,
$2 \theta=\frac{A B}{A C}=\tan \theta$,
when the two shaded regions have the same area.