869

已知 ${\displaystyle \mathrm{\Delta }ABC}$, ${\displaystyle D}$ 是 ${\displaystyle AB}$ 的中点，沿直线 ${\displaystyle CD}$ 将 ${\displaystyle \mathrm{\Delta }ACD}$ 折成 ${\displaystyle \mathrm{\Delta }{A}^{\prime }CD}$，所成二面角 ${\displaystyle A^{\prime }-CD-B}$ 的平面角为 ${\displaystyle \alpha }$, 则

• A. ${\displaystyle \mathrm{\angle }{A}^{\prime }DB\le \alpha }$
• B. ${\displaystyle \mathrm{\angle }{A}^{\prime }DB\ge \alpha }$
• C. ${\displaystyle \mathrm{\angle }{A}^{\prime }CB\le \alpha }$
• D. ${\displaystyle \mathrm{\angle }{A}^{\prime }CB\le \alpha }$

作 ${\displaystyle A^{\prime }{H}^{\prime },BH\perp DC}$ 于 ${\displaystyle H^{\prime },H}$，则 ${\displaystyle \mathrm{Rt}\mathrm{\Delta }{A}^{\prime }D{H}^{\prime }\cong \mathrm{Rt}\mathrm{\Delta }BDH}$.剥离出这个局部图形如图，并补全底部的矩形 ${\displaystyle H{H}^{\prime }EB}$.设 ${\displaystyle A^{\prime }{H}^{\prime }=H^{\prime }E=y,HD=H^{\prime }D=x,A^{\prime }E=t\le 2y}$，则由 ${\displaystyle BE\perp \text{平面}{A}^{\prime }{H}^{\prime }E}$ 有 ${\displaystyle A^{\prime }B=\sqrt{t^2+4x^2}}$.这样$$\begin{array}{rl}& \cos \mathrm{\angle }{A}^{\prime }{H}^{\prime }E-\cos \mathrm{\angle }{A}^{\prime }DB\\ =& \frac{2y^2-t^2}{2y^2}-\frac{2(x^2+y^2)-(4x^2+t^2)}{2(x^2+y^2)}=\frac{x^2(4y^2-t^2)}{2y^2(x^2+y^2)}\ge 0,\end{array}$$
因此 ${\displaystyle \mathrm{\angle }{A}^{\prime }{H}^{\prime }E=\alpha \le \mathrm{\angle }{A}^{\prime }DB}$, 故 A 错 B 对.

固定其它点不动，沿 ${\displaystyle CD}$ 方向移动 ${\displaystyle D}$ 使 ${\displaystyle CD\to \mathrm{\infty }}$，则 ${\displaystyle \mathrm{\angle }{A}^{\prime }CB\to 0<\alpha }$;使 ${\displaystyle CD\to 0}$，则 ${\displaystyle \mathrm{\angle }{A}^{\prime }CB\to A^{\prime }DB\ge \alpha }$，故 C,D 均不恒成立.

综上，答案选 B.