25~26高二上·青岛四区期中联考·第14题

$\frac{1}{2}$

平面四边形$ABCD$，$AB = BC = 2$，$CD = 1$，$AD = \sqrt{3}$，$\angle ADC = 90^\circ$，则$AC = 2$，
过$D$作$DE \perp AC$于$E$，取$AC$中点$O$，连接$BO$，则$BO \perp AC$，$BO = \sqrt{3}$，
$DE = \frac{AD \cdot CD}{AC} = \frac{\sqrt{3}}{2}$，$OE = CE = \frac{1}{2}$，过$O$作平面$ABC$的垂线$Oz$，则直线$OC, OB, Oz$两两垂直，
以$O$为原点，直线$OC, OB, Oz$分别为$x, y, z$轴建立空间直角坐标系.

设二面角$D’-AC-B$的大小为$\theta(0 \leq \theta \leq \pi)$，在$\triangle ACD$翻折过程中$DE$始终垂直于$AC$，
则$C(1,0,0)$，$B(0, \sqrt{3}, 0)$，$A(-1,0,0)$，$D’\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\cos\theta, \frac{\sqrt{3}}{2}\sin\theta\right)$，
$\overrightarrow{AC} = (2, 0, 0)$，$\overrightarrow{BD’} = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\cos\theta – \sqrt{3}, \frac{\sqrt{3}}{2}\sin\theta\right)$，设直线$AC$与直线$BD’$所成角为$\alpha$，
因此$\cos\alpha = \left|\cos \langle \overrightarrow{AC}, \overrightarrow{BD’} \rangle\right| = \frac{|\overrightarrow{AC} \cdot \overrightarrow{BD’}|}{|\overrightarrow{AC}||\overrightarrow{BD’}|}$$= \frac{1}{2\sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\cos\theta – \sqrt{3}\right)^2 + \left(\frac{\sqrt{3}}{2}\sin\theta\right)^2}}$
$= \frac{1}{2\sqrt{4 – 3\cos\theta}} \leq \frac{1}{2}$，当且仅当$\theta = 0$时取等号，
所以直线$AC$与直线$BD’$所成角的余弦值的最大值为$\frac{1}{2}$.

故答案为：$\frac{1}{2}$.