易证$ EF\perp AD,EF\perp BC $,过$ F $作$ D_1A_1\px DA $且$ D_1A_1=DA $,连接$ BD_1CA_1 $,易知\[ AF=\dfrac{3}{2}\\EF=1 \]有\[ A_1B=CD_1=\sqrt{3}\\A_1C=BD_1=\sqrt{2}\\CD_1\perp BD_1\\CA_1\perp BA_1 \]
$ PQMN $为截面,有\[ PQ\px AD\px MN\\MQ\px BC\px NP \]即$ PQMN $为平行四边形。在矩形$ A_1CD_1B $中令\[ CP_1=x \]则\[ CQ_1=\dfrac{\sqrt{3}}{\sqrt{2}}x \\S_{P_1Q_1M_1N_1}=\sqrt{6}-\dfrac{\sqrt{3}}{\sqrt{2}}x^2-(\sqrt{2}-x)(\sqrt{3}-\dfrac{\sqrt{3}}{\sqrt{2}}x)=\dfrac{\sqrt{6}}{2}-\sqrt{6}(x-\dfrac{\sqrt{2}}{2})^2\geqslant \dfrac{\sqrt{6}}{2}\] |