对于$A$：因为$A_{1}D_{1}$∥$AD$∥$BC$，且$A_{1}D_{1}=AD=BC$，所以四边形$A_{1}D_{1}CB$是平行四边形，所以$A_{1}B$∥$CD_{1}$，又因为$CD_{1}$⊄平面$A_{1}C_{1}B$，$A_{1}B\subset $平面$A_{1}C_{1}B$，所以$CD_{1}$∥平面$A_{1}C_{1}B$，同理$D_{1}A$∥平面$A_{1}C_{1}B$，又$CD_{1}\cap D_{1}A=D_{1}$，$CD_{1}\subset $平面$ACD_{1}$，$D_{1}A\subset $平面$ACD_{1}$，所以平面$ACD_{1}$∥平面$A_{1}C_{1}B$，故$A$正确；对于$B$：易得$AB\bot $平面$ADD_{1}A_{1}$，所以$AB\bot A_{1}D$，又$A_{1}D\bot AD_{1}$，又$AB\cap AD_{1}=A$，所以$A_{1}D\bot $平面$ABD_{1}$，所以$A_{1}D\bot BD_{1}$，同理可证$C_{1}D\bot BD_{1}$，又$A_{1}D\cap C_{1}D=D$，所以$BD_{1}\bot $平面$A_{1}C_{1}D$，故$B$正确；对于$C$：由正方体$ABCD-A_{1}B_{1}C_{1}D_{1}$棱长为$a$，可得$AB_{1}=B_{1}C=AC=\sqrt{2}a$，所以$\triangle AB_{1}C$为等边三角形，所以$\angle AB_{1}C=60^{\circ}$，$AB_{1}$与$B_{1}C$成$60^{\circ}$角，故$C$正确；对于$D$：${V}_{A-BD{D}_{1}}={V}_{{D}_{1}-ABD}=\frac{1}{3}\times \frac{1}{2}\times AB\times AD\times DD_{1}=\frac{1}{6}a^{3}$，故$D$错误.故选：$D$.
标签：正方体,ABCD,棱长
