$(1)$证明：在三棱柱$ABC-A_{1}B_{1}C_{1}$中，取$AB_{1}$的中点$F$，连结$DF$，$CF$，如图，
因$D$为边$A_{1}B_{1}$中点，则$DF$∥$AA_{1}$，且$DF=\frac{1}{2}AA_{1}$，又$AA_{1}$∥$CC_{1}$，$AA_{1}=CC_{1}$，
则$DF$∥$CE$，且$DF=\frac{1}{2}CC_{1}=CE$，
于是得四边形$CEDF$为平行四边形，即$DE$∥$FC$，因$A_{1}B_{1}\bot DE$，则$A_{1}B_{1}\bot CF$，
因平面$ACC_{1}A_{1}\bot $平面$ABB_{1}A_{1}$，平面$ACC_{1}A_{1}$⋂平面$ABB_{1}A_{1}=AA_{1}$，$AC\subset $平面$ACC_{1}A_{1}$，侧面$ACC_{1}A_{1}$为矩形，即$AC\bot AA_{1}$，
从而得$AC\bot $平面$ABB_{1}A_{1}$，又$A_{1}B_{1}\subset $平面$ABB_{1}A_{1}$，则$AC\bot A_{1}B_{1}$，又$AC\cap CF=C$，$AC$，$CF\subset $平面$AB_{1}C$，
所以$A_{1}B_{1}\bot $平面$AB_{1}C$；
$(2)$由（1）知：$DF$∥$CC_{1}$，而$DF$⊄平面$BCC_{1}B_{1}$，$CC_{1}\subset $平面$BCC_{1}B_{1}$，
则$DF$∥平面$BCC_{1}B_{1}$，
于是得$D$、$F$到平面$BCC_{1}B_{1}$的距离相等，设此距离为$d$，而$F$为$AB_{1}$的中点，
则点$A$到平面$BCC_{1}B_{1}$的距离为$2d$，
由（1）知：$AC\bot AB$，$AC\bot AB_{1}$，$A_{1}B_{1}\bot AB_{1}$，而$AB$∥$A_{1}B_{1}$，则$AB\bot AB_{1}$，
又$AC=1$，$AB=AB_{1}=2$，$V_{C-ABB_{1}}=\frac{1}{3}×\frac{1}{2}×AB×AB_{1}×AC=\frac{1}{6}×2×2×1=\frac{2}{3}$，
$BC=B_{1}C=\sqrt{5}$，$BB_{1}=2\sqrt{2}$，$S_{△BB_{1}C}=\frac{1}{2}BB_{1}×\sqrt{BC^{2}-(\frac{1}{2}BB_{1})^{2}}=\sqrt{2}×\sqrt{3}=\sqrt{6}$，
由$V_{A-BB_{1}C}=V_{C-ABB_{1}}$得$\frac{1}{3}S_{△BB_{1}C}×2d=\frac{2\sqrt{6}}{3}d=\frac{2}{3}$，解得：$d=\frac{\sqrt{6}}{6}$，
所以点$D$到平面$BCC_{1}B_{1}$的距离为$\frac{\sqrt{6}}{6}$.
标签：bot,侧面,ABC
