$(1)$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