Trong `(SBD)`, kẻ `NH` // `MD`

`=>MD` // `(NHC)`

`=>d(MD,CN)=d(MD,(NHC))=d(D,(NHC))=d(S,(NHC))`

`SA=sqrt(SB^2 -AB^2) = 2a`

`V_(S,ABCD) = 1/3 . SA. AB.AD = 2/3 a^3`

`V_(S.BCD) = 1/2 V_(S.ABCD) = 1/3 a^3`

`NH` // `DM` ; `N` là trung điểm `SD` `=>` `HN` là đường trung bình trong `triangle SMD`

`=> H` là trung điểm `SM`

`(V_(S.HNC))/(V_(S.BDC)) = (SH)/(SB) . (SN)/(SD) = 1/4 . 1/2 = 1/8`

`=>V_(S.HNC) = 1/8 V_(S.BDC) = 1/24 a^3`

`CB⊥SB;CB⊥SA=>CB⊥SB`

`SC=sqrt(SB^2+BC^2) = a sqrt6`

`CD⊥SD;CD⊥SA=>CD⊥SD`

`SD=sqrt(SC^2 -CD^2) = a sqrt5`

`BD = AB sqrt2 = a sqrt2`

`SH = 1/4 SB = sqrt5/4 SB`

`Cos (hat(BSC)) = (BS^2 +SC^2 -CB^2)/(2.BS.SC)= sqrt30/6`

`HC=sqrt(SH^2 + SC^2 -2SH.SC.cos (hat(BSC)) = sqrt61/4a`

`CN = sqrt((CS^2+CD^2)/2 - (SD^2)/4)= 3/2 a`

`SN = 1/2 SD = sqrt5/2 a`

`DM = sqrt((SD^2 +BD^2)/2 - (SB^2)/4) = 3/2 a`

`HN = 1/2 DM = 3/4a`

`HN =  3/4a;HC=sqrt61/4a; CN= 3/2 a`

`=>S_(CNH) = sqrt((HN+HC+CN)/2((HN+HC+CN)/2 -HN)((HN+HC+CN)/2 -HC)((HN+HC+CN)/2 -CN)) = sqrt65/16a^2`

Ta có:

`=>V_(S.HNC) =1/3 . d_(S,(NHC)). S_(CNH) `

`=>1/24 a^3 = 1/3 . d_(S,(NHC)) . sqrt65/16a^2`

`=> d_(S,(NHC)) = 2sqrt65/65`

`=>d(MD,CN) =  2sqrt65/65a`