`AB` // `CD`

`=>AB` // `(SCD)`

`+>d(AB,SC)=d(AB,(SCD))=d(A,(SCD))=2d(O,(SCD))`

Trong `(ABCD)`, kẻ `OE⊥CD`

Trong `(SOE)` , kẻ `OH⊥SE`

`CD⊥OE;CD⊥SO=>CD⊥(SOE)=>CD⊥OH`

`OH⊥CD;OH⊥SE=>OH⊥(SCD)`

`=>d(O,(SCD))=OH`

`OE` là đường trung bình trong `triangle ACD=>OE= 1/2 AD= a/2`

Ta có:

`1/(OE^2)+1/(SO^2)=1/(OH^2)`

`=>1/((a/2)^2)+1/(a^2)=1/(OH^2)`

`=>OH=sqrt5/5 a`

`=>d(SC,AB) = (2sqrt5)/5 a`