`B'C'` // `BC`

`=>B'C'` // `(A'BC)`

`=>d(B'C',(A'BC))=d(B'(A'BC))=d(A,(A'BC))`

Gọi `D` là trung điểm `BC =>AD⊥BC;AD= sqrt3/2`

Trong `(A'AD)`, kẻ `AH⊥A'D`

`BC⊥AD;BC⊥A'A=>BC⊥(A'AD)=>BC⊥AH`

`AH⊥A'D;AH⊥BC=>AH⊥(A'BC)

`=>d(A,(A'BC))=AH=sqrt3/3`

Ta có:

`1/(A'A^2)+1/(AD^2)=1/(AH^2)`

`=> 1/(A'A^2)+1/( (sqrt3/2)^2)=1/((sqrt3/3)^2)`

`=>A'A=sqrt15/5`

V_(ABC.A'B'C') = A'A. S_(ABC) = sqrt15/5 . sqrt3/4 = (3sqrt5)/20`