`(SAB)⊥(ABCD),(SAC)⊥(ABCD),(SAB)∩(SAC)=SA=>SA⊥(ABCD)`

Ta có:

`d(M,(SBC))= 1/2d(D,(SBC))=1/2d(A,(SBC))`

Trong `(SAB)`, kẻ `AH⊥SB`

`BC⊥AB,BC⊥SA=>BC⊥(SAB)=>BC⊥AH`

`AH⊥BC,AH⊥SB=>AH⊥(SBC)`

`=>d(A,(SBC))=AH`

Ta có:

`1/(SA^2) +1/(AB^2) = 1/(AH^2)`

`=>1/(1^2) +1/(1^2) = 1/(AH^2)`

`=>AH = sqrt2/2`

`=>d(M,(SBC))= 1/2 . sqrt2/2 = sqrt2/4`