`a, d(I,(SBD))=1/2 d(C,(SBD))=1/2d(A,(SBD))`

Trong `(ABCD)`, kẻ `AF⊥BD`

Trong`(SAF)`, kẻ `AH⊥SF`

`BD⊥AF;BD⊥SA=>BD⊥AH`

`AH⊥SF;AH⊥BD=>AH⊥(SBD)`

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

Ta có:

`1/(AB^2)+1/(AD^2)=1/(AF^2)=>AF=(2 sqrt5)/5 a`

`1/(AF^2)+1/(AS^2)=1/(AH^2)=>AH=2/3 a`

=>d(I,(SBD))=1/2 . 2/3 a = 1/3 a`

`b,` Trong `(ABCD)`, kẻ `AM⊥BE`

Trong`(SAM)`, kẻ `AN⊥SM`

`BE⊥AM;BE⊥SA=>BE⊥AN`

`AN⊥SM;AN⊥BE=>AB⊥(SBE)`

`=>d(A,(SBE))=AN`

Ta có: `AB=a`

`BE= sqrt(CE^2+BC^2)=sqrt17/2 a

`AE=sqrt(AD^2+DE^2) = sqrt17/2a`

`=>S_(ABE)=a^2`

`=>AM =( 2 S_(ABE))/(BE)=(4sqrt17)/17`

`1(SA^2)+1/(AM^2)=1/(AN^2)=>AN=(4sqrt33)/33 a`

`=> d(A,(SBE))=(4sqrt33)/33 a`