1) d(M, AB)=d(M, AC)=MA=12
2) d (M, DC)=MD 9T3p, vezi demo pt BC mai jos)
d(M,BC)=MB demo:
MA⊥(ABC)
BC⊂(ABC)
AB⊂(ABC)
AB⊥BC din ultimele 4 rel ⇒(T3p)MB⊥BC⇔ MB=d(M, BC)
MB=√(MA²+AB²)=√(12²+16²)=20
ΔMAB≡ΔMAD (CC)⇒MD=MB=20
Fie AC⊥BD ( in patrat diagonalele sunt ⊥)
fie AC∩BD={O}
MA⊥(ABC)
AO siBD⊂(ABC)
AO⊥BD
din celede mai sus⇒MO⊥BD⇔d( M, BD)=MO
MO=√(MA²+AO²)
AO=16√2/2=8√2
MO=√(12²+(8√2)²)=4√(3²+(2√2)²)=4√(9+8)=4√17
