d( M, AB)=MA pt ca MA este ⊥(ABC)⇒MA⊥AB⊂(ABC)
MA=12 (ipoteza)
d (M, AD) = (analog)= MA=12
d(M, BC)=MB pt ca MB⊥BC pt ca
AM⊥( ABC), AB⊥BC⊂(ABC) Teorema celor 3 perpendiculare
deci MB=√(12²+16²)=20
analog MA⊥( ABC) DC⊂(ABC), AD⊥DC⇒(T3p)MD⊥DC,
d(M, DC)=MD
MD=√(12²+9²)=15
