d(M,AB)=MA=12
d(M,AD)=MA=12
MB²=MA²+AB²=12²+16²=144+256=400 => MB=20
d(M,BC)=MB=20
MD²=MA²+AD²=12²+16²=144+256=400 => MD=20
d(M,CD)=MD=20
Fara sa calculam, puteam demonstra si ca MD=MB pentru ca ΔMAD≡ΔMAB (pentru ca sunt triunghiuri dreptunghice, si AD=AB, MA=latura comuna)
AC=AD√2=16√2
fie AC intersectat cu BD={O}
AO=AC/2=16√2/2=8√2
MO²=OA²+MA²=8²*2+12²=128+144=272=4√17
d(M, BD)=MO=4√17
