a)AM⊥(ABC) ipoteza⇒AM⊥DB⊂(ABC),DB⊥AM (1)
DB⊥AC (diagonale in patrat) (2)
din (1) si (2)⇒BD⊥(AMC)⇒BD⊥BN⊂(AMNC), cerinta a)
b)AM=9( ipoteza) AO=12√2/2=6√2
NC=8
OC=AO=6√2
ΔMAO si ΔOCN tr dr
MA/OC=9/6√2=3/2√2=3√2/4
AO/CN=6√2/8=3√2/4=MA/OC
caz de asemanare LUL ( ∡MAO≡∡OCN=90°)
⇒ΔMAO≈ΔOCN
verrificare MO=√(MA²+AO²)=3√17
ON=√(CN²+OC²)=2√34
OM/NO=3√17/2√34=3/2√2=3√2/4
MB=MD=√AB²+AM²=√AD²+AM² cum AB≡AD⇒MB≡ΔMDB isoscel
OD≡OB⇒OM mediana, OM inaltime OM⊥BD (3)
OM=3√17
ON=2√34
fie NN'⊥AM, N'∈AM⇒MN'=AC=12√2
si MN'=9-8=1
MN=√(MN'²+N'N²)= √((12√2)²+1)
MN²=144*2+1=289
OM²+ON²=9*17+8*17=17*17=289
deci
MN²=OM²+ON²⇒( Rec Teo Pitagora) OM⊥ON (4)
OM⊥BD (3)
OM⊥ON (4)
din (3) si (4)⇒OM⊥(BND)⇒OM⊥BN⊂(BND), cerinta c)
