1)
BM≡MN≡NP≡PC⇒BM+MN=NP+PC⇒BN≡NC
NN'||AB⇒NN' l.m in ΔADC
BD=AB/2 (CD mediana)=24/2=12
NP≡PC⇒NN' l,m in Δ ADC, NN'=12/2=6
PP' l.m inΔ ANN', PP'=NN'/2=6/2=3
DN'=(Thales) (BN/BC)*DC=(1/2)DC=36/2=18*
CM'=(Thales)=(CM/CB)*CD=(3/4)*CD=(3/4) *36=27
2)
Fie d⊥AM, A∈d
BN||AM||CP⇒BNCPtrapez (dreptunghic) si AM linie mijlocie (BM≡MC)⇒AN≡CP
altfel
BN||AM||CP⇒(fascicol de drepte paralele) AN/NP=BM/MC=1⇒AN=NP