Răspuns:
Explicație pas cu pas:
a) MA⊥(ABC), deci MA⊥AD
din ΔMAD, AD²=MD²-MA²=12²-6²=6²·2²-6²=6²·(2²-1)=6²·3, deci AD=√(6²·3)=6√3 cm
b) d(M,C)=MC. MA⊥(ABC), deci MA⊥AC.
ΔABC dreptunghic in B, deci AC²=AB²+BC²=12²+AD²=12²+6²·3=6²·2²+6²·3=6²·(2²+3)=6²·7.
Atunci MC²=MA²+AC²=6²+6²·7=6²·(1+7)=6²·8=6²·4·2, deci MC=√(6²·4·2)=6·2·√2=12√2cm= d(M,C).
c) m(∡(BC,MD))=??? Deoarece BC║AD, atunci ∡(BC,MD)=∡(AD,MD).
Deoarece in ΔMAD, MA=(1/2)·MD, ⇒∡(ADM)=30°=m(∡(AD,MD))=m(∡(BC,MD)).
