Question

Teorrema lui menelaus

Answer 1

$Pentru~inceput~sa~denumim~notiunea~de~\underline{transversala}: este~o \\ dreapta~care~intersecteaza~toate~cele~trei~drepte~suport~ale \\ unui~triunghi. \\ \\ \underline{Teorema~lui~Menalaus~(Teorema~transversalei)} \\ \\ Fie~triunghiul~ABC~si~o~dreapta~care~intersecteaza~laturile~in \\ punctele~M,~N,~P.~(M \in BC,~N \in (AB),~P \in (AC)). \\ \\ Atunci:~ \frac{MC}{MB} \cdot \frac{NB}{NA} \cdot \frac{PA}{PC}=1 . \\ ------------------------------ \\ In~acest~caz~transversala~este~M-N-P.$

$Iti~voi~explica~si~teorema~lui~Ceva,~deoarece~este~o~teorema \\ asemanatoare~cu~cea~a~lui~Menelaus. \\ \\ \underline{Teorema~lui~Ceva} \\ \\ Fie~triunghiul~ABC~si~M \in (BC),~N \in (AC)~P \in (AB). ~Daca~AM, \\ BN~si~PC~sunt~concurente,~atunci~are~loc~relatia: \\ \\ \frac{AP}{PB} \cdot \frac{BM}{MC} \cdot \frac{CN}{NA}=1 . \\ ------------------------------ \\ Segmentele~[AM],~[BN]~si~[CN]~se~numesc~ceviene. \\ \\ Observatie:~Reciprocele~acestor~teoreme~sunt~valabile.$

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