dau coroana va rog!!!!!
Răspunsuri la întrebare
Răspuns:
a. cum F e mijl. [BD] => [CF] mediana in ΔCBD => cf. teor. medianei in plan vectorial ca CF=CB+CD/2 ( rel. vectoriala)
b. cum M mijl. [EF] => [CM] mediana in Δ CEF => cf. teor. medianei in plan vectorial ca CM= CE+CF/2 ( rel. vectoriala)
c. CM= CB+CA+CD/4= 1/2* ( CB+CA+CD/2) ( rel. vectoriala)
<=> CM= 1/2* (CB+CD/2 +CA/2) ( am impartit relatia in 2) ( rel. vectoriala)
<=> CM= 1/2* (CF + CA/2) ( rel. vectoriala)
Cum E mijl [AC] => AE=EC=AC/2, in plan vectorial, CE=EA=CA/2 => CE=CA/2 ( rel. vectoriala)
==> CM=1/2*(CF + CE)=CE+CF/2 si este adv cf. b) => rel. initiala e adv. ( rel. vectoriala)
d) aplic succesiv teor. medianei in plan vectorial, tinand cont ca M e mijl. [EF].
avem [AM] mediana in ΔAEF => AM=AE+AF/2 ( rel. vectoriala)
[BM] mediana in ΔBEF => BM=BE+BF/2 ( rel. vectoriala)
[CM] mediana in ΔCEF => CM=CE+CF/2 ( rel. vectoriala)
[DM] mediana in ΔDEF => DM=DE+DF/2 ( rel. vectoriala)
cu [EF] baza a triunghiurilor.
AM+BM+CM+DM= ( rel. vectoriala)
= AE+AF/2 + BE+BF/2 + CE+CF/2 + DE+DF/2 ( rel. vectoriala)
= AE+AF+BE+BF+CE+CF+DE+DF/2 ( rel. vectoriala)
din desen, obs ca AE, CE v. opuși ( E mijl.[AC]) => AE=-CE => AE+CE=0
si BF,DF v.opuși (F mijl[BD]) => BF= -DF => BF+DF=0 ( rel. vectoriala)
= AF+BE+CF+DE/2 ( rel. vectoriala)
in continuare, [AF] e mediana in Δ ABD => AF= AD+AB/2 ( rel. vectoriala)
[ BE] mediana in ΔBAC => BE=BA+BC/2 ( rel. vectoriala)
[CF] mediana in ΔCBD => CF=CB+CD/2 (a) ) ( rel. vectoriala)
si [DE] mediana in ΔDAC => DE=DA+DC/2 ( rel. vectoriala)
=AD+AB/2 + BA+BC/2 + CD+CB/2 + DA+DC/2 ( rel. vectoriala)
= AD+AB+BA+BC+CD+CB+DA+DC/2 ( rel. vectoriala)
= AD+DA + AB+BA + BC+CB + CD+DC /2 ( rel. vectoriala)
= AA+ AA+ BB+ CC/2 ( rel. vectoriala)
=0/2 =0 ( rel. vectoriala)
