Question

ad=dc=cb=6
ab=12
dc paralel ab

Answer 1

Aplicam teorema fundamentala a asemanarii:
Avand DC || AB => ΔMDC ~ ΔMAB
(Notez / = impartire ) => DM/AM = DC/AB = MC/MB
Luam primul caz : 
DM/AM = DC/AB => DM/AD+DM = 6/12
     (Pentru ca pe AM il putem scrie ca AD+DM) => DM/DM+6 = 6/12
=> 12DM = 6DM + 36
DM = 6 cm
Al doilea caz : MC/MC+6 = 6 /12 => MC = 6 cm (de asemenea)
=>ΔABM are AM=MB=AB (12 cm ; de ex. AM = MD+AD = 12)
ΔABM = echilateral