loik/6=324/6=fie A,B,C,D masurile unghiurilor
atunci
A+B+C+D=360
(A.B,C,D) i.p. (2,3,6,9)
A/(1/2)=B/(1/3)=C/(1/6) =D/(1/9)=k
2A=3B=6C=9D=k
A=k/2
B=k/3
C=k/6
D=k/9
k(1/2+1/3+1/6+1/9)=360
k(9+6+3+2)/18=360
20k/18=360
2k/18=36
k/18=18
k=324
A=k/2=324/2=162°
B=k/3=324/3=108°
C=324/6=54°
D=324/9=36°
analog, caz b)
A=2B=3C=6D=k
A=k
B=k/2
C=k/3
D=k/6
k(1+1/2+1/3+1/6)=360°
k*(6+3+2+1)/6=360
12k/6=360
2k=360
k=180
A=180 DATE GRESITE!!!!!!!!!!!!!!!!!!!
B=90
C=60
D=30
numerele 180;90;60 si 30 verifica acele conditii , dar Nu pot fi unghiurile unui patrulater convex..pt ca A=180°, unghi alungit
180*1=90*2=60*3=30*6 adica (180,90,60,30) sunt i.p. cu (1;2,3,6) si au ca suma 360 dar Nu pot fi masurile unghiurilor unui patrulater, datorita valorii lui A
de fapt exista triunghiul BCD
