AB=a , CD=b cu a>b
MN linie mijlocie
demonstrați că Aria ABMN/Aria NMCD=(3a+b)/(a+3b)
demonstrație
MN=c=(a+b)/2
Aria ABMN=[(a+c)×h/2]/2=[²a+(a+b)/2]×h/4=
(2a+a+b)×h/8=(3a+b)×h/8
Aria NMCD=[(b+c)×h/2]/2=[²b+(b+a)/2]×h/4=
(2b+b+a)×h/8=(a+3b)×h/8
Aria ABMN/Aria NMCD=(3a+b)×(h/8)(a+3b)×(h/8)=
=(3a+b)/(a+3b)
[tex].[/tex]
