1. BM/MA=3/4 (BM+MA)/MA=(3+4)/4 AB/MA=7/4 MA/AB=4/7
MF//BC deci pot aplica Thales: MA/AB=FA/AC inlocuind: 4/7=FA/14 FA=8cm
atunci, FC=AC-FA=14-8=6cm
BM/MA=3/4 BM/(BM+MA)=3/(3+4) BM/AB=3/7
ME//AC conform Thales: BM/AB=BE/BC deci 3/7=BE/21 BE=9cm
si EC=BC-BE=21-9=12 cm
MECF=paralelogram P=2*(FC+EC)=2*(6+12)=36 cm
2. a) DE//AB din Thales: CD/BC=CE/AC
CD=BC-DB=12-9=3 cm si CE=AC-AE deci 3/12=(AC-AE)/AC AC=4*(AC-AE)
4*AE=3*AC
dar, AE+AC=28 rezolvand se obtine AE=12 cm si AC=16 cm
b) in triunghiul DCE avem AB//DE conform Thales: CB/CD=AC/EC
inlocuind 5/8=AC/16 AC=10cm AE=EC-AC=16-10=6cm