Răspuns :
OM║AB si OM║CD⇒T.Thales
in ΔACD si ΔDAB urmatoarele rapoarte:
[tex]\frac{AO}{AC} =\frac{AM}{AD} =\frac{OM}{10} \\\\\frac{DM}{DA} =\frac{DO}{DB} =\frac{OM}{15} \\\\\frac{DM}{DA} =\frac{OM}{15} \\\\\frac{AD-DM}{DA} =\frac{15-OM}{15} \\\\\frac{AM}{DA} =\frac{15-OM}{15}\\\\dar\ \frac{AM}{AD} =\frac{OM}{10}\\\\\frac{OM}{10}=\frac{15-OM}{15}\\\\150-10OM=15OM\\\\150=25OM\\\\OM=6cm[/tex]
Răspuns:
MN-linie mijlocie
MO-jumatate din linia mijlocie
MN=(B+b)/2
MN=(15+10)/2
MN=25/2
MN=12,5
MO=12,5/2
MO=6,25 cm