Răspuns :
I.p:{-fie ΔABC si M∈AB , N∈AC ⇔AM=2CM, AB=8CM si NC=7,5cm}
C.{AN=?}
Dem.{-folosind teorema lui Thales in ΔABC obtinem:
AM/AB=AN/AC -dar AM=2CM si AB=8CM si observam ca AC=AN+NC adica AC=AN+7,5cm si inlocuim: 2CM/8CM=AN/AN+7,5 ⇔1/4=AN/AN+7,5 ⇒ 4AN=AN+7,5 ⇒3AN=7,5 ⇒ AN=2,5cm}
C.{AN=?}
Dem.{-folosind teorema lui Thales in ΔABC obtinem:
AM/AB=AN/AC -dar AM=2CM si AB=8CM si observam ca AC=AN+NC adica AC=AN+7,5cm si inlocuim: 2CM/8CM=AN/AN+7,5 ⇔1/4=AN/AN+7,5 ⇒ 4AN=AN+7,5 ⇒3AN=7,5 ⇒ AN=2,5cm}
(1). MB=AB-AM=8cm-2cm=6cm;
daca MN || BC => conf.t.Thales ca
AM/MB=AN/NC adica:
2/6=AN/7,5 => AN=2*7,5:6=2,5cm ;
daca MN || BC => conf.t.Thales ca
AM/MB=AN/NC adica:
2/6=AN/7,5 => AN=2*7,5:6=2,5cm ;