[tex] \frac{BM}{AM} = \frac{3}{5} [/tex] si
AN=AC-NC=32-12=20 cm, deci:
[tex] \frac{CN}{AN} = \frac{12}{20} = \frac{3}{5} [/tex]
Observam ca [tex] \frac{BM}{AM} = \frac{CN}{AN} = \frac{3}{5} [/tex], deci MN determina pe AB si AC segmente proportionale, deci MN || BC si deci ΔAMN≈ΔABC, iar raportul ariilor celor doua triunghiuri este egal cu patratul raportului de asemanare, adica:
[tex] \frac{Aria AMN}{Aria ABC} = ( \frac{AN}{AC} )^{2} [/tex]
[tex] \frac{Aria AMN}{Aria ABC} = ( \frac{20}{32} )^{2} [/tex]
[tex] \frac{Aria AMN}{Aria ABC} = ( \frac{5}{8} )^{2} [/tex]
[tex] \frac{Aria AMN}{Aria ABC} = \frac{25}{64} [/tex]
