AA'⊥BC
MM'⊥BC
NN'⊥BC
PP'⊥BC⇒ AA'║MM'║NN'║PP'
AC=4x
AN=2x
AP=3x
MC=3x
- Aplicam Teorema lui Thales in ΔACA'
[tex]\frac{AM}{MC} =\frac{A'M'}{M'C'}[/tex]
[tex]\frac{x}{3x} =\frac{A'M'}{M'C'}=k\\\\\frac{1}{3}=\frac{A'M'}{M'C'} =k[/tex]
A'M'=k
M'C'=3k
Daca A'M'=k⇒M'N'=N'P'=P'C'=k
Deci CN'=2k
M'P'=2k
A'C=4k
- Formam rapoartele cerute si obtinem:
[tex]\frac{A'M'}{M'N'}=\frac{k}{k} =1\\\\\\\frac{P'C}{P'N'} =\frac{k}{k} =1\\\\\\\frac{A'N'}{CN'} =\frac{2k}{2k} =1\\\\\\\frac{M'N'}{M'C}=\frac{k}{3k} =\frac{1}{3} \\\\\\\frac{A'C}{M'P'} =\frac{4k}{2k} =2[/tex]