Răspuns:
Explicație pas cu pas:
Notam {O}=AD∩CB
DO=x
CO=y
DC║AB⇒Thales ⇒[tex]\frac{x}{AD} =\frac{y}{CB}\\\\\frac{AD}{BC} =\frac{x}{y}[/tex]
[tex]\frac{DM}{AM} =\frac{CN}{BN}[/tex] facem proportii derivate
[tex]\frac{DM}{AD} =\frac{CN}{BC}\\\\frac{AD}{BC} =\frac{DM}{CN} =\frac{x}{y}[/tex]
[tex]\frac{AM}{BN}=\frac{DM}{CN} =\frac{x}{y}=k\\AM=xk=DM\\CN=yk=CN[/tex]
Demonstram ca este adevarata relatia
[tex]\frac{AM}{MA} =\frac{AN}{BN} \\\\\\\frac{x+DM}{AM} =\frac{y+CN}{BN}[/tex]
Inlocuim
[tex]\frac{x+xk}{xk} =\frac{y+yk}{yk} \\\frac{x(1+k)}{xk} =\frac{y(1+k)}{yk} \\\frac{1}{k} =\frac{1}{k}[/tex]⇒Thales ca MN║AB
