Explicație pas cu pas:
D ∈ BC, BD ≡ DC
M ∈ BC, ME || AD, E ∈ BA, ME ∩ AC = {F}
=> ΔACD ~ ΔFCM
[tex]\frac{AC}{FC} = \frac{DC}{MC} \iff \frac{AC}{AC - FC} = \frac{DC}{DC - MC} \\ \implies \frac{AC}{AF} = \frac{DC}{DM}\iff \frac{AC}{AF} = \frac{BC}{2DM}[/tex]
=> ΔEBM ~ ΔABD
[tex]\frac{EB}{AB} = \frac{BM}{BD} \iff \frac{EB - AB}{AB} = \frac{BM - BD}{BD} \\ \implies \frac{AE}{AB} = \frac{DM}{BD} \iff \frac{AE}{AB} = \frac{2DM}{BC}[/tex]
[tex]\frac{AC}{AF} = \frac{BC}{2DM}\iff \frac{AF}{AC} = \frac{2DM}{BC} \\ [/tex]
[tex]\frac{AE}{AB} = \frac{AF}{AC} \iff \frac{AE}{AF} = \frac{AB}{AC} \\ [/tex]
q.e.d.
