[tex]Deoarece~CD~||~BP~si~unghiurile~\angle DMC~si ~ \angle PMB~sunt~opuse \\ \\ la~varf,~rezulta~ca~ \Delta DCM \sim \Delta PBM \Rightarrow \frac{CM}{BM}= \frac{CD}{BP}= \frac{AB}{BP}. \\ \\ Deoarece~CN~||~AB~si~unghiurile ~\angle NMC~si~ \angle AMB~sunt~opuse \\ \\ la~varf,~rezulta~ca ~ \Delta NMC \sim \Delta AMB \Rightarrow \frac{CM}{BM} = \frac{CN}{AB}. [/tex]
[tex]Din~ \frac{CM}{BM}= \frac{AB}{BP}~si~ \frac{CM}{BM}= \frac{CN}{AB}~rezulta~ca~ \frac{AB}{BP}= \frac{CN}{AB} \Rightarrow \\ \\ \Rightarrow \boxed{AB^2=BP \cdot CN}~.[/tex]
