Iz slike imamo:
A E → = A B → + B E → \overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE} A E = A B + B E C M → = C D → + D M → \overrightarrow{CM}=\overrightarrow{CD}+\overrightarrow{DM} C M = C D + D M
Dovoljno je pokazati da je skalarni produkt vektora A E → \overrightarrow{AE} A E C M → \overrightarrow{CM} C M
A E → ⋅ C M → = ( A B → + B E → ) ⋅ ( C D → + D M → ) = A B → ⋅ C D → ⏟ = 0 + B E → ⋅ C D → + A B → ⋅ D M → + B E → ⋅ D M → ⏟ = 0 = C D → ⋅ ( B D → + D E → ) + D M → ⋅ ( 2 D B → ) ( uz 2 D M → = D E → ) = C D → ⋅ B D → ⏟ = 0 + C D → ⋅ D E → + D E → ⋅ D B → = D E → ⋅ ( C D → + D B → ) = D E → ⋅ C B → = 0. \begin{aligned}\overrightarrow{AE}\cdot\overrightarrow{CM}&=(\overrightarrow{AB}+\overrightarrow{BE})\cdot(\overrightarrow{CD}+\overrightarrow{DM})\\&=\underbrace{\overrightarrow{AB}\cdot\overrightarrow{CD}}_{=0}+\overrightarrow{BE}\cdot\overrightarrow{CD}+\overrightarrow{AB}\cdot\overrightarrow{DM}+\underbrace{\overrightarrow{BE}\cdot\overrightarrow{DM}}_{=0}\\&=\overrightarrow{CD}\cdot(\overrightarrow{BD}+\overrightarrow{DE})+\overrightarrow{DM}\cdot(2\overrightarrow{DB})\quad(\text{uz }2\overrightarrow{DM}=\overrightarrow{DE})\\&=\underbrace{\overrightarrow{CD}\cdot\overrightarrow{BD}}_{=0}+\overrightarrow{CD}\cdot\overrightarrow{DE}+\overrightarrow{DE}\cdot\overrightarrow{DB}\\&=\overrightarrow{DE}\cdot(\overrightarrow{CD}+\overrightarrow{DB})=\overrightarrow{DE}\cdot\overrightarrow{CB}=0.\end{aligned} A E ⋅ C M = ( A B + B E ) ⋅ ( C D + D M ) = = 0 A B ⋅ C D + B E ⋅ C D + A B ⋅ D M + = 0 B E ⋅ D M = C D ⋅ ( B D + D E ) + D M ⋅ ( 2 D B ) ( uz 2 D M = D E ) = = 0 C D ⋅ B D + C D ⋅ D E + D E ⋅ D B = D E ⋅ ( C D + D B ) = D E ⋅ C B = 0. 
