\(\)Xét tam giác ABC có
D\(\in BC\)
F\(\in AB\)
\(DE//AC\left(gt\right)\)
\(\Rightarrow\frac{AE}{AB}=\frac{CD}{CB}\)( định lí Talet)
Xét tam giác ABC có
D\(\in BC\)
F\(\in AC\)
\(DE//AB\left(gt\right)\)
\(\Rightarrow\frac{AF}{AC}=\frac{BD}{CB}\)( đt Talet)
Do đó \(\frac{AE}{AB}+\frac{AF}{CB}=\frac{CD}{CB}+\frac{BD}{CB}=1\)
Hay \(\frac{AE}{AB}+\frac{AF}{AC}=1\)