Xét \(\Delta ABC\) có \(DE//AC\left(gt\right)\)
\(\dfrac{AE}{AB}=\dfrac{CD}{BC}\) (định lý Ta lét)
Xét \(\Delta ABC\) có \(DF//AB\left(gt\right)\)
\(\dfrac{AF}{AC}=\dfrac{BD}{BC}\) (định lý Ta lét)
\(\dfrac{CD}{BC}+\dfrac{BD}{BC}=\dfrac{CD+BD}{BC}=\dfrac{BC}{BC}=1\\ \Rightarrow\dfrac{AE}{AB}+\dfrac{AF}{AC}=1\left(đpcm\right)\)
Xét ΔABC có
D∈BC(gt)
E∈AB(gt)
DE//AC(gt)
Do đó: \(\dfrac{AE}{AB}=\dfrac{DC}{CB}\)(Định lí Ta lét)
Xét ΔABC có
D∈BC(gt)
F∈AC(Gt)
DF//AB(gt)
Do đó: \(\dfrac{AF}{AC}=\dfrac{BD}{BC}\)(Định lí Ta lét)
Ta có: \(\dfrac{AE}{AB}+\dfrac{AF}{AC}\)
\(=\dfrac{CD}{BC}+\dfrac{BD}{BC}\)
\(=\dfrac{CD+BD}{BC}=\dfrac{BC}{BC}=1\)(đpcm)