Xét ΔABC có
M∈BC(gt)
D∈AB(gt)
MD//AC(gt)
Do đó: \(\dfrac{AD}{AB}=\dfrac{CM}{CB}\)(Định lí Ta lét)
Xét ΔABC có
M∈BC(gt)
E∈AC(gt)
ME//AB(gt)
Do đó: \(\dfrac{AE}{AC}=\dfrac{BM}{BC}\)(Định lí Ta lét)
Ta có: \(\dfrac{AD}{AB}+\dfrac{AE}{AC}=\dfrac{MC}{BC}+\dfrac{MB}{BC}\)
\(\Leftrightarrow\dfrac{AD}{AB}+\dfrac{AE}{AC}=\dfrac{MB+MC}{BC}=\dfrac{BC}{BC}=1\)(đpcm)