*BM cắt CD tại E. Giả sử \(AB< CD\)
△ABM có: AB//DE.
\(\Rightarrow\dfrac{MA}{MD}=\dfrac{MB}{ME}=k\) (định lí Ta-let)
Mà \(\dfrac{MA}{MD}=\dfrac{NB}{NC}\).
\(\Rightarrow\dfrac{MB}{ME}=\dfrac{NB}{NC}\)
△BCE có: \(\dfrac{MB}{ME}=\dfrac{NB}{NC}\)
\(\Rightarrow\)MN//CE (định lí Ta-let đảo).
\(\Rightarrow\dfrac{CE}{MN}=\dfrac{BE}{BM}\) (định lí Ta-let).
.\(\Rightarrow\dfrac{CE}{MN}-1=\dfrac{BE}{BM}-1=\dfrac{ME}{MB}=\dfrac{1}{k}\)
\(\Rightarrow\dfrac{CE}{MN}=\dfrac{k+1}{k}\Rightarrow MN=\dfrac{k.CE}{k+1}\)
△ABM có: AB//DE.
\(\Rightarrow\dfrac{ED}{AB}=\dfrac{MD}{MA}=\dfrac{1}{k}\) (hq định lí Ta-let)
\(\Rightarrow DE=\dfrac{AB}{k}\)
\(MN=\dfrac{k.CE}{k+1}=\dfrac{k.\left(DE+DC\right)}{k+1}=\dfrac{k.DE+k.DC}{k+1}=\dfrac{k.\dfrac{AB}{k}+k.DC}{k+1}=\dfrac{AB+k.DC}{k+1}\)
