Ta có:
\(\frac{BC}{MD}=\frac{BD+DC}{MD}=\frac{BD}{MD}+\frac{DC}{MD}=\cot \widehat{B_1}+\cot\widehat{C_1}\)
Mà \(\widehat{B_1}=\widehat{A_1}; \widehat{C_1}=\widehat{A_2}\) (theo tính chất của tứ giác nội tiếp)
Do đó : \(\frac{BC}{MD}=\cot \widehat{A_1}+\cot \widehat{A_2}(*)\)
\(\frac{AC}{ME}=\frac{AE+EC}{ME}=\frac{AE}{ME}+\frac{EC}{ME}=\cot \widehat{A_1}+\cot\widehat{MCE}\)
\(\frac{AB}{MF}=\frac{AF-BF}{MF}=\frac{AF}{MF}-\frac{BF}{MF}=\cot \widehat{A_2}-\cot \widehat{FBM}\)
\(\Rightarrow \frac{AC}{ME}+\frac{AB}{MF}=\cot \widehat{A_1}+\widehat{A_2}+\cot \widehat{MCE}-\cot \widehat{FBM}\)
Mà \(\widehat{MCE}=\widehat{FBM}(=180^0-\widehat{MBA})\)
Do đó: \(\frac{AC}{ME}+\frac{AB}{MF}=\cot \widehat{A_1}+\cot \widehat{A_2}(**)\)
Từ $(*)$ và $(**)$ ta có đpcm.
