a. Từ C kẻ đường thẳng song song với Ax và By, cắt AB tại I,

 => ACI = CAx 

 => ICB = CBy

=> ACB = ACI + ICB = xAC + CBy 

b. Từ C kẻ đường thẳng song song với Ax, cắt AB tại I,

=> ACI = CAx  

 ACB = xAC + CBy => ACI + ICB = xAC + CBy 

mà  ACI = CAx  nên ICB = CBy => CI // By

=> Ax //By

\(\frac{1}{2}\)\(+\)\(\left(\frac{16}{21}+\frac{27}{13}\right)\)\(-\)\(\left(\frac{14}{13}-\frac{5}{21}\right)\)

\(=\)\(\frac{1}{2}\)\(+\)\(\frac{16}{21}\)\(+\)\(\frac{27}{13}\)\(-\)\(\frac{14}{13}\)\(+\)\(\frac{5}{21}\)

\(=\)\(\frac{1}{2}\)\(+\)\(\left(\frac{16}{21}+\frac{5}{21}\right)\)\(+\)\(\left(\frac{27}{13}-\frac{14}{13}\right)\)

\(=\)\(\frac{1}{2}\)\(+\)\(\frac{21}{21}\)\(+\)\(\frac{13}{13}\)

\(=\)\(\frac{1}{2}\)\(+\)\(1\)\(+\)\(1\)

\(=\)\(\frac{5}{2}\)

\(\frac{7}{2}\)\(-\)\(\left(\frac{-1}{5}\right)\)\(-\)\(\frac{5}{6}\)\(+\)\(\frac{2}{3}\)\(+\)\(\left(\frac{-1}{5}\right)\)

\(=\)\(\frac{7}{2}\)\(+\)\(\frac{1}{5}\)\(-\)\(\frac{5}{6}\)\(+\)\(\frac{2}{3}\)\(-\)\(\frac{1}{5}\)

\(=\)\(\left(\frac{7}{12}-\frac{5}{6}+\frac{2}{3}\right)\)\(+\)\(\left(\frac{1}{5}-\frac{1}{5}\right)\)

\(=\)\(\frac{7}{2}\)\(-\)\(\frac{5}{6}\)\(+\)\(\frac{2}{3}\)\(+\)\(0\)

\(=\)\(\frac{5}{12}\)\(+\)\(0\)

\(=\)\(\frac{5}{12}\)