Xét diện tích tam giác ABC:
\(S_{ABC}=\frac{AM.BC}{2}=\frac{CP.AB}{2}=\frac{BN.AC}{2}\)
=> \(AM.BC=CP.AB=BN.AC\)
=> \(AM=\frac{CP.AB}{BC}\); \(BN=\frac{CP.AB}{AC}\)
Theo gt, ta có:
\(BC+AM=AB+CP\)
\(\Leftrightarrow BC+\frac{CP.AB}{BC}=AB+CP\)
\(\frac{\Leftrightarrow CP.AB}{BC}-AB=CP-BC\)
\(\frac{\Leftrightarrow\left(CP.AB-AB.BC\right)}{BC}=\frac{\left(CP.BC-BC^2\right)}{BC}\)
\(\frac{\Leftrightarrow AB.\left(CP-BC\right)}{BC}=\frac{BC.\left(CP-BC\right)}{BC}\)
\(\Rightarrow AB=BC\)(1)
Theo gt, ta lại có:
\(AC+BN=AB+CP\)
\(\Leftrightarrow AC+\frac{AB.PC}{AC}=AB+CP\)
\(\frac{\Leftrightarrow AB.PC}{AC}-AB=PC-AC\)
\(\frac{\Leftrightarrow\left(AB.PC-AB.AC\right)}{AC}=\frac{\left(CP.AC-AC^2\right)}{AC}\)
\(\frac{\Leftrightarrow AB.\left(PC-AC\right)}{AC}=\frac{AC.\left(CP-AC\right)}{AC}\)
\(\Rightarrow AB=AC\)(2)
Từ (1) và (2) suy ra \(AB=BC=AC\)
=> ĐPCM
