a: Xét ΔABC có AP là phân giác
nên \(\frac{PB}{PC}=\frac{AB}{AC}\)
=>\(\frac{PB}{6}=\frac48=\frac12\)
=>\(PB=\frac62=3\left(\operatorname{cm}\right)\)
BC=BP+CP=3+4=7(cm)
b: Xét ΔBFC có AP//FC
nên \(\frac{AP}{FC}=\frac{BP}{BC}\)
Xét ΔCBE có AP//BE
nên \(\frac{AP}{BE}=\frac{CP}{CB}\)
\(\frac{AP}{BE}+\frac{AP}{CF}=\frac{BP}{BC}+\frac{CP}{BC}=\frac{BP+CP}{BC}=\frac{BC}{BC}=1\)
=>\(AP\left(\frac{1}{BE}+\frac{1}{CF}\right)=1\)
=>\(\frac{1}{BE}+\frac{1}{CF}=\frac{1}{AP}\)