a) -Xét △ABC có: AM, BN, CP lần lượt là ba đường phân giác (gt).
\(\Rightarrow\dfrac{MB}{MC}=\dfrac{AB}{AC};\dfrac{NC}{NA}=\dfrac{BC}{AB};\dfrac{PA}{PB}=\dfrac{AC}{BC}\) (định lí đường phân giác trong tam giác).
\(\Rightarrow\dfrac{MB}{MC}.\dfrac{NC}{NA}.\dfrac{PA}{PB}=\dfrac{AB}{AC}.\dfrac{BC}{AB}.\dfrac{AC}{BC}=1\)
b) Ta có:\(\dfrac{MB}{MC}=\dfrac{AB}{AC}\) (cmt)
\(\Rightarrow\dfrac{MB}{AB}=\dfrac{MC}{AC}=\dfrac{MB+MC}{AB+AC}=\dfrac{BC}{AB+AC}\)
\(\Rightarrow MC=\dfrac{BC.AC}{AB+AC}\)
-Tương tự: \(NC=\dfrac{BC.AC}{AB+BC}\) ; \(BP=\dfrac{BC.AB}{AC+BC}\)
-Xét △AMC có: CI là đường phân giác (gt)
\(\Rightarrow\dfrac{AI}{MI}=\dfrac{AC}{MC}\) (định lí đường phân giác trong tam giác)
\(\Rightarrow\dfrac{AI}{MI}+1=\dfrac{AC}{MC}+1\)
\(\Rightarrow\dfrac{MA}{MI}=\dfrac{AC}{\dfrac{AC.BC}{AB+AC}}+1\)
\(\Rightarrow\dfrac{MA}{MI}=\dfrac{1}{\dfrac{BC}{AB+AC}}+1\)
\(\Rightarrow\dfrac{MA}{MI}=\dfrac{AB+AC}{BC}+1=\dfrac{AB+AC+BC}{BC}\)
\(\Rightarrow\dfrac{MI}{MA}=\dfrac{BC}{AB+AC+BC}\)
-Tương tự: \(\dfrac{NI}{NB}=\dfrac{AC}{AB+AC+BC};\dfrac{PI}{PC}=\dfrac{AB}{AB+AC+BC}\)
\(\Rightarrow\dfrac{MI}{MA}+\dfrac{NI}{NB}+\dfrac{PI}{PC}=\dfrac{AB+AC+BC}{AB+AC+BC}=1\)