Giải thích các bước giải:

Qua $A$ kẻ $AD//BC, D\in MP$

$\to \dfrac{AD}{BP}=\dfrac{MA}{MB}=\dfrac12$

Ta có:

$CP=\dfrac15BC$

$\to BP=BC+CP=\dfrac65BC$

$\to \dfrac{AD}{\dfrac65BC}=\dfrac12$

$\to AD=\dfrac12\cdot\dfrac65BC$

$\to AD=\dfrac35BC$

$\to \dfrac{AD}{CP}=\dfrac35BC:(\dfrac15BC)=3$

$\to \dfrac{AD}{CP}=\dfrac{NA}{NC}$

Mà $\widehat{NAD}=\widehat{NCP}$ do $AD//BC$

$\to \Delta NAD\sim\Delta NCP(c.g.c)$

$\to \widehat{AND}=\widehat{PNC}$

$\to D, N, P$ thẳng hàng

$\to M, N, P$ thẳng hàng