Từ $M$ kẻ $MK//AC \, (K\in AC)$
$\Rightarrow \dfrac{MK}{AC} = \dfrac{BM}{BA} = \dfrac{3}{4}$
$\Rightarrow \dfrac{MK}{NC} = \dfrac{3}{2}$
Ta lại có: $\dfrac{PC}{PK} =\dfrac{NC}{MK}$
$\Rightarrow \dfrac{PC}{PK} = \dfrac{2}{3}$
$\Rightarrow \dfrac{PC}{KC} = \dfrac{2}{1} = 2$
$\Rightarrow PC = 2KC$ $(1)$
Mặt khác: $\dfrac{BK}{BC} = \dfrac{BM}{BA} = \dfrac{3}{4}$
$\Rightarrow \dfrac{BK}{KC} = \dfrac{3}{1} = 3$
$\Rightarrow BK = 3KC$
$\Rightarrow BC = 4KC$ $(2)$
$(1)(2) \Rightarrow PC = \dfrac{1}{2}BC$