Take any point $P_{1}$ on the side $BC$ of a triangle $ABC$ and draw the following chain of lines:
$P_{1}P_{2}\parallel AC$,$P_{2}P_{3}\parallel BC$,$P_{3}P_{4}\parallel AB$,$P_{4}P_{5}\parallel CA$,$P_{5}P_{6}\parallel BC$.Here,$P_{2},P_{5}$ lie on $AB$;$P_{3},P_{6}$ lie on $CA$ and $P_{4}$ on $BC$.Show that,$P_{6}P_{1}$ $\parallel$ $AB$
Show it parallel
"Questions we can't answer are far better than answers we can't question"
Re: Show it parallel
Hint:
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$