Show it parallel

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Show it parallel

Unread post by tanmoy » Thu Nov 06, 2014 5:25 pm

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,$P_{6}P_{1}$ $\parallel$ $AB$
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Re: Show it parallel

Unread post by SANZEED » Thu Nov 06, 2014 10:14 pm

In $\triangle ABC$, $D\in AB, E\in AC, DE\parallel BC\Rightarrow \dfrac{AD}{DB}=\dfrac{AE}{EB}$.
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