## A nice problem

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### A nice problem

(From Geometry Revisited) $ABC$ is a triangle and let $G$ is its centroid.A line through $G$ meets $AB$ and $AC$ at point $P$ and $Q$, respectively.Prove that (with respect to sign) $\frac {BP} {PA} + \frac { CQ } {QA} =1$
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### Re: A nice problem

Note that $PA=BP+2PE \Rightarrow \frac{BP}{PA}+\frac{2PE}{PA}=1$

Now via menelaus we have $\frac{QA.CG.PE}{CQ.EG.PA}=1 \Rightarrow \frac{CQ}{QA}=\frac{2PE}{AP}$.
$\frac{1}{0}$