A nice problem

[Phlembac Adib Hasan]

(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) \[ \frac {BP} {PA} + \frac { CQ } {QA} =1 \]

[Nadim Ul Abrar]

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

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

[Phlembac Adib Hasan]

My proof was also the same. This problem taught me Manelaus.