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In a $\triangle ABC,$ If $D,E,F$ are the mid points of $BC,CA,AB$ respectively and $P,Q,R$ are the

only points on $AD,BE,CF$ such that $\displaystyle \frac{AP}{AG} = \alpha$ and $\displaystyle \frac{BQ}{BG} = \beta$ and $\displaystyle \frac{CR}{CG}=\gamma$

then ratio of $\triangle PQR$ to $\triangle ABC$ is , where $\alpha,\beta,\gamma \in (0,1)$ and $G$ is centroid of $\triangle ABC$

could some help me with this, thanks

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    The [lines of fixed distance ratio](https://en.wikipedia.org/wiki/Circles_of_Apollonius#Apollonius.27_definition_of_a_circle) from two points are circles (unless of course the ratio is $1$). – 2017-01-07

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It is easy to show that if two triangles $ABC$ and $ABD$ share the same base $AB$, with line $CD$ crossing line $AB$ at $H$, then their areas are in the ratio $CH/DH$. You can immediately apply that in your case, to obtain the well-known result (here and in the following I'll write $ABC$ meaning the area of triangle $ABC$): $$ ABG=BCG=CAG={1\over3}ABC. $$

For the same reason we have $$ PGQ={PG\over AG}AGQ=(1-\alpha)AGQ \quad\hbox{and}\quad AGQ={QG\over BG}AGB=(1-\beta)AGB, $$ so that in the end we get: $PGQ=(1-\alpha)(1-\beta){1\over3}ABC$. In the same way we also obtain: $QGR=(1-\beta)(1-\gamma){1\over3}ABC$ and $RGP=(1-\gamma)(1-\alpha){1\over3}ABC$.

By summing these three equalities we finally get: $$ PQR=PGQ+QGR+RGP=[(1-\alpha)(1-\beta)+(1-\beta)(1-\gamma)+(1-\gamma)(1-\alpha)]{1\over3}ABC. $$