A point inside triangle $ABC$. Point $P$ inside $\triangle ABC$. Suppose $\triangle PAB : \triangle PBC : \triangle PCA = 2 : 3 : 5$. $BP$ intersect $AC$ at $Q$.
Find $AQ : QC$ 
Can you give me the solution to this or some hint? Because i tried to find $AC$ and $BQ$ using $\triangle PAB : \triangle PBC : \triangle PCA = 2 : 3 : 5$. But it gives many equations.
 A: Consider this figure.

If the area of $\triangle BPA:\triangle BPC = 2:3$ then the heights are in the are in the same proportion as they have the same base.
Extending $BP$ to $Q,$ $\triangle PQA$ has the same height as $\triangle BPA$
We can say the same thing about the heights of  $\triangle PQC$ and $\triangle BPC$
$\triangle PQA:\triangle PQC = 2:3$
Which means that $Q$ splits $AC$ in a ratio of $2:3$
That we have been given that $\triangle PAC:\triangle PCB:\triangle PBA = 5:3:2$ is additional information that is not entirely necessary.  But it does imply that $BP:PQ = 1:1$
A: Let there be a triangles $\Delta,\Delta_1,\Delta_2,$ such that $\Delta_1+\Delta_2=\Delta$. Let deltas also deonote their areas. if $b_1$ and $b_2$ are the respective bases small triangles then $\frac{\Delta_1}{\Delta_2}=\frac{b1}{b_2}$
Let the area of $\Delta PAQ=x, \Delta PQC=y$ Then
$x+y=5k$ Next $\frac{BP}{PQ}=\frac{2}{x}=\frac{3}{y}.$
This means $x=2k, y=3k$, finally $\frac{AQ}{QC}=\frac{x}{y}=\frac{2}{3}$
A: $$\frac{AQ}{QC}=\frac{S_{\Delta PAQ}}{S_{\Delta PQC}}=\frac{\frac{S_{\Delta PAQ}}{S_{\Delta ABP}}}{\frac{S_{\Delta PQC}}{S_{\Delta CBP}}}\cdot\frac{S_{\Delta ABP}}{S_{\Delta CBP}}=\frac{\frac{QP}{BP}}{\frac{QP}{BP}}\cdot\frac{2}{3}=\frac{2}{3}.$$
