​ Geometry problem im stuck on.

[BQC]/[CRQ] = 4/15.

(BQC)/(CRQ) = 5/9  

Geometry problem

$\text{Let $RA=\dfrac{2}{5}AC $} \\ \text{Let $CR=\dfrac{3}{5}AC $} \\ \text{Let $AP=\dfrac{1}{4}AB $} \\ \text{Let $AQ=\dfrac{3}{4}AB $}$

$\begin{array}{|rcll|} \hline 2[ABC] &=& AB*AC*\sin(A) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 2[ACQ] &=& AC*\dfrac{3}{4}AB*\sin(A) \\ 2[ACQ] &=& \dfrac{3}{4}AB*AC*\sin(A) \\ 2[ACQ] &=& \dfrac{3}{4}*2[ABC] \\ \mathbf{2[ACQ]} &=& \mathbf{\dfrac{3}{2}[ABC]} \\ \hline \end{array} \begin{array}{|rcll|} \hline [BQC] &=& [ABC]-[ACQ] \quad | \quad * 2 \\ 2[BQC] &=& 2[ABC]-2[ACQ] \\ 2[BQC] &=& 2[ABC]-2[ACQ] \quad | \quad \mathbf{2[ACQ]=\dfrac{3}{2}[ABC]} \\ 2[BQC] &=& 2[ABC]- \dfrac{3}{2}[ABC] \quad | \quad : 2 \\ [BQC] &=& [ABC]- \dfrac{3}{4}[ABC] \\ \mathbf{[BQC]} &=& \mathbf{\dfrac{1}{4}[ABC]} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 2[ARQ] &=& \dfrac{2}{5}AC*\dfrac{3}{4}AB*\sin(A) \\ 2[ARQ] &=& \dfrac{3}{10}AB*AC*\sin(A) \\ 2[ARQ] &=& \dfrac{3}{10}*2[ABC] \\ \mathbf{2[ARQ]} &=& \mathbf{\dfrac{3}{5}[ABC]} \\ \hline \end{array} \begin{array}{|rcll|} \hline [CRQ] &=& [ACQ]-[ARQ] \quad | \quad * 2 \\ 2[CRQ] &=& 2[ACQ]-2[ARQ] \\ 2[CRQ] &=& 2[ACQ]-2[ARQ] \\ 2[CRQ] &=& \dfrac{3}{2}[ABC]- \dfrac{3}{5}[ABC] \quad | \quad : 2 \\ [CRQ] &=& \dfrac{3}{4}[ABC]- \dfrac{3}{10}[ABC] \\ [CRQ] &=& \left(\dfrac{3}{4}- \dfrac{3}{10}\right) [ABC] \\ [CRQ] &=& \left(\dfrac{30-12}{40}\right) [ABC] \\ [CRQ] &=& \left(\dfrac{18}{40}\right) [ABC] \\ \mathbf{[CRQ]} &=& \mathbf{\dfrac{9}{20}[ABC]} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \dfrac{[BQC]}{[CRQ]} &=& \dfrac{\dfrac{1}{4}[ABC]}{\dfrac{9}{20}[ABC]} \\\\ \dfrac{[BQC]}{[CRQ]} &=& \dfrac{\dfrac{1}{4}}{\dfrac{9}{20}} \\\\ \dfrac{[BQC]}{[CRQ]} &=& \dfrac{1}{4} *\dfrac{20}{9} \\\\ \mathbf{\dfrac{[BQC]}{[CRQ]}} &=& \mathbf{\dfrac{5}{9} } \\ \hline \end{array}$