In triangle $ABC,$ points $P$ and $Q$ are on side $\overline{AB},$ and point $R$ is on side $\overline{AC}.$ If $\frac{AP}{2} = \frac{PQ}{5} = \frac{QB}{11}$ and $\frac{AR}{8} = \frac{RC}{13},$ then find $\frac{[QBC]}{[CRQ]}.$
B
(11/5)PQ
Q
P
(2/5)PQ
A R (13/8)AR C
[ QBC ] =
[ BAC ] - [ QAC ] =
(18/5)PQ * (21/8)AR - (7/5)PQ * (21/8)AR = [231/40 ] PQ * AR
[CRQ ] =
[QAC ] - [QAR ]
(7/5)PQ * (21/8)AR - (7/5)PQ * AR = [91/40 ] PQ*AR
[QBC ] / [CRQ ] = ( 231/ 40 ) / ( 91/40) = 231 / 91
+400 
