In triangle ABC, points P and Q are on side ¯AB, and point R is on side ¯AC. If AP2=PQ5=QB11 and AR8=RC13, then find [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 
