+273 In triangle ABC, \(AP = \frac{PQ}{2} = BQ\) and \(\frac{CR}{RA} = \frac{3}{2}\). Find \(\frac{[QBC]}{[CRQ]}.\)
+9673 Let AR = 2k, RC = 3k, AP = p, PQ = 2p, QB = p.
Then
\(\begin{array}{rcl} [QBC] &=& [ABC] - [ACQ]\\ &=& \dfrac12 (5k)(4p) \sin \angle A - \dfrac12 (5k)(3p) \sin \angle A\\ &=& \dfrac{5kp}2 \sin \angle A\\ [CRQ] &=& [ACQ] - [AQR]\\ &=& \dfrac12 (5k)(3p)\sin \angle A - \dfrac12 (2k)(3p)\sin \angle A\\ &=& \dfrac{9kp}2\sin \angle A \end{array}\)
Therefore,
\(\dfrac{[QBC]}{[CRQ]} = \dfrac{\dfrac{5kp}2\sin \angle A}{\dfrac{9kp}2\sin \angle A} = \dfrac5{9}\)
