IMC 2011 Q25

Let triangle \(CEF\) have area \(a\). Note that \(\angle AFD = \angle CFE\) (vertically opposite angles) and \(\angle DAF = \angle ECF\) (alternate angles), so triangles \(ADF\) and \(CEF\) are similar.
Note also that the side \(AD\) is twice the length of the corresponding side \(CE\).
Hence:
(i) triangle \(ADF\) has area 4\(a\) and;
(ii) \(AF\) has twice the length of the corresponding side \(CF\).
View \(AF\) and \(CF\) as bases of the triangles \(ADF\) and \(CDF\) (which then share the same height). Therefore, by (ii), triangle \(ADF\) has twice the area of triangle \(CDF\) (area P), which thus is 2\(a\).
The area of triangle \(ACD\) is 6\(a\); so that of triangle \(ABC\) is also 6\(a\) and that of area Q is 5\(a\). So the required ratio is 2 : 5.