Ratios of Line Segments in Parallelogram ABCD: Divide BC & AE

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In parallelogram ABCD, segment DC is extended to point E with a ratio of DE to EC being 3:2. The line AE intersects BC at point F, and the goal is to find the ratios in which F divides both BC and AE. The division-point theorem is mentioned as a potential method for solving the problem, but there is confusion regarding its application. There is uncertainty about the interpretation of the ratio, particularly regarding the relationship between segments ED and EC. Clarification on the use of external division and the correct approach to the problem is needed.
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Homework Statement


in the parallelogram ABCD, DC is extended to E so that DE:EC= 3:2. The line AE meets BC at F. Determine the ratios in which F divides BC and F divides AE

The Attempt at a Solution



so i figured i could use the division-point theorem

OP=[b/(a+b)]*OA +[a/(a+b)]OB

but i don't know how to use this equation in this case..i really don't know how to get started T-T
 
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I'm not sure I understand, ED < EC so how can the ratio be 3:2?? I am familiar with external division of a point but are you sure that is it?
 

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