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ABCD is a parallelogram. E is a point on AB such that 234×AE=EB. Let DE intersect AC at F. What is the ratio AC:AF?

Given:
Let r=AE:EB
Draw a diagram of the parallelogram ABCD.
Insert E on AB.
Extend DE to join CA at F.
(The above is as given by the question).
Examine the triangles FAE and DBE.
Show that they are similar noting that FA is parallel to DB.
Therefore the ration FE:ED = AE:EB
Similarly, show that triangles FAE and FCD are similar, noting again that AE is parallel to CD.
Hence show that
FE:AC = FE:ED
or equivalently,
AC:AF = ED:FD. 
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