Posted by urgently needed!!! on Wednesday, February 27, 2013 at 1:56am.
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.
23
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