Concurrent and Parallel Lines in Parallelogram

What Might This Be About?

25 May 2014, Created with GeoGebra

Problem

Through point $F$ on the diagonal $AC$ of parallelogram $ABCD$ draw to lines $EH$ and $GI$ where $E\in BC,$ $I\in CD,$ $H\in AD,$ $G\in AB.$

A configuration related to Bottema's theorem

Prove that $GH\parallel EI.$

Solution

There is a good deal of similar triangles, e.g., $\Delta AHF\sim\Delta CEH,$ $\Delta AGF\sim\Delta CIF.$ From these we derive proportions:

$\displaystyle\frac{AH}{EC}=\frac{AF}{CF}=\frac{AG}{CI}.$

Now, since also $\angle BAD=\angle BCD,$ $\Delta AGH\sim\Delta CIE,$ implying that one is obtained from the other with a homothety centered at $F$ and the coefficient $\displaystyle -\frac{AF}{CF}.$ This insures that $GH$ and $EI$ are indeed parallel.

Acknowledgment

This problem has been posted by Dao Thanh Oai (Vietnam) at the CutTheKnotMath facebook page.

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