Reim's Similar Coins III
What Might This Be About?
Let two circles cross at points $A$ and $B;$ $E,F,E'F'$ are four points on one of the circles, $EA$ and $E'B$ meet the other circle at $H$ and $H',$ respectively; $FB,F'A$ meet it at $G$ and $G',$ respectively. Assume $EF\parallel E'F'.$
Then $GH\parallel EF\parallel E'F'\parallel G'H'.$
The problem can be solved directly, as suggested by the diagram below.
But, in fact, it is a direct consequence of Reim's Similar Coins theorem.