There are four collinear spots: $ A,B,C,D $ $ \bar{AB}=\bar{BC}=\frac{\bar{CD}}{4}=\sqrt{5} $ There are two circles; One which has $ \bar{AC} $ as a diameter, and the other having $ \bar{BD} $ as a diameter. Let's put $ \odot (AC) \cap \odot (BD) = E,F $ Let's put the area of $ EAFD $ $ S $ Find $ S^2 $.