Question 1149198
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There are undoubtedly many different paths to the solution; here is what I came up with.<br>
AB = AD = 4, so extend DC to H so that ABHD is a square.  Extend AF to H, completing diagonal AH of the square.  Note triangle FCH is a 45-45-90 right triangle.<br>
Triangles ABE and HCE are similar; because CH = 2 and AB = 4, the ratio of similarity is 1:2.  Then<br>
{{{EH/EA = 1/2}}}
{{{EH/AH = 1/3}}}
{{{AH = 4*sqrt(2)}}}
{{{EH = (1/3)AH = (4/3)*sqrt(2)}}}
{{{FH = 2*sqrt(2)}}}
{{{EF = FH-EH = (2/3)*sqrt(2)}}}<br>