Ken's Puzzle of the Week

A Set of Triangles

In rectangle ABCD, point E is on AB and point F is on AD. Angle CEF is a right angle.  Draw triangle CEF.  The rectangle is divided into four triangles (AEF, BEC, DCF, CEF).  All three sides of each triangle are integers, and CF is the longest line in the diagram.  What is the smallest possible length of CF?

Extension: What solutions exist for CF < 100?  In particular what solutions exist which aren't multiples of smaller solutions?
 
Source: Original.

Solutions were received from Philippe Fondanaiche, David Madfes, Joseph DeVincentis, Bernie R. Erickson, Alex Ang, Claudio Baiocchi, Alex Doskey, Anthony Nguyen. Alex Doskey found 11 solutions meeting the requirements (12 if we allow CF=100). He also found that the first solution which doesn't use a 3-4-5 based triangle has CF=221. 
(AE,AF,EF=H1),(BE,BC,CE=H2),(DF,CD,CF=H3),(H1,H2,H3)
1 =  (12,9,15),(12,16,20), (7,24,25),(15,20,25)
2 = (24,18,30),(24,32,40),(14,48,50),(30,40,50) = #1x2
3 = (15,20,25),(48,36,60),(16,63,65),(25,60,65)
4 = (20,15,25),(36,48,60),(33,56,65),(25,60,65)
5 = (48,20,52),(15,36,39),(16,63,65),(52,39,65)
6 = (36,15,39),(20,48,52),(33,56,65),(39,52,65)
7 = (36,27,45),(36,48,60),(21,72,75),(45,60,75) = #1x3
8 = (32,24,40),(45,60,75),(36,77,85),(40,75,85)
9 = (45,24,51),(32,60,68),(36,77,85),(51,68,85)
10= (24,32,40),(60,45,75),(13,84,85),(40,75,85)
11= (60,32,68),(24,45,51),(13,84,85),(68,51,85)
12= (48,36,60),(48,64,80),(28, 96,100),(80,80,100) = #1x4
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