Austin draws \(\triangle ABC\) with \(AB=3\) cm, \(BC=4\) cm, and \(\angle ABC=90\degree\). Lachlan then draws \(\triangle DBF\) on top of \(\triangle ABC\) so that \(D\) lies on \(AB\), \(F\) lies on the extension of \(BC\), \(DB=2\) cm, and sides \(AC\) and \(DF\) meet at \(E\). If \(AE=3\) cm and \(EC=2\) cm, determine the length of \(CF\).