$(1)证明:∵DF//AC,DE//BC$
$∴四边形ECFD是平行四边形$
$∵∠C=90°$
$∴四边形ECFD是矩形$
$(2)解:过点C作CH⊥EF 于点H,$
$则点C到EF 的距离即为CH的长$
$∵在Rt△ECF 中,CF=2,CE=4$
$∴EF= \sqrt {CE²+CF²} = \sqrt {20} =2 \sqrt {5}$
$∵S_{△ECF}= \frac {1}{2}\ \mathrm {CF} \cdot CE= \frac {1}{2}\ \mathrm {EF} \cdot CH$
$∴CH= \frac {CF \cdot CE}{EF} =\frac {2×4}{2\sqrt {5}} = \frac {4\sqrt {5}}{5}$
$∴点C到EF 的距离为 \frac {4\sqrt {5}}{5}$