$解:設(shè)AE=x，則BE=3x，BC=4x，AM=2x，CD = 4x,$
$∴EC=\sqrt{(3x)^{2}+(4x)^{2}}= 5x,$
$EM=\sqrt{x^{2}+(2x)^{2}}=\sqrt{5}x,$
$CM=\sqrt{(2x)^{2}+(4x)^{2}}= 2\sqrt{5}x,$
$∴EM^{2}+CM^{2}=CE^{2}，$
$∴△CEM是直角三角形,$
$∴tan∠ECM=\frac {EM}{CM}=\frac {\sqrt{5}x}{2\sqrt{5}x}=\frac {1}{2}，$
$sin∠ECM=\frac {EM}{CE}=\frac {\sqrt{5}x}{5x}=\frac {\sqrt{5}}{5}，$
$cos∠ECM=\frac {CM}{CE}=\frac {2\sqrt{5}x}{5x}=\frac {2\sqrt{5}}{5}.$
