$解:∵ BC//EF, AB⊥BC,CE⊥EF,∴ ∠ACB=∠CDE, ∠ABC=∠CED=90°$
$\ ∴△ABC∽△CED,∴\frac{AB}{CE}=\frac{BC}{ED}$
$∴ED=\frac{CE×BC}{AB}=\frac{0.8×2.4}{1.6}=1.2m$
$∵CE⊥EF,FH⊥ EF, ∴∠CED=∠HFD=90°$
$∵ ∠CDE=∠HDF, ∴△CED∽△HFD$
$∴\frac{CE}{HF}=\frac{DE}{DF},∴ DF=\frac{HF×DE}{CE}= \frac{4×1.2}{0.8}=6m$
$∴EF=ED+DF=1.2+6=7.2m$
$∴河的宽度EF为7.2m$
