$解：過(guò)點(diǎn)?E?作?EM⊥AB，?垂足為點(diǎn)?M?$
$過(guò)點(diǎn)?G?作?GN⊥CD，?垂足為點(diǎn)?N，?如圖所示$
$由平行投影可知，?\frac {AM}{ME}=\frac {CN}{NG}?$
$∵?AB=10m，??MB=EF=2m?$
$∴?AM=AB-MB=8m?$
$∵?ME=BF=10m，??NG=DH=5m?$
$∴?\frac 8{10}=\frac {CN}5?$
$∴?CN=4m?$
$∵?GH=DN=3m?$
$∴?CD=CN+DN=7m?$
$∴電線桿的高度為?7?米$
