解：设$AH、$$DE$分别与$BC$交于点$M、$$N，$分别与$GF $交于点$Q，$$P$
$∵△AMB$的内角和为$180°$
$∴∠A+∠B+∠AMB= 180° $
$∵∠AMN + ∠AMB =180°$
$∴∠AMN =∠A+∠B$
同理可得$∠DNM =∠C+∠D，$$∠NPF =∠E+∠F，$$∠PQH =∠G+ ∠H$
$∴∠A +∠B+ ∠C+ ∠D+ ∠E+∠F+∠G+ ∠H$
$=∠AMN+∠DNM+ ∠NPF+ ∠PQH$
∵四边形$ MNPQ $的外角和为$360°$
$∴∠A + ∠B+∠C+∠D+∠E+∠F+ ∠G+ ∠H=360°$
