∵∠EAF=45°,∴∠C=360°-∠AEC-∠AFC-∠EAF=135°,∴∠B=∠D=180°-∠C=45°,∴AE=BE,AF=DF,设AE=x,则AF=2√2-x,根据勾股定理可得,AB=√（AE²+BE²)=√2X同理可得AD=√2(2√2-X)平行四边形ABCD的周长是2...