10.
(2018·曲靖)
如图,在正方形ABCD中,连接AC,以点A为圆心,适当长为半径画弧,交AB,AC于点M,N,分别以M,N为圆心,大于MN长的一半为半径画弧,两弧交于点H,连结AH并延长交BC于点E,再分别以A,E为圆心,以大于AE长的一半为半径画弧,两弧交于点P,Q,作直线PQ,分别交CD,AC,AB于点F,G,L,交CB的延长线于点K,连接GE,下列结论:①∠LKB=22.5°,②GE∥AB,③tan∠CGF=
![](//math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmfrac%3E%3Cmrow%3E%3Cmi%3EK%3C%2Fmi%3E%3Cmi%3EB%3C%2Fmi%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmi%3EL%3C%2Fmi%3E%3Cmi%3EB%3C%2Fmi%3E%3C%2Fmrow%3E%3C%2Fmfrac%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
,④S
△CGE:S
△CAB=1:4.其中正确的是( )
![](//tikupic.21cnjy.com/58/56/5856c3cad96191134b67d03bb0982909.png)
A . ①②③
B . ②③④
C . ①③④
D . ①②④