3.
如图,△ABC和△ADE均为等腰直角三角形,∠BAC=∠DAE=90°,点B,D,E在同一直线上,AG是∠DAE的平分线,分别交DE,BC于点F,G,连接CE,∠GAC=25°,所给结论:①∠BAD=∠CAE;②tan∠ABE=$\frac{\sqrt{3}}{3}$;③AG∥CE;④2AF+CE=BE;⑤AD=CG中,正确的有( )
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| A. | ①③⑤ | B. | ②③④ | C. | ①②④ | D. | ①③④ |