17.
如图,四边形ABCD是正方形,以B点为圆心,BD的长为半径画弧交BC延长线于点E,以DE为边作正方形DEFG,作FH⊥BE交BE的延长线于点H,连接AE、CG,则下列结论中正确的有( )
①∠CDE=22.5°;②S正方形DEFG-S正方形ABCD=FH2;③AE⊥CG;④DC2=CP•CG;⑤S△DCE:S△BCD=$\sqrt{2}$-1.
0 292345 292353 292359 292363 292369 292371 292375 292381 292383 292389 292395 292399 292401 292405 292411 292413 292419 292423 292425 292429 292431 292435 292437 292439 292440 292441 292443 292444 292445 292447 292449 292453 292455 292459 292461 292465 292471 292473 292479 292483 292485 292489 292495 292501 292503 292509 292513 292515 292521 292525 292531 292539 366461
①∠CDE=22.5°;②S正方形DEFG-S正方形ABCD=FH2;③AE⊥CG;④DC2=CP•CG;⑤S△DCE:S△BCD=$\sqrt{2}$-1.
| A. | ①②③④⑤ | B. | ①②③④ | C. | ①②③ | D. | ①②③⑤ |