Question

如图,正方形ABCD的边长为4,点M在BC上,BM=1.过点A作AO⊥DM,垂足为O,AO交DC于N.

Answer 1

(1)由图∵AO⊥DM ∴∠MDC+∠DNA=90°,∵正方形ABCD,即∠MDC+∠DMC=90°∴∠DMC=∠DNA 在正方形ABCD 中,AB=DC,∠DCM＝∠ADN=90° ∴⊿ADN≌⊿DCM（角角边）∴AN=DM

(2)∵AB=DC=4,BM=1∴CM=3,在直角⊿DON和直角⊿DMC中,∠MDC=∠NDO,故∠DNO=∠DMC；在直角⊿AND和直角⊿ADO中,∠DAN=∠DAO,故∠ADO=∠AND

∵⊿ADN≌⊿DCM ∴∠DAN=∠MDC ;DN=CM=3∴⊿ADO∽⊿DCM∽⊿DON ,DM=AN=5(MC=3,CD=4,RT⊿CDM)

即AD/AO=DM/DC即4/AO=5/4 ∴AO=16/5

故AO=16/5；AN=5