ADADOFBECOMFEBC
(2)证明:∵BE平分∠CBD,
∴∠CBE=∠DBE. ............................................................................................... 2 ∵正方形ABCD的对角线AC,BD交于点O, ∴∠BOC=∠BCD=90°. ∵∠CBE+∠CEB=90°, ∠DBE+∠BFO=90°,
∴∠CEB=∠BFO. ............................................................................................... 3 ∵∠EFC=∠BFO, ∴∠EFC=∠CEB.
∴CF=CE. ............................................................................................................ 4 (3)证明:取BE的中点M,连接OM. .............................................................................. 5 ∵O为AC的中点,
∴OM∥DE, DE=2OM. ................................................................................... 6 ∴∠OMF=∠CEF.
∵∠OFM=∠EFC=∠CEF, ∴∠OMF=∠OFM. ∴OF=OM.
∴DE=2OF. .......................................................................................................... 7
28.解:(1)○1P1 ,P2 2; ..................................................................................................... ○2当直线y=x+b与O相切时,b?22或?22; ................................ 3 ∴?22?b?22. .................................................................................. 5 (2)当直线y=4与M相切时,m=2或6. .......................................................... 6 ∴2≤m≤6. ...................................................................................................... 7
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