25．解：
（1）过点O作OH⊥CD，垂足为点H，联结OC．
在Rt△POH中，∵sinP＝，PO?6，∴OH?2．········································ （1分） ∵AB＝6，∴OC＝3． ···························································································· （1分） 由勾股定理得 CH?5． ······················································································· （1分）
∵OH⊥DC，∴CD?2CH?25． ································································· （1分） （2）在Rt△POH中，∵sinP＝，PO ＝m，∴OH＝21313m． ··································· （1分） 3?m?在Rt△OCH中，CH2＝9???． ····································································· （1分）
?3?m??在Rt△O1CH中，CH＝36??n??． ···························································· （1分）
3??223n2?81m???m?可得 36??n??＝9???，解得m＝． ········································ （2分）
3?2n??3?（3）△POO1成为等腰三角形可分以下几种情况：
● 当圆心O1、O在弦CD异侧时
223n2?81①OP＝OO1，即m＝n，由n＝解得n＝9． ········································ （1分）
2n即圆心距等于⊙O、⊙O1的半径的和，就有⊙O、⊙O1外切不合题意舍去．（1分） ②O1P＝OO1，由(n?m2m2)?m2?()＝n， 332293n2?81m＝nnn＝15． ·，即，········································ （1分） ＝解得解得
3352n81?3n2● 当圆心O1、O在弦CD同侧时,同理可得 m＝．
2n981?3n2∵?POO1是钝角，∴只能是m?n，即n＝，解得n＝5． ··········· （2分）
52n995或15．综上所述，n的值为55
青浦区
25.（本题满分14分，第（1）小题4分，第（2）小题6分，第（3）小题4分）
如图9-1，已知扇形MON的半径为2，∠MON=90o，点B在弧MN上移动，联结BM，作OD?BM，垂足为点D，C为线段OD上一点，且OC=BM，联结BC并延长交半径OM于点A，设OA= x，∠COM的正切值为y．
（1）如图9-2，当AB?OM时，求证：AM =AC； （2）求y关于x的函数关系式，并写出定义域； （3）当△OAC为等腰三角形时，求x的值.
25．解：（1）∵OD⊥BM，AB⊥OM，∴∠ODM =∠BAM =90°． ······································· （1分）
∵∠ABM +∠M =∠DOM +∠M，∴∠ABM =∠DOM． ································· （1分） ∵∠OAC=∠BAM，OC =BM，
∴△OAC≌△ABM， ························································································· （1分） ∴AC =AM． ······································································································· （1分） （2）过点D作DE//AB，交OM于点E． ······························································· （1分）
∵OB＝OM，OD⊥BM，∴BD＝DM． ···························································· （1分） ∵DE//AB， ∴
OCADMOCDMOMNBNBN
A图9-1 图9-2
备用图
MDME，∴AE＝EM， ?DMAE1∵OM=2，∴AE＝······························································· （1分） 2?x． ·
2??∵DE//AB， ∴
OAOC2DM， ················································································ （1分） ??OEODOD∴
DMOA， ?OD2OEx．（0?x?2） ······································································ （2分）
x?2∴y?（3）（i） 当OA=OC时， ∵DM?111BM?OC?x， 222OM2?DM2?2?在Rt△ODM中，OD?DM12x．∵y?， 4OD1x14?2?14?2x2∴．解得x?，或x?（舍）．（2分） ?221x?22?x24（ii）当AO=AC时，则∠AOC =∠ACO，
∵∠ACO >∠COB，∠COB =∠AOC，∴∠ACO >∠AOC，
∴此种情况不存在． ························································································ （1分） （ⅲ）当CO=CA时，
则∠COA =∠CAO=?，
∵∠CAO >∠M，∠M=90???，∴?>90???，∴?>45?，
∴?BOA?2??90?，∵?BOA?90?，∴此种情况不存在． ··········· （1分）
松江区
25．（本题满分14分，第（1）小题4分，第（2）小题每个小题各5分）
如图，已知Rt△ABC 中，∠ACB=90°，BC=2，AC=3，以点C为圆心、CB为半径的圆交AB于点D，过点A作AE∥CD，交BC延长线于点E. （1）求CE的长；
（2）P是 CE延长线上一点，直线AP、CD交于点Q.
① 如果△ACQ ∽△CPQ，求CP的长；
② 如果以点A为圆心，AQ为半径的圆与⊙C相切，求CP的长.
D A D A B
C
E B
C
E
25．（本题满分14分，第（1）小题4分，第（2）小题每个小题各5分） 解：（1）∵AE∥CD
∴BCDCBE?AE…………………………………1分 ∵BC=DC
∴BE=AE …………………………………1分 设CE=x 则AE=BE=x+2 ∵ ∠ACB=90°， ∴AC2?CE2?AE2
即9?x2?(x?2)2………………………1分
∴x?54 即CE?54…………………………………1分
（2）①
∵△ACQ ∽△CPQ，∠QAC>∠P ∴∠ACQ=∠P…………………………………1分 又∵AE∥CD ∴∠ACQ=∠CAE
∴∠CAE=∠P………………………………1分 ∴△ACE ∽△PCA，…………………………1分 ∴AC2?CE?CP…………………………1分
即32?54?CP ∴CP?365 ……………………………1分
A D B
C
E (第25题图)
Q A D B
C E P