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解（1）如图1，过点E作EG?BC于点G． ··························· 1分
∵E为AB的中点，
1∴BE?AB?2．
2在Rt△EBG中，∠B?60?，∴∠BEG?30?． ·············· 2分
122∴BG?BE?1，EG?2?1?3．
2即点E到BC的距离为3．··············································· 3分
A E B
D F C
图1
G
（2）①当点N在线段AD上运动时，△PMN的形状不发生改变． ∵PM?EF，EG?EF，∴PM∥EG． ∵EF∥BC，∴EP?GM，PM?EG?3．
同理MN?AB?4． ······································································································· 4分 如图2，过点P作PH?MN于H，∵MN∥AB， ∴∠NMC?∠B?60?，∠PMH?30?． N
A D ∴PH?13 PM?．22B
E P H F C
图2
3cos30??．∴MH?PM?
235则NH?MN?MH?4??．
222G M ?5??3?22Rt△PNH在中，PN?NH?PH????? ?7．???22????∴△PMN的周长=PM?PN?MN?3?7?4． ················································· 6分 ②当点N在线段DC上运动时，△PMN的形状发生改变，但△MNC恒为等边三角
形．
当PM?PN时，如图3，作PR?MN于R，则MR?NR．
23． 2∴MN?2MR?3． ········································································································· 7分 ∵△MNC是等边三角形，∴MC?MN?3．
此时，x?EP?GM?BC?BG?MC?6?1?3?2． ············································· 8分
类似①，MR?A E B
P R
G
M
图3
C
B
G
图4
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D N F
A E P
D F N C
B
A E D F（P） N C
M G
图5
M
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当MP?MN时，如图4，这时MC?MN?MP?3．
此时，x?EP?GM?6?1?3?5?3．
当NP?NM时，如图5，∠NPM?∠PMN?30?．
则∠PMN?120?，又∠MNC?60?， ∴∠PNM?∠MNC?180?．
因此点P与F重合，△PMC为直角三角形．
．∴MC?PM?tan30??1
此时，x?EP?GM?6?1?1?4．
综上所述，当x?2或4或5?3时，△PMN为等腰三角形． ·························· 10分
??9（09兰州）如图①，正方形 ABCD中，点A、B的坐标分别为（0，10），（8，4）， 点C在第一象限．动点P在正方形 ABCD的边上，从点A出发沿A→B→C→D匀速运动，
同时动点Q以相同速度在x轴正半轴上运动，当P点到达D点时，两点同时停止运动，
设运动的时间为t秒．
(1)当P点在边AB上运动时，点Q的横坐标x（长度单位）关于运动时间t（秒）的函数图象如图②所示，请写出点Q开始运动时的坐标及点P运动速度；
(2)求正方形边长及顶点C的坐标；
(3)在（1）中当t为何值时，△OPQ的面积最大，并求此时P点的坐标； (4)如果点P、Q保持原速度不变，当点P沿A→B→C→D匀速运动时，OP与PQ能否相等，若能，写出所有符合条件的t的值；若不能，请说明理由．
解：（1）Q（1，0） ·············································································································· 1分 点P运动速度每秒钟1个单位长度． ····································································································································· 2分 （2） 过点B作BF⊥y轴于点F，BE⊥x轴于点E，则BF＝8，OF?BE?4． ∴AF?10?4?6．
y 在Rt△AFB中，AB?82?62?10 3分 过点C作CG⊥x轴于点G，与FB的延长线交于点H． ∵?ABC?90?,AB?BC ∴△ABF≌△BCH．
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∴BH?AF?6,CH?BF?8． ∴OG?FH?8?6?14,CG?8?4?12．
∴所求C点的坐标为（14，12）． 4分 （3） 过点P作PM⊥y轴于点M，PN⊥x轴于点N， 则△APM∽△ABF． ∴
APAMMPtAMMP． ?． ????ABAFBF10683434 ∴AM?t，PM?t． ∴PN?OM?10?t,ON?PM?t．
5555设△OPQ的面积为S（平方单位）
13473∴S??(10?t)(1?t)?5?t?t2（0≤t≤10） ······························································ 5分
251010说明:未注明自变量的取值范围不扣分．
473<0 ∴当t??时， △OPQ的面积最大． ································ 6分 ?36102?(?)104710 ∵a?? 此时P的坐标为（
9453，） ． ························································································ 7分 15105295（4） 当 t?或t?时， OP与PQ相等． ······························································ 9分
31310（09临沂）数学课上，张老师出示了问题：如图1，四边形ABCD是正方形，点E是边BC的中点．?AEF?90?，且EF交正方形外角?DCG的平行线CF于点F，求证：AE=EF．
经过思考，小明展示了一种正确的解题思路：取AB的中点M，连接ME，则AM=EC，易证△AME≌△ECF，所以AE?EF．
在此基础上，同学们作了进一步的研究：
（1）小颖提出：如图2，如果把“点E是边BC的中点”改为“点E是边BC上（除B，C外）的任意一点”，其它条件不变，那么结论“AE=EF”仍然成立，你认为小颖的观点正确吗？如果正确，写出证明过程；如果不正确，请说明理由；
（2）小华提出：如图3，点E是BC的延长线上（除C点外）的任意一点，其他条件不变，结论“AE=EF”仍然成立．你认为小华的观点正确吗？如果正确，写出证明过程；如果不正确，请说明理由．
A
D
F
B E C 图1
G
B
E C 图2 A
D
F G
B 图3
C E G
F A
D
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解：（1）正确． ·································································· （1分） 证明：在AB上取一点M，使AM?EC，连接ME． （2分）
D A
?BM?BE．??BME?45°，??AME?135°．
F M ?CF是外角平分线，
??DCF?45°，
B E C G ??ECF?135°．
??AME??ECF．
??AEB??BAE?90°，?AEB??CEF?90°， ??BAE??CEF．
?△AME≌△BCF（ASA）． ···················································································· （5分） ?AE?EF． ················································································································ （6分） （2）正确． ··································································· （7分） 证明：在BA的延长线上取一点N． 使AN?CE，连接NE． ············································ （8分） N F ?BN?BE． D A ??N??PCE?45°． ?四边形ABCD是正方形， ?AD∥BE．
B C E G ??DAE??BEA．
??NAE??CEF．
?△ANE≌△ECF（ASA）． ··················································································· （10分） ?AE?EF． ·············································································································· （11分）
11（09天津）已知一个直角三角形纸片OAB，其中?AOB?90°，OA?2，OB?4．如图，将该纸片放置在平面直角坐标系中，折叠该纸片，折痕与边OB交于点C，与边AB交于点D． （Ⅰ）若折叠后使点B与点A重合，求点C的坐标； y
B
x O A （Ⅱ）若折叠后点B落在边OA上的点为B?，设OB??x，OC?y，试写出y关于x的函数解析式，并确定y的取值范围；
B y x O A
（Ⅲ）若折叠后点B落在边OA上的点为B?，且使B?D∥OB，求此时点C的坐标． y B 京翰教育1对1家教 http://www.zgjhjy.com/ O A x 初中数学辅导网 http://www.shuxuefudao.cn
解（Ⅰ）如图①，折叠后点B与点A重合， 则△ACD≌△BCD.
设点C的坐标为?0，m??m?0?. 则BC?OB?OC?4?m. 于是AC?BC?4?m.
在Rt△AOC中，由勾股定理，得AC?OC?OA，
22即?4?m??m?2，解得m?22223. 2?3?········································································································ 4分 ?点C的坐标为?0，?. ·2??（Ⅱ）如图②，折叠后点B落在OA边上的点为B?,
则△B?CD≌△BCD. 由题设OB??x，OC?y， 则B?C?BC?OB?OC?4?y，
在Rt△B?OC中，由勾股定理，得B?C?OC?OB?.
222??4?y??y2?x2，
212x?2 ···················································································································· 6分 8由点B?在边OA上，有0≤x≤2，
1? 解析式y??x2?2?0≤x≤2?为所求.
8即y??? ?当0≤x≤2时，y随x的增大而减小，
3?y的取值范围为≤y≤2. ······················································································· 7分
2（Ⅲ）如图③，折叠后点B落在OA边上的点为B??，且B??D∥OB. 则?OCB????CB??D.
??OCB????CBD，有CB??∥BA. 又??CBD??CB??D，?Rt△COB??∽Rt△BOA. OB??OC?有，得OC?2OB??. ···················································································· 9分 OAOB在Rt△B??OC中，
设OB???x0?x?0?，则OC?2x0. 由（Ⅱ）的结论，得2x0??12x0?2， 8京翰教育1对1家教 http://www.zgjhjy.com/
初中数学几何的动点问题专题练习 - 图文
