初三数学几何的动点问题专题练习(2)

8时，以⊙P 与直线l 的两个交点和圆心P 为顶点的三角形是正三角形.

10 4.

5.解：（1）1，8

5

；

11 （2）作QF ⊥AC 于点F ，如图3， AQ = CP = t ，∴3AP t =-． 由△AQF ∽△ABC

，4BC ==， 得

45QF t =．∴4

5

QF t =． ∴14(3)25

S t t =-?， 即22655

S t t =-+．

（3）能．

①当DE ∥QB 时，如图4．

∵DE ⊥PQ ，∴PQ ⊥QB ，四边形QBED 是直角梯形． 此时∠AQP =90°． 由△APQ ∽△ABC ，得AQ AP AC AB

=

， 即335t t -=

． 解得9

8

t =． ②如图5，当PQ ∥BC 时，DE ⊥BC ，四边形QBED 是直角梯形．

此时∠APQ =90°． 由△AQP ∽△ABC ，得

AQ AP

AB AC

=

， 即353t t -=． 解得158

t =．

（4）52t =

或45

14

t =． ①点P 由C 向A 运动，DE 经过点C ．

连接QC ，作QG ⊥BC 于点G ，如图6．

PC t =，222QC QG CG =+2234

[(5)][4(5)]55

t t =-+--．

由2

2

PC QC =，得2

2234

[(5)][4(5)]55

t t t =-+--，解得52t =．

②点P 由A 向C 运动，DE 经过点C ，如图7． 22234

(6)[(5)][4(5)]55t t t -=-+--，4514

t =】

6.解（1）①30，1；②60，1.5；

（2）当∠α=900

时，四边形EDBC 是菱形. ∵∠α=∠ACB=900

，∴BC //ED .

∵CE //AB , ∴四边形EDBC 是平行四边形. ……………………6分 在Rt △ABC 中，∠ACB =900

，∠B =600

,BC =2,

∴∠A =300.

∴AB =4,AC 图4

P

图5

12 ∴AO =12

AC

……………………8分 在Rt △AOD 中，∠A =300，∴AD =2.

∴BD =2.

∴BD =BC .

又∵四边形EDBC 是平行四边形，

∴四边形EDBC 是菱形 ……………………10分

7.解：（1）如图①，过A 、D 分别作AK BC ⊥于K ，DH BC ⊥于H ，则四边形ADHK

是矩形

∴3KH AD ==． ················································································ 1分 在Rt ABK △中，sin 4542

AK AB =?== ．

cos 454BK AB =?== ·························································· 2分 在Rt CDH △

中，由勾股定理得，3HC

∴43310BC BK KH HC =++=++= ················································· 3分

（2）如图②，过D 作DG AB ∥交BC 于G 点，则四边形ADGB 是平行四边形 ∵MN AB ∥

∴MN DG ∥

∴3BG AD ==

∴1037GC =-= ············································································· 4分 由题意知，当M 、N 运动到t 秒时，102CN t CM t ==-，．

∵DG MN ∥

∴NMC DGC =∠∠

又C C =∠∠

∴MNC GDC △∽△ ∴

CN CM CD CG

= ··················································································· 5分 即10257

t t -= 解得，5017t = ···················································································· 6分 （图①） A D C B K H （图②） A D C B G M N

13 （3）分三种情况讨论：

①当NC MC =时，如图③，即102t t =- ∴103

t = ·························································································· 7分

②当MN NC =时，如图④，过N 作NE MC ⊥于E 解法一：

由等腰三角形三线合一性质得()11

102522

EC MC t t ==-=- 在Rt CEN △中，5cos EC t

c NC t -== 又在Rt DHC △中，3

cos 5

CH c CD =

= ∴53

5

t t -= 解得25

8

t = ······················································································· 8分

解法二：

∵90C C DHC NEC =∠=∠=?∠∠， ∴NEC DHC △∽△

∴

NC EC

DC HC = 即553t t -= ∴258

t = ·························································································· 8分

③当MN MC =时，如图⑤，过M 作MF CN ⊥于F 点.11

22

FC NC t ==

解法一：（方法同②中解法一）

1

3

2cos 1025t

FC C MC t ===-

解得60

17

t =

解法二：

∵90C C MFC DHC =∠=∠=?∠∠， ∴MFC DHC △∽△ ∴

FC MC

HC DC

= A

D

C

B M

N

（图③）

（图④）

A

D C

B

M N

H E

（图⑤）

A D

C

B

H N M

F

14 即1102235

t t -= ∴6017

t = 综上所述，当103

t =、258t =或6017t =时，MNC △为等腰三角形 ··············· 9分 8.解（1）如图1，过点E 作EG BC ⊥于点G ． ··················· 1分

∵E 为AB 的中点， ∴122BE AB ==． 在Rt EBG △中，60B =?∠，∴30BEG =?∠． ············ 2分

∴112BG BE EG ====， 即点E 到BC

····································· 3分

（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 ==

同理4MN AB ==． ·················································································· 4分 如图2，过点P 作PH MN ⊥于H ，∵MN AB ∥，

∴6030NMC B PMH ==?=?∠∠，∠．

∴12PH PM == ∴3cos302MH PM =?= ． 则35422

NH MN MH =-=-=． 在Rt PNH △

中，PN == ∴PMN △的周长

=4PM PN MN ++． ······································· 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 =． 类似①，32

MR =

． ∴23MN MR ==． ··················································································· 7分 ∵MNC △是等边三角形，∴3MC MN ==．

此时，6132x EP GM BC BG MC ===--=--=． ··································· 8分 图1 A D E B F C G

图2 A D E B F C

P N H

15 当MP MN =时，如图4

，这时MC MN MP ===

此时，615x EP GM ===-=

当NP NM =时，如图5，30NPM PMN ==?∠∠．

则120PMN =?∠，又60MNC =?∠， ∴180PNM MNC +=?∠∠．

因此点P 与F 重合，PMC △为直角三角形．

∴tan 301MC PM =?= ．

此时，6114x EP GM ===--=．

综上所述，当2x =或4

或(5-时，PMN △为等腰三角形． ···················· 10分 9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 ＝8，4OF BE ==． ∴1046AF =-=．

在Rt △AFB

中，10AB 3分 过点C 作CG ⊥x 轴于点G ，与FB 的延长线交于点H ． ∵90,ABC AB BC ∠=?= ∴△ABF ≌△BCH ． ∴6,8BH AF CH BF ====． ∴8614,8412OG FH CG ==+==+=．

∴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 ． ∴

AP AM MP AB AF BF ==． 1068

t A M M P

∴==

． ∴3455AM t PM t ==，． ∴34

10,55

PN OM t ON PM t ==-==．

设△OPQ 的面积为S （平方单位）

∴213473

(10)(1)5251010

S t t t t =?-+=+-（0≤t ≤10） ················································· 5分

说明:未注明自变量的取值范围不扣分．

∵3

10a =-

<0 ∴当474710

362()10

t =-=

?-时， △OPQ 的面积最大． ························· 6分 图3

A D E B

F

C

P

N M 图4

A D E

B

F C

P

M N 图5

A D E

B

F （P ）

C

M

N G

G

R

G

16 此时P 的坐标为（9415，5310

） ． ····································································· 7分 （4） 当 53t =或29513

t =时， OP 与PQ 相等． ················································· 9分

10.解：（1）正确． ················································· （1分）

证明：在AB 上取一点M ，使AM EC =，连接ME ． （2分） BM BE ∴=．45BME ∴∠=°，135AME ∴∠=°． CF 是外角平分线，

45DCF ∴∠=°， 135ECF ∴∠=°．

AME ECF ∴∠=∠．

90AEB BAE ∠+∠= °，90AEB CEF ∠+∠=°，

∴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） BN BE ∴=． 45N PCE ∴∠=∠=°．

四边形ABCD 是正方形，

AD BE ∴∥． DAE BEA ∴∠=∠．

NAE CEF ∴∠=∠．

ANE ECF ∴△≌△（ASA ）． ································································· （10分） AE EF ∴=． （11分）

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