动点问题专题训练
1、如图,已知ABC
△中,10
AB AC
==厘米,8
BC=厘米,点D为AB的中点.(1)如果点P在线段BC上以3厘米/秒的速度由B点向C点运动,同时,点Q 在线段CA上由C点向A点运动.
①若点Q的运动速度与点P的运动速度相等,经过1秒后,BPD
△与
CQP
△是否全等,请说明理由;
②若点Q的运动速度与点P的运动速度不相等,当点Q的运动速度
为多少时,能够使BPD
△与CQP
△全等?
(2)若点Q以②中的运动速度从点C出发,点P以原来的运动速度
从点B同时出发,都逆时针沿ABC
△三边运动,求经过多长时间点P
与点Q第一次在ABC
△的哪条边上相遇?
2、直线
3
6
4
y x
=-+与坐标轴分别交于A B
、两点,动点P Q
、同时从O点出发,
同时到达A点,运动停止.点Q沿线段OA运动,速度为每秒1个单位长度,
点P沿路线O→B→A运动.
(1)直接写出A B
、两点的坐标;
(2)设点Q的运动时间为t秒,OPQ
△的面积为S,求出S
与t之间的函数关系式;
(3)当
48
5
S=时,求出点P的坐标,并直接写出以点
O P Q
、、为顶点的平行四边形的第四个顶点M的坐标.
3如图,在平面直角坐标系中,直线l:y=-2x-8分别与x轴,y轴相交于A,B 两点,点P(0,k)是y轴的负半轴上的一个动点,以P为圆心,3为半径作⊙P.
(1)连结P A,若P A=PB,试判断⊙P与x轴的位置关系,并说明理由;
(2)当k为何值时,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?
4 如图1,在平面直角坐标系中,点O是坐标原点,四边形ABCO是菱形,点A 的坐标为(-3,4),
点C在x轴的正半轴上,直线AC交y轴于点M,AB边交y轴于点H.(1)求直线AC的解析式;
(2)连接BM,如图2,动点P从点A出发,沿折线ABC方向以2个单位/秒的速度向终点C匀速运动,设△PMB的面积为S(S≠0),点P的运动时间为t秒,求S与t之间的函数关系式(要求写出自变量t的取值范围);
(3)在(2)的条件下,当t为何值时,∠MPB与∠BCO互为余角,并求此时直线OP与直线AC所夹锐角的正切值.
5在Rt △ABC 中,∠C =90°,AC = 3,AB = 5.点P 从点C 出发沿CA 以每秒1个单位长的速度向点A 匀速运动,到达点A 后立刻以原来的速度沿AC 返回;点Q 从点A 出发沿AB 以每秒1个单位长的速度向点B
匀速运动.伴随着P 、Q 的运动,DE 保持垂直平分PQ ,且交PQ 于点D ,交折线QB -BC -CP 于点E .点P 、Q 同时出发,当点Q 到达点B 时停止运动,点P 也随之停止.设点P 、Q 运动的时间是t 秒(t >0).
(1)当t = 2时,AP = ,点Q 到AC 的距离是 ;
(2)在点P 从C 向A 运动的过程中,求△APQ
的面积S 与
t 的函数关系式;(不必写出t 的取值范围)
(3)在点E 从B 向C 运动的过程中,四边形QBED 能否成
为直角梯形?若能,求t 的值.若不能,请说明理由; (4)当DE 经过点C 时,请直接..
写出t 的值.
6如图,在Rt ABC △中,9060ACB B ∠=∠=°,°,2BC =.点O 是AC 的中点,过点O 的直线l 从与AC 重合的位置开始,绕点O 作逆时针旋转,交AB 边于点D .过点C 作CE AB ∥交直线l 于点E ,设直线l 的旋转角为α. (1)①当α= 度时,四边形EDBC 是等腰梯形,此时AD 的长为 ;②当α= 度时,四边形EDBC 是直角梯形,此时AD 的长为 ;
(2)当90α=°时,判断四边形EDBC 是否为菱形,并说明理由.
图
16
(备用图)
7如图,在梯形ABCD
中,3545AD BC AD DC AB B ====?∥,,,.动点M 从B 点出发沿线段BC 以每秒2个单位长度的速度向终点C 运动;动点N 同时从C 点出发沿线段CD 以每秒1个单位长度的速度向终点
D
运动.设运动的时间为t 秒. (1)求BC 的长.
(2)当MN AB ∥时,求t 的值. (3)试探究:t 为何值时,MNC △为等腰三角形.
8如图1,在等腰梯形ABCD 中,AD BC ∥,E 是AB 的中点,过点E 作EF BC ∥交CD 于点F .46AB BC ==,,60B =?∠. (1)求点E 到BC 的距离;
(2)点P 为线段EF 上的一个动点,过P 作PM EF ⊥交BC 于点M ,过M 作MN AB ∥交折线ADC 于点N ,连结PN ,设EP x =. ①当点N 在线段AD 上时(如图2),P M N △的形状是否发生改变?若不变,求出PMN △的周长;若改变,请说明理由; ②当点N 在线段DC 上时(如图3),是否存在点P ,使PMN △为等腰三角形?若存在,请求出所有满足要求的x 的值;若不存在,请说明理由.
C M A
D
E B
F C
图4(备用)
A
D
E B
F C
图5(备用)
A D E B
F C
图1 图2
A D
E
B
F C P
N M 图3
A D E
B
F
C
P
N M
(第25题)
9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.
10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.90AEF ∠=,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
C G
B
图1
A
D
F
C G B 图2
A
D
F
C G
B
图3
11已知一个直角三角形纸片OAB ,其中9024AOB OA OB ∠===°,,.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;
(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;
(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.
12如图(1),将正方形纸片ABCD 折叠,使点B 落在CD 边上一点E (不与点C ,D 重合),压平后得到折痕MN .当12CE CD =
时,求AM
BN 的值.
类比归纳
在图(1)中,若13CE CD =,则AM BN 的值等于 ;若14CE CD =,
则AM
BN 的值等于 ;若1CE CD n =(n 为整数)
,则AM
BN
的值等于 .(用含n 的式子表示) 联系拓广 如图(2),将矩形纸片ABCD 折叠,使点B 落在CD 边上一点E (不与点C D
,重合),压平后得到折痕MN ,设()111AB CE m BC m CD n =>=,,
则AM
BN
的值等于 .(用含m n ,的式子表示)
方法指导: 为了求得AM BN 的值,可先求BN 、AM 的长,不妨设:AB =2 图(2) A
B C D E
F M 图(1) A B C D E F
M N
12..如图所示,在直角梯形ABCD中,AD//BC,∠A=90°,AB=12,BC=21,AD=16。动点P从点B出发,沿射线BC的方向以每秒2个单位长的速度运动,动点Q同时从点A出发,在线段AD上以每秒1个单位长的速度向点D运动,当其中一个动点到达端点时另一个动点也随之停止运动。设运动的时间为t(秒)。
(1)设△DPQ的面积为S,求S与t之间的函数关系式;
(2)当t为何值时,四边形PCDQ是平行四边形?
(3)分别求出出当t为何值时,①PD=PQ,②DQ=PQ ?
13.三角形ABC中,角C=90度,角CBA=30度,BC=20根号3。一个圆心在A点、半径为6的圆以2个单位长度/秒的速度向右运动,在运动的过程中,圆心始终都在直线AB上,运动多少秒时,圆与△ABC的一边所在的直线相切。
1.解:(1)①∵1t =秒, ∴313BP CQ ==?=厘米,
∵10AB =厘米,点D 为AB 的中点, ∴5BD =厘米.
又∵8PC BC BP BC =-=,厘米, ∴835PC =-=厘米, ∴PC BD =. 又∵AB AC =, ∴B C ∠=∠,
∴BPD CQP △≌△. ············································································· (4分) ②∵P Q v v ≠, ∴BP CQ ≠,
又∵BPD CQP △≌△,B C ∠=∠,则45BP PC CQ BD ====,, ∴点P ,点Q 运动的时间4
33
BP t ==秒, ∴515
443
Q CQ v t =
==厘米/秒. ·
································································· (7分) (2)设经过x 秒后点P 与点Q 第一次相遇, 由题意,得15
32104
x x =+?, 解得80
3
x =
秒. ∴点P 共运动了80
3803
?=厘米.
∵8022824=?+,
∴点P 、点Q 在AB 边上相遇, ∴经过
80
3
秒点P 与点Q 第一次在边AB 上相遇. ········································· (12分) 2.解(1)A (8,0)B (0,6) ············· 1分 (2)86OA OB ==, 10AB ∴=
点Q 由O 到A 的时间是
8
81
=(秒) ∴点P 的速度是
610
28
+=(单位/秒) ·
1分 当P 在线段OB 上运动(或03t ≤≤)时,2OQ t OP t ==,
2S t = ·
········································································································· 1分 当P 在线段BA 上运动(或38t <≤)时,6102162OQ t AP t t ==+-=-,, 如图,作PD OA ⊥于点D ,由
PD AP BO AB =
,得4865
t
PD -=, ······························ 1分 21324255
S OQ PD t t ∴=?=-+ ·
······································································ 1分 (自变量取值范围写对给1分,否则不给分.)
(3)82455P ?? ???
, ···························································································· 1分
1238241224122455555
5I M M 2??????-- ? ? ???????,,,,, ···················································· 3分
3.解:(1)⊙P 与x 轴相切.
∵直线y =-2x -8与x 轴交于A (4,0),
与y 轴交于B (0,-8), ∴OA =4,OB =8. 由题意,OP =-k , ∴PB =P A =8+k .
在Rt △AOP 中,k 2+42=(8+k )2, ∴k =-3,∴OP 等于⊙P 的半径, ∴⊙P 与x 轴相切.
(2)设⊙P 与直线l 交于C ,D 两点,连结PC ,PD 当圆心P
在线段OB 上时,作PE ⊥CD 于E .
∵△PCD 为正三角形,∴DE =12CD =3
2
,PD =3,
∴PE . ∵∠AOB =∠PEB =90°, ∠ABO =∠PBE , ∴△AOB ∽△PEB ,
∴
2,
AO PE AB PB PB =,
∴PB =
∴8PO BO PB =-=
∴8
k=-.
当圆心P在线段OB延长线上时,同理可得P(0,-8),
∴k=-8,
∴当k-8或k=-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.
4.
;
5
(2)作QF⊥AC于点F,如图3,AQ = CP= t,∴3
AP t
=-.
BC==,
得
45QF t =.∴4
5
QF t =. ∴14(3)2
5
S t t =-?, 即2265
5
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 . ∴AO =
1
2
AC ……………………8分 在Rt △AOD 中,∠A =300
,∴AD =2.
P
图4
图5
∴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 =?==.
2
cos 4542
4BK 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,50
17
t = ···················································································· 6分
(3)分三种情况讨论:
①当NC MC =时,如图③,即102t t =-
(图①) A D C B K H (图②) A D C B G M
N
∴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 ==
∴535
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
即1102235t
t -= ∴6017
t =
综上所述,当10
3
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,
∴1
22
BE AB ==.
在Rt EBG △中,60B =?∠,∴30BEG =?∠. ············ 2分
∴1
12
BG 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 =
= ∴3
cos302
MH PM =?=.
则35
422
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①,3
2
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
M
G H
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. ∴tan301MC 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
此时P 的坐标为(
9415,53
10
) . ····································································· 7分 (4) 当 53t =或295
13
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)
11.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()()00m m >,. 则4BC OB OC m =-=-. 于是4AC BC m ==-.
在Rt AOC △中,由勾股定理,得2
2
2
AC OC OA =+, 即()2
2
2
42m m -=+,解得32
m =
. ∴点C 的坐标为302??
???
,. ··················································································· 4分
(Ⅱ)如图②,折叠后点B 落在OA 边上的点为B ',
则B CD BCD '△≌△.
A
D F C G B M A D F C G B N
由题设OB x OC y '==,, 则4B C BC OB OC y '==-=-,
在Rt B OC '△中,由勾股定理,得2
2
2
B C OC OB ''=+.
()2
224y y x ∴-=+,
即2
128
y x =-
+ ·
··························································································· 6分 由点B '在边OA 上,有02x ≤≤,
∴ 解析式21
28
y x =-+()02x ≤≤为所求.
∴ 当02x ≤≤时,y 随x 的增大而减小,
y ∴的取值范围为3
22
y ≤≤. ·
···································································· 7分 (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 ''''∠=∠. 又CBD CB D OCB CBD ''''∠=∠∴∠=∠,,有CB BA ''∥. Rt Rt COB BOA ''∴△∽△. 有OB OC OA OB
''=,得2OC OB ''=. ·································································· 9分 在Rt B OC ''△中,
设()00OB x x ''=>,则02OC x =. 由(Ⅱ)的结论,得2
001228
x x =-
+,
解得000808x x x =-±>∴=-+,∴点C
的坐标为()
016. ··································································· 10分
12解:方法一:如图(1-1),连接BM EM BE ,,.
由题设,得四边形ABNM 和四边形FENM 关于直线MN 对称.
∴MN 垂直平分BE .∴BM EM BN EN ==,. ···································· 1分 ∵四边形ABCD 是正方形,∴902A D C AB BC CD DA ∠=∠=∠=====°,.
N 图(1-1)
A B
C E
F M
∵
1
12
CE CE DE CD =∴==,.设BN x =,则NE x =,2NC x =-.
在Rt CNE △中,2
2
2
NE CN CE =+.
∴()2
2221x x =-+.解得54x =
,即54
BN =. ········································· 3分 在Rt ABM △和在Rt DEM △中,
222AM AB BM +=, 222DM DE EM +=,
∴2222AM AB DM DE +=+. ····························································· 5分
设AM y =,则2DM y =-,∴()2222
221y y +=-+.
解得14y =,即1
4AM =. ····································································· 6分
∴1
5
AM BN =.
····················································································· 7分 方法二:同方法一,5
4
BN =. ································································ 3分
如图(1-2),过点N 做NG CD ∥,交AD 于点G ,连接BE .
∵AD BC ∥,∴四边形GDCN 是平行四边形.
∴NG CD BC ==.
同理,四边形ABNG 也是平行四边形.∴54
AG BN ==. ∵90MN BE EBC BNM ⊥∴∠+∠=,°. 90NG BC MNG BNM EBC MNG ⊥∴∠+∠=∴∠=∠,°,. 在BCE △与NGM △中
90E B C M N G B C N G C N G M ∠=∠??
=??∠=∠=?
,
,°.∴BCE NGM EC MG =△≌△,. ························· 5分
∵1
14
AM AG MG AM =--=5,=.4 ····················································· 6分 ∴
1
5
AM BN =. ··················································································· 7分 类比归纳
N
图(1-2)
A B
C D
E
F
M
G