内容正文:
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AB=AC,
∴.PB=PC.在△ABP和△ACP中,PB=PC,∴.△ABP
AP=APR:△ADP
(2)在R△ADP和R△AEP中,{DP=EP,
AP-AP,
≌R△AEP(HL).AD=AE.,AB=6cm,AC=10cm,
≌△ACP(SSS)..∠ABP=∠ACP
∴.6+AD=10-AE,即6+AD=10-AD..AD=2cm.
15.解:(1)点M与点P关于OA对称,.OA垂直平分MP
16.解:(1)6-x6+x(2)2(3)线段DE的长不变,DE=
.EM=EP,同理可得FP=FN..△PEF的周长为PE
3.理由如下:过点Q作QF⊥AB,交AB的延长线于点F
+EF+PF=ME+EF+FV=MN=20cm.(2)连接OM,
PE⊥AB,∴∠BFQ=∠AEP-90°.:点P,Q速度相
ON,OP.OA垂直平分MP,∴.OM=OP.又OE=OE,
同,.AP=BQ.:△ABC是等边三角形,.∠A=∠ABC
EM=EP,∴.△MOE2△POE(SSS).同理可得△POF2
=∠FBQ=60.∴.△APE≌△BQF(AAS).∴.AE=BF,
△NOF(SSS)..∠MOE=∠POE,∠OME=∠OPE,
PE=QF.又,∠DEP=∠DFQ=90°,∠PDE=∠QDF,
∠POF=∠NOF,∠OPF=∠ONF.∴.∠MON=2∠AOB
∴△PDE≌△QDF(AAS).∴.DE=DF=DB+BF=DB
=70°.∴.∠EPF=∠OPE+∠OPF=∠OME+∠ONF=
180°-∠M0N=110°.
+AE.DE-号AB.“等边三角形ABC的边长为6,
16.解:(1)图略.(2)P(0,一1),P(一1,-1),P(2,1).
DE-3..在运动过程中,线段DE的长不变,DE-3.
周测(15.3~15.4)
单元测试(一)平面直角坐标系
1.D2.C3.B4.A5.C6.C7.608.29.80或50
1.B2.D3.D4.A5.A6.C7.C8.A9.B
10.52°11.812.(3,-1)或(-3,2)
10.D11.(8,7)12.413.4或0
13.解:连接AE,:AB=AC,∠BAC=120°,∴∠B=∠C=
号180-∠BAC)=30.“直线ED是线段AC的垂直平
14.11,-10或(-1)(2(-33)
15.解:图略
分线,,.EA=EC=6..∠EAC=∠C=30°..∠BAE=
16.解:(1)图略.(2)1一2(3)图略.
∠BAC-∠EAC=90°..BE=2AE=12.
17.解:(1)A(0,4),B(一3,一1),C(3,3).(2)图略.B(一2,
14.解:(1)图略.(2)过点D作DE⊥AB于点E,:AD平分
∠BAC,DELAB,∠C=90°,∴.DE=CD=4cm.Sam
-3).C(4,1).
=7AB.DE=号×10X4=20(m).
18.解:(1)(2,3)(一2,3)(-4,0)(-3,-3)(2,-3)
15.解:(1)AB=AC,AD⊥BC,∴.∠BAD=∠CAD,∠ADC
(3.0)(2)六边形ABCDEF的面积为6×7-号×2×
=90°.又,∠C=42°,.∠BAD=∠CAD=90°-42°=
48°.(2)证明:EF∥AC,∴∠F=∠CAD.∠BAD=
3-×1×3-7×1×3-7×1×3=34.5.
∠CAD.∴∠BAD=∠F.AE=FE.
19.解:(1)根据题意,得2m十4=0,解得m=一2.所以m一1
16.证明:(1):△ABC和△CDE都是等边三角形,.BC=
=一3.所以点P的坐标为(0,一3).(2)根据题意,得2m+
AC,CE=CD,∠BCA=∠ECD=60°.∴·∠BCA+∠ACE
4=m一1或2m+4十m-1=0,解得m=一5或m=一1.
=∠ECD+∠ACE,即∠BCE=∠ACD.在△BCE和
当m=一5时,2m十4=一6