7.2.2 平行线的判定 同步练习(学生版+答案版) 2025-2026学年数学人教版七年级下册

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7.2.2 平行线的判定 同步练习(学生版+答案版) 2025-2026学年数学人教版七年级下册

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7.2.2 平行线的判定
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1.经历探索两直线平行条件的过程,理解两直线平行的条件,会运用条件判定两直线平行.2.掌握平行线基本事实Ⅱ:两条直线被第三条直线所截,如果同位角相等,那么这两条直线平行.3.探索并证明平行线的判定定理:两条直线被第三条直线所截,如果内错角相等(或同旁内角互补),那么这两条直线平行.4.进一步掌握两直线平行的条件,并能解决一些简单的问题.5.初步了解推理论证的方法,会正确地书写简单的推理过程.
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知识点一 平行线的判定方法
1.两条直线被第三条直线所截,如果同位角相等,那么这两条直线平行.简单说成:同位角相等,两直线平行.
2.两条直线被第三条直线所截,如果内错角相等,那么这两条直线平行.简单说成:内错角相等,两直线平行.
3.两条直线被第三条直线所截,如果同旁内角互补,那么这两条直线平行.简单说成:同旁内角互补,两直线平行.
练习1 如图,点E在BC的延长线上,下列条件中,不能推断AB∥CD的是( B ).
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A.∠1=∠2
B.∠3=∠4
C.∠B=∠5
D.∠B+∠BCD=180°
知识点二 添加条件,判定平行
练习2 如图,点E是AD延长线上一点,如果添加一个条件,使BC∥AD,则可添加的条件为( A ).
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A.∠C+∠ADC=180°
B.∠A+∠ABD=180°
C.∠CBD=∠ADC
D.∠C=∠CDA
知识点三 平行线判定的实际应用
练习3 一辆汽车在公路上行驶,两次拐弯后,仍在原来的方向上行驶,那么两次拐弯的方向可能是( D ).
A.先右转50°,后左转40°
B.先右转50°,后左转130°
C.先右转50°,后右转40°
D.先右转50°,后左转50°
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1.已知a,b,c是直线,下列说法正确的是( D ).
A.若a⊥b,b∥c,则a∥c
B.若a⊥b,b⊥c,则a⊥c
C.若a∥b,b⊥c,则a∥c
D.若a∥b,b∥c,则a∥c
2.如图,小明在地图上量得∠1=∠2,由此判断幸福大街与平安大街互相平行,他判断的依据是( B ).
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A.同位角相等,两直线平行
B.内错角相等,两直线平行
C.同旁内角互补,两直线平行
D.对顶角相等
3.如图,把三角尺的直角顶点放在直线b上.若∠1=50°,则当∠2=________时,a∥b.
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【答案】40°
4.如图,已知∠ABC=∠ADC,BF,DE分别平分∠ABC与∠ADC,且∠1=∠3.试说明:AB∥DC.(请根据条件进行推理,得出结论,并在括号内注明理由)
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解:因为BF,DE分别平分∠ABC与∠ADC(已知),
所以∠1=∠ABC,
∠2=∠ADC(__________________).
因为∠ABC=∠ADC(____________),
所以∠____________=∠____________(等量代换).
因为∠1=∠3(____________),
所以∠2=∠__________(__________).
所以____________∥____________(__________________________).
【答案】角平分线的定义 已知 1 2 已知 3 等量代换 AB DC 内错角相等,两直线平行
5.如图,已知三角形ABC中,∠ACB=80°,点E,F分别在AB,AC上,ED交AC于点G,交BC的延长线于点D,∠FEG=32°,∠CGD=48°.试猜想EF与BC的位置关系,并说明理由.
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【解】EF∥BC.理由如下:
因为∠CGD=48°,
所以∠EGF=∠CGD=48°.
因为∠FEG=32°,
所以∠GFE=180°-∠EGF-∠FEG=180°-48°-32°=100°.
因为∠ACB=80°,
所以∠GFE+∠ACB=180°,
所以EF∥BC.
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6.如图,在下列四组条件:①∠1=∠2,②∠3=∠4,③∠BAD+∠ABC=180°,④∠BAD+∠ADC=180°中,能判定AD∥BC的是________(填序号).
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【答案】①②③
7.如图,已知点O在直线AB上,射线OD平分∠BOC,过点O作OE⊥OD,G是射线OB上一点,连接DG,满足∠ODG+∠DOG=90°.
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(1)试说明:∠AOE=∠ODG;
(2)若∠ODG=∠C,请说明:CD∥OE.
【解】(1)因为OE⊥OD,
所以∠DOE=90°.
因为∠DOE+∠AOE+∠DOG=180°,
所以∠AOE+∠DOG=90°.
因为∠ODG+∠DOG=90°,
所以∠AOE=∠ODG.
(2)因为OD平分∠BOC,
所以∠DOG=∠COD=∠BOC.
因为OE⊥OD,所以∠DOE=90°,
所以∠COE+∠COD=90°.
因为∠ODG+∠DOG=90°,
所以∠ODG=∠COE.
因为∠ODG=∠C,
所以∠C=∠COE,
所以CD∥OE.
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8.实践探究题
【学习新知】射到平面镜上的光线(入射光线)和反射后的光线(反射光线)与平面镜所夹的角相等.如图1,若入射光线与水平镜面夹角为∠1,反射光线与水平镜面夹角为∠2,则∠1=∠2.
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(1)【初步应用】如图2,有两块平面镜AB,BC,入射光线DO1经过两次反射,得到反射光线O2E,若∠B=90°,说明:DO1∥O2E;
(2)【拓展探究】如图3,有三块平面镜AB,BC,CD,入射光线EO1经过三次反射,得到反射光线O3F,已知∠1=36°,∠B=120°,若要使EO1∥O3F,则∠C为多少度?
【解】(1)因为∠B=90°,∠B+∠2+∠3=180°,所以∠2+∠3=90°.
由题意知∠1=∠2,∠3=∠4,
所以∠1+∠2+∠3+∠4=180°.
又因为∠1+∠DO1O2+∠2=180°,
∠3+∠O1O2E+∠4=180°,
所以∠DO1O2+∠O1O2E=180°,
所以DO1∥O2E.
(2)如图,过点O2作O2M∥EO1.
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因为∠1=36°,∠B=120°,∠1=∠2,
所以∠2=∠1=36°,
所以∠3=180°-∠B-∠2=180°-120°-36°=24°,∠EO1O2=180°-∠1-∠2=180°-36°-36°=108°.
又∠3=∠4,
所以∠4=∠3=24°,
所以∠O1O2O3=180°-∠3-∠4=180°-24°-24°=132°.
因为O2M∥O1E,
所以∠O1O2M=180°-∠EO1O2=180°-108°=72°,
所以∠MO2O3=∠O1O2O3-∠O1O2M=132°-72°=60°.
因为O2M∥EO1,EO1∥O3F,
所以O2M∥O3F,
所以∠O2O3F=180°-∠MO2O3=180°-60°=120°.
又∠5=∠6,
所以∠5=∠6=×(180°-∠O2O3F)=×(180°-120°)=30°,
所以∠C=180°-∠4-∠5=180°-24°-30°=126°.7.2.2 平行线的判定
1.经历探索两直线平行条件的过程,理解两直线平行的条件,会运用条件判定两直线平行.2.掌握平行线基本事实Ⅱ:两条直线被第三条直线所截,如果同位角相等,那么这两条直线平行.3.探索并证明平行线的判定定理:两条直线被第三条直线所截,如果内错角相等(或同旁内角互补),那么这两条直线平行.4.进一步掌握两直线平行的条件,并能解决一些简单的问题.5.初步了解推理论证的方法,会正确地书写简单的推理过程.
eq \o(\s\up7(),\s\do5(       ))
知识点一 平行线的判定方法
1.两条直线被第三条直线所截,如果同位角相等,那么这两条直线平行.简单说成: .
2.两条直线被第三条直线所截,如果内错角相等,那么这两条直线平行.简单说成: .
3.两条直线被第三条直线所截,如果同旁内角互补,那么这两条直线平行.简单说成: .
练习1 如图,点E在BC的延长线上,下列条件中,不能推断AB∥CD的是( ).
A.∠1=∠2
B.∠3=∠4
C.∠B=∠5
D.∠B+∠BCD=180°
知识点二 添加条件,判定平行
练习2 如图,点E是AD延长线上一点,如果添加一个条件,使BC∥AD,则可添加的条件为( ).
A.∠C+∠ADC=180°
B.∠A+∠ABD=180°
C.∠CBD=∠ADC
D.∠C=∠CDA
知识点三 平行线判定的实际应用
练习3 一辆汽车在公路上行驶,两次拐弯后,仍在原来的方向上行驶,那么两次拐弯的方向可能是( ).
A.先右转50°,后左转40°
B.先右转50°,后左转130°
C.先右转50°,后右转40°
D.先右转50°,后左转50°
eq \o(\s\up7(),\s\do5(       ))
基础巩固
1.已知a,b,c是直线,下列说法正确的是( ).
A.若a⊥b,b∥c,则a∥c
B.若a⊥b,b⊥c,则a⊥c
C.若a∥b,b⊥c,则a∥c
D.若a∥b,b∥c,则a∥c
2.如图,小明在地图上量得∠1=∠2,由此判断幸福大街与平安大街互相平行,他判断的依据是( ).
A.同位角相等,两直线平行
B.内错角相等,两直线平行
C.同旁内角互补,两直线平行
D.对顶角相等
3.如图,把三角尺的直角顶点放在直线b上.若∠1=50°,则当∠2=________时,a∥b.
4.如图,已知∠ABC=∠ADC,BF,DE分别平分∠ABC与∠ADC,且∠1=∠3.试说明:AB∥DC.(请根据条件进行推理,得出结论,并在括号内注明理由)
解:因为BF,DE分别平分∠ABC与∠ADC(已知),
所以∠1=∠ABC,
∠2=∠ADC(__________________).
因为∠ABC=∠ADC(____________),
所以∠____________=∠____________(等量代换).
因为∠1=∠3(____________),
所以∠2=∠__________(__________).
所以____________∥____________(__________________________).
5.如图,已知三角形ABC中,∠ACB=80°,点E,F分别在AB,AC上,ED交AC于点G,交BC的延长线于点D,∠FEG=32°,∠CGD=48°.试猜想EF与BC的位置关系,并说明理由.
6.如图,在下列四组条件:①∠1=∠2,②∠3=∠4,③∠BAD+∠ABC=180°,④∠BAD+∠ADC=180°中,能判定AD∥BC的是________(填序号).
7.如图,已知点O在直线AB上,射线OD平分∠BOC,过点O作OE⊥OD,G是射线OB上一点,连接DG,满足∠ODG+∠DOG=90°.
(1)试说明:∠AOE=∠ODG;
(2)若∠ODG=∠C,请说明:CD∥OE.
8.实践探究题
【学习新知】射到平面镜上的光线(入射光线)和反射后的光线(反射光线)与平面镜所夹的角相等.如图1,若入射光线与水平镜面夹角为∠1,反射光线与水平镜面夹角为∠2,则∠1=∠2.
eq \o(\s\up7(),\s\do5(图1))  eq \o(\s\up7(),\s\do5(图2))
eq \o(\s\up7(),\s\do5(图3))
(1)【初步应用】如图2,有两块平面镜AB,BC,入射光线DO1经过两次反射,得到反射光线O2E,若∠B=90°,说明:DO1∥O2E;
(2)【拓展探究】如图3,有三块平面镜AB,BC,CD,入射光线EO1经过三次反射,得到反射光线O3F,已知∠1=36°,∠B=120°,若要使EO1∥O3F,则∠C为多少度?

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