题目内容

探索与发现:
(1)已知A1B∥A2C,如图1所示,则∠A1+∠A2=
180°
180°

(2)已知A1B∥A3C,如图2所示,则∠A1+∠A2+∠A3=
360°
360°

(3)已知A1B∥A4C,如图3所示,则∠A1+∠A2+∠A3+∠A4=
540°
540°

(4)已知A1B∥AnC,如图4所示,则∠A1+∠A2+…+∠An=
180°(n-1)
180°(n-1)

(5)写出图2所得结论的推理过程.
分析:(1)由A1B∥A2C,根据两直线平行,同旁内角互补,即可求得答案;
(2)首先过点A2作A2D∥A1B,由A1B∥A3C,即可得A2D∥A1B∥A3C,继而可求得答案;
(3)过点A2作A2D∥A1B,过点A3作A3E∥A1B,由A1B∥A4C,即可得A3E∥A2D∥A1B∥A4C,同理可求得∠A1+∠A2+∠A3+∠A4的值;
(4)同理,可求得∠A1+∠A2+…+∠An=180°(n-1).
(5)首先过点A2作A2D∥A1B,由A1B∥A3C,即可得A2D∥A1B∥A3C,继而可求得答案.
解答:解:(1)∵A1B∥A2C,
∴∠A1+∠A2=180°;

(2)过点A2作A2D∥A1B,
∵A1B∥A3C,
∴A2D∥A1B∥A3C,
∴∠A1+∠A1A2D=180°,∠DA2A3+∠A3=180°,
∴∠A1+∠A1A2A3+∠A3=360°;

(3)过点A2作A2D∥A1B,过点A3作A3E∥A1B,
∵A1B∥A4C,
∴A3E∥A2D∥A1B∥A4C,
∴∠A1+∠A1A2D=180°,∠DA2A3+∠A2A3E=180°,∠EA3A4+∠A4=180°;
∴∠A1+∠A2+∠A3+∠A4=540°;

(4)过点A2作A2D∥A1B,过点A3作A3E∥A1B,…
∵A1B∥AnC,
∴A3E∥A2D∥…A1B∥AnC,
∴∠A1+∠A1A2D=180°,∠DA2A3+∠A2A3E=180°,∠EA3A4+∠A4=180°,…;
∴∠A1+∠A2+…+∠An=180°(n-1).
故答案为:(1)180°;(2)360°;(3)540°;(4)180°(n-1).

(5)过点A2作A2D∥A1B,
∵A1B∥A3C,
∴A2D∥A1B∥A3C,
∴∠A1+∠A1A2D=180°,∠DA2A3+∠A3=180°,
∴∠A1+∠A1A2A3+∠A3=360°.
点评:此题考查了平行线的性质.此题难度适中,注意掌握辅助线的作法,注意数形结合思想的应用.
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