题目内容
探索与发现:
(1)已知A1B∥A2C,如图1所示,则∠A1+∠A2=______;
(2)已知A1B∥A3C,如图2所示,则∠A1+∠A2+∠A3=______;
(3)已知A1B∥A4C,如图3所示,则∠A1+∠A2+∠A3+∠A4=______;
(4)已知A1B∥AnC,如图4所示,则∠A1+∠A2+…+∠An=______;
(5)写出图2所得结论的推理过程.
(1)已知A1B∥A2C,如图1所示,则∠A1+∠A2=______;
(2)已知A1B∥A3C,如图2所示,则∠A1+∠A2+∠A3=______;
(3)已知A1B∥A4C,如图3所示,则∠A1+∠A2+∠A3+∠A4=______;
(4)已知A1B∥AnC,如图4所示,则∠A1+∠A2+…+∠An=______;
(5)写出图2所得结论的推理过程.
(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°.
∴∠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°.
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