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  • Method of determination of 4 parallel and parallel lines 1. Doc

       2026-03-05 NetworkingName950
    Key Point:(i) 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 2 = 1 = 2 = 1 = 1 = 1 = 2 = 1 = 2 = 1 = 2 = 1 = 1 = 2 = 2 = 1 = 2 = 1 = 2 = 1 = 2 = 2 = 1 = 1 = 2 = 1 = 1 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2, eg equals meb, then ab//cd. Answer: (1)ll; ll; 4; 1(2)c(3)25 hint: when meb =mffd, ab∥cd, i. E. 1 = 2

    Parallel line determination method

    (i) 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 2 = 1 = 2 = 1 = 1 = 1 = 2 = 1 = 2 = 1 = 2 = 1 = 1 = 2 = 2 = 1 = 2 = 1 = 2 = 1 = 2 = 2 = 1 = 1 = 2 = 1 = 1 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2°, eg equals ∠meb, then ab//cd. Answer: (1)ll; ll; ∠4; ∠1(2)c(3)25° hint: when ∠meb = ∠ mffd, ab∥cd, i. E. ∠1 = ∠2 + 3. And eg evenly ∠meb, ∴∠2 = ∠3, ∴2 = ∠1 = 50° 2 = 25°. Example 3, (1) as shown, ∠2 = 3 ∠1 and ∠1 + 3 = 90°, test description ab//cd. ∵∠ 1 + 2 = 180° and 2 = 3 ∠1, 1 + 3 ∠1 = 180° 1 = 45° 1 + 3 = 90° 3 = 45° 1 = ∠ 3, ∴ ab/cd. (2) as graph, known ∠ 1 = ∠ 2, af split ∠ abn, bc split ∠ abn. Deposition: ∵ = q, ∠ q q q q q ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (3) (3) (3) (3) m (3) m m m (3) (3) (3) (3) (3) (3) (3) (3) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

     
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