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Special training of junior high school mathematics circle
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Answer: As an auxiliary line: the vertical line O2Q 1)∫∠EBF =∠FCE (circle angle) connecting BF, CE, MN, NO2 and MO2 with the center of MN; And ∠ CBF =180-45-∠ EBF ∠ BCE =180-45-∠ FCE; ∴∠CBF=∠BCE (circumferential angle on the same circle), ∴ Be = cf.2) ∵∠ P = 45 (circumferential angle of vertical diameter), ∠ PNB = 30 (known), then ∞ O2 is the center of circle 2.