(1)∥;平行(2)∵AF∥BE∴∠A+∠ABE=180°∵CD∥BE∴∠C+∠CBE=180°∵∠ABC=∠ABE+∠CBE∴∠A+∠ABC+∠C=180°+180°=360°(3)∵AF∥BE,∠F=50°∴∠BEF=180°-50°=130°同理可得,∠BED=130°∴∠POQ=360°-130°-130°=100°
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