如图1,已知△ABC中,∠ABC为钝角,BE平分∠ABC,CE平分△ABC的外角∠ACD,BE、CE交于点E,(1)试说明∠E=∠A; (2)如图2,延长AB、EC交于点F,若∠AGE=85°,∠F=25°,求∠A的度数; (3)如图3,将△ABC的∠A沿着直线M
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![如图1,已知△ABC中,∠ABC为钝角,BE平分∠ABC,CE平分△ABC的外角∠ACD,BE、CE交于点E,(1)试说明∠E=∠A; (2)如图2,延长AB、EC交于点F,若∠AGE=85°,∠F=25°,求∠A的度数; (3)如图3,将△ABC的∠A沿着直线M](/uploads/image/z/8896541-5-1.jpg?t=%E5%A6%82%E5%9B%BE1%2C%E5%B7%B2%E7%9F%A5%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0ABC%E4%B8%BA%E9%92%9D%E8%A7%92%2CBE%E5%B9%B3%E5%88%86%E2%88%A0ABC%2CCE%E5%B9%B3%E5%88%86%E2%96%B3ABC%E7%9A%84%E5%A4%96%E8%A7%92%E2%88%A0ACD%2CBE%E3%80%81CE%E4%BA%A4%E4%BA%8E%E7%82%B9E%2C%281%29%E8%AF%95%E8%AF%B4%E6%98%8E%E2%88%A0E%EF%BC%9D%E2%88%A0A%EF%BC%9B+%282%29%E5%A6%82%E5%9B%BE2%2C%E5%BB%B6%E9%95%BFAB%E3%80%81EC%E4%BA%A4%E4%BA%8E%E7%82%B9F%2C%E8%8B%A5%E2%88%A0AGE%3D85%C2%B0%2C%E2%88%A0F%3D25%C2%B0%2C%E6%B1%82%E2%88%A0A%E7%9A%84%E5%BA%A6%E6%95%B0%EF%BC%9B+%283%29%E5%A6%82%E5%9B%BE3%2C%E5%B0%86%E2%96%B3ABC%E7%9A%84%E2%88%A0A%E6%B2%BF%E7%9D%80%E7%9B%B4%E7%BA%BFM)
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