已知三角形ABC的三个外角,角DAC、角CBG和角ACF,AE和CE平分角DAC和角ACF.试说明角CBG=2角E
问题描述:
已知三角形ABC的三个外角,角DAC、角CBG和角ACF,AE和CE平分角DAC和角ACF.试说明角CBG=2角E
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答
角CBG=180°-角ABC角E=180°-1/2(角DAC+角ACF)角DAC=180°-角CAB角ACF=180°-角ACB角ABC=180°-(角CAB+角ACB)所以角E=180°-1/2(角DAC+角ACF)=180°-1/2(180°-角CAB+180°-角ACB)=1/2角CAB+角ACB所以角CAB+角AC...