初中数学.各位帮帮忙
来源:百度知道 编辑:UC知道 时间:2024/05/27 13:23:54
已知:三角形ABC的角B和角C的平分线BE,CF交于I. 求证:角BIC=90度+二分之一角A. 谢谢!~~~~~~
证明:
角BIC=180-角IBC-角ICB
=180-1/2角ABC-1/2角ACB
=180-1/2(角ABC+角ACB)
=180-1/2(180-角A)
=90+1/2角A
求证:角BIC=90度+二分之一角A:
角BIC=180-角IBC-角ICB
=180-1/2角ABC-1/2角ACB
=180-1/2(角ABC+角ACB)
=180-1/2(180-角A)
=90+1/2角A
投我一票!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
连接AI,延长AI交BC于D
角BIC=角BID+角CID
=角BAC+1/2角ABC+1/2角ACB
=1/2(角ABC+角ACB+角BAC)+1/2角BAC
=(1/2)*180+1/2角BAC
=90+1/2角BAC
角BIC=角BID+角CID
=角BAC+1/2角ABC+1/2角ACB
=1/2(角ABC+角ACB+角BAC)+1/2角BAC
=(1/2)*180+1/2角BAC
=90+1/2角BAC
证明:
角BIC=180-角IBC-角ICB
=180-1/2角ABC-1/2角ACB
=180-1/2(角ABC+角ACB)
=180-1/2(180-角A)
=90+1/2角A
连接AI,延长AI交BC于D
角BIC=角BID+角CID
=角BAC+1/2角ABC+1/2角ACB
=1/2(角ABC+角ACB+角BAC)+1/2角BAC
=(1/2)*180+1/2角BAC
=90+1/2角BAC