三角形ABC中,角B与角A的平分线交于点D,则角BDA等于( )A.180-角CB.90+角CC.180-1/2角CD.90+1/2角C
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![三角形ABC中,角B与角A的平分线交于点D,则角BDA等于( )A.180-角CB.90+角CC.180-1/2角CD.90+1/2角C](/uploads/image/z/14283081-9-1.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%A7%92B%E4%B8%8E%E8%A7%92A%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9D%2C%E5%88%99%E8%A7%92BDA%E7%AD%89%E4%BA%8E%EF%BC%88+%EF%BC%89A.180-%E8%A7%92CB.90%EF%BC%8B%E8%A7%92CC.180-1%2F2%E8%A7%92CD.90%EF%BC%8B1%2F2%E8%A7%92C)
三角形ABC中,角B与角A的平分线交于点D,则角BDA等于( )A.180-角CB.90+角CC.180-1/2角CD.90+1/2角C
三角形ABC中,角B与角A的平分线交于点D,则角BDA等于( )
A.180-角C
B.90+角C
C.180-1/2角C
D.90+1/2角C
三角形ABC中,角B与角A的平分线交于点D,则角BDA等于( )A.180-角CB.90+角CC.180-1/2角CD.90+1/2角C
d
A+B+C=180
(A+B)/2=90-C/2
180-(A+B)/2=90+C/2
180-(A+B)/2=角BDA
C+A+B=180
C'+A/2+B/2=180
连立的 D
∠DAB+∠DBA=(A+B)/2=(180-C)/2
∠DBA=180-∠DAB-∠DBA=180-90+C/2=90+C/2