Question

一道初一的图形题!如图,BD,CD分别是△ABC的内角∠ABC、∠ACB的平分线,请说明∠BDC与∠A之间的等量关系是∠

Answer 1

∠BDC=180°-1/2∠ABC-1/2∠ACD

=180°-1/2(∠ABC+∠ACD)

=180°-1/2(180°-∠A)

=90°+1/2∠A

因为BD CD分别为∠ABC ∠ACB角平分线

所以1/2∠ABC+1/2∠ACB=1/2（∠ABC+∠ACB)=∠DBC+∠DCB

∠DBC+∠DCB=1/2(180-∠A）=1/2(180-60）=60

因为三角形内角和为180

所以∠BDC=180-∠BCD-∠DBC=180-60=120