Question

如图,BD、CD平分∠EBC与∠FCB,求证:∠BDC=90°-1/2∠A

Answer 1

解根据三角形外角定理，
∠EBC+∠FCB=(∠ACB+∠A)+(∠ABC+∠A)
=∠A+∠ACB+∠ABC+∠A
=2*90°+∠A
又因为BD,CD分别为角平分线
所以∠DBC+∠DCB=90°+1/2∠A

:∠BDC=2*90°-(90°+1/2∠A)
=90°-1/2∠A

Answer 2

∠BDC=180-∠DBC-∠DCB
=180-1/2∠CBE-1/2∠BCF
=180-1/2(A+ACD)-1/2(A+ABC)
=180-(2A+ABC+ACD)/2
=180-(2A+180-A)/2
=90-A/2

Answer 3

∠BDC=180-∠DBC-∠DCB=180-1/2∠CBE-1/2∠BCF=180-1/2(A+C)-1/2(A+B)=180-(2A+B+C)/2=90-1/2A