求：在三角形ABC中,角B的平分线和角C的外角平分线相交于点D,若角D等于40度,则

估计想求:∠A的度数.
设∠C的外角为∠ACE;设∠ABD=∠DBC=X(度),∠ACD=∠ECD=Y(度).
∠ACE=∠A+∠ABC,即2Y=∠A+2X,则2X=2Y-∠A.--------------------(1)
∠D+∠CBD=∠DCE,即∠D+X=Y,则2∠D+2X=2Y;-------------(2)
所以,2∠D+(2Y-∠A)=2Y,2∠D=∠A,∠A=80度.