【1】如图1,在三角形ABC中.∠ABC、∠ACB的平分线相交于点O,∠A=40°,求∠BOC的度数

(1) ∠BOC=180°-(∠1+∠2)=180°-(∠ABC+∠ACB)/2=180°-(180°-∠A)/2=90°+∠A/2=90°+20°=110°(2)∠B'O'C'=180°-(∠1+∠2)=180°-(∠DB'C'+∠EC'B')/2=180°-(180°-∠A'B'C'+180°-∠A'C'B')/2=(∠A'B'C'+∠A'C'B')/2=(180°-∠A')/2=70°(3)∠BOC+∠B'O'C'=180°即∠BOC与∠B'O'C'互补若∠A=∠A'=n°,∠BOC与∠B'O'C'之间仍然具有这样的关系移动右边三角形使A',B',C'分别与A,B,C重合∵∠OBC+∠OCB=(∠ABC+∠ACB)/2∠O'BC+∠O'CB=(∠DBC+∠ECB)/2∴∠OBC+∠OCB+∠O'BC+∠O'CB=(∠ABC+∠ACB+∠DBC+∠DCB)/2又∠ABC+∠DBC=180,∠ACB+∠ECB=180°∴∠OBC+∠OCB+∠O'BC+∠O'CB=180°在四边形OBO'C中,内角和为360°∴∠BOC+∠BO'C=180°即∠BOC+∠B'O'C'=180°