已知如图四边形abcd中,角abc和角dcb的平分线交于点o,试说明:角BOC=二分之一(角A+D)

Question

已知如图四边形abcd中,角abc和角dcb的平分线交于点o,试说明:角BOC=二分之一(角A+
D)

Answer 1

证明：∵BO、CO分别平分∠ABC、∠BCD,
∴∠OBC+∠OCB=1/2(∠ABC+∠BCD),
∵∠A+∠ABC+∠BCD+∠D=360°,
∴∠BOC=180°-(∠OBC-∠OCB)
=180°-1/2［360°-(∠A+∠D)］
=180°-180°+1/2(∠A+∠D)
=1/2(∠A+∠D).