Question

如图，AB、AC是圆O的两条弦，延长CA到D，使AD=AB，若∠ADB=40°，求∠BOC的度数

Answer 1

解：

∵AB=AD

∴∠BAD=∠D

∵∠D=40°

∴∠ADB=40°

∴∠CAB=40°+40°=80°

∵∠BOC是弧BC对的圆心角

∴∠BOC=2∠BAC=160°

Answer 2

因为AD=AB，所以三角形ADB为等腰三角形，∠BAC=∠ADB+∠ABD=80°
如果你学了圆心角与圆周角的知识，那么直接得出∠BOC=2∠BAC=160°
如果没学，连接AO并延长交圆于E点，连CE BE，由O是两个直角三角形ABE ACE斜边中点即可得证。

Answer 3

解证： 如图; 连AO并延长交圆O

                  ∵ AD=AB 

               ∴ ∠ABD=∠ADB=40°

                ∴ ∠CAB=∠ABD+∠ADB=80°

                    即：∠1+∠2=80°

                     在圆O中 ∵OA=OB

                            ∴ ∠OBA=∠1

                               ∴ ∠3=2∠1

                               同理∠4=2 ∠2

                           ∴∠3+∠4=2（∠1+∠2）=2×80°=160°

                                   即：∠BOC=160°

Answer 4

角BOC＝2＊角BAC＝4＊角ADB＝160

Answer 5

也是40度 角CAB与之互补