Вписанный угол равен половине дуги, на которую он опирается..
1.BO=АO=r ⇒ ΔAOB - равнобедренный ⇒ ∠ A {\displaystyle \angle A} = ∠ B {\displaystyle \angle B} ∠ A O C {\displaystyle \angle AOC} - внешний для ΔAOB ⇒ ∠ A O C {\displaystyle \angle AOC} = ∠ B {\displaystyle \angle B} + ∠ A {\displaystyle \angle A} = 2 ∠ B {\displaystyle \angle B} ⇒ ∠ B {\displaystyle \angle B} = 1/2 ∪ AC
2. ∠ A B C {\displaystyle \angle ABC} = ∠ A B D {\displaystyle \angle ABD} + ∠ C B D {\displaystyle \angle CBD} ∠ A B D {\displaystyle \angle ABD} = 1/2 ∪ AD ∠ C B D {\displaystyle \angle CBD} = 1/2 ∪ CD ⇒ ∠ A B C {\displaystyle \angle ABC} = 1/2 ∪ AD+1/2 ∪ CD = 1/2 ∪ AC
3. ∠ A B C {\displaystyle \angle ABC} = ∠ A B D {\displaystyle \angle ABD} - ∠ C B D {\displaystyle \angle CBD} ∠ A B D {\displaystyle \angle ABD} = 1/2 ∪ AD ∠ C B D {\displaystyle \angle CBD} = 1/2 ∪ CD ∠ A B C {\displaystyle \angle ABC} = 1/2 ∪ AD-1/2 ∪ CD = 1/2 ∪ AC