Заполни пропуски в доказательстве

На рисунке \(\angle AOB = \angle COD\) , \(\angle AOC = \angle COE\) . Докажи, что \(\angle BOC = \angle DOE\) .

\(\angle AOC = \angle AOB + \angle \) [ \(AOC\) | \(AOD\) | \(AOE\) | \(BOC\) | \(BOD\) | \(BOE\) | \(COE\) ]

\(\angle COE = \angle COD + \angle \) [ \(AOC\) | \(AOD\) | \(AOE\) | \(BOD\) | \(BOE\) | \(COE\) | \(DOE\) ]

\(\angle AOC = \angle COE\) (по условию)

Следовательно, \(\angle AOB + \angle \) [ \(AOC\) | \(AOD\) | \(AOE\) | \(BOC\) | \(BOD\) | \(BOE\) | \(COE\) ] \(= \angle COD + \angle \) [ \(AOC\) | \(AOD\) | \(AOE\) | \(BOD\) | \(BOE\) | \(COE\) | \(DOE\) ]

Т. к. \(\angle AOB = \angle COD\) (по условию), то \(\angle BOC=\angle DOE\)