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Равные отрезкт ab и сd точкой пересечения о делятся в отношении ао: ов=со: do=2: 1 а) докажите равенство треугольников bad и dcb б) найдите угол obc, если угол oda=40

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samira57
1) в треугольниках BAD и DCB AB и DC равны по условию,DB- общая,DOB - равносторонний и его углы при основании равны.значит,BAD=DCB по двум сторонам и углу между ними

2)треугольники oad и ocb тоже равны,по двум сторонам и углу между: углы doa и boc равны как вертикальные,do=ob, ao=oc.углы obc и oda соответственные углы в равных треугольниках => равны. угол OBC=40°
Равные отрезкт ab и сd точкой пересечения о делятся в отношении ао: ов=со: do=2: 1 а) докажите равен
Apresov

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mihailpolehin8

ΔОСВ равносторонний. В нем углы при вершинах С и В равны.т.к. ОС=ОВ= радиусы одной окружности. Т.е.  равнобедренный получается. но поскольку углы С и В еще и по 60°в, то и угол О в этом треугольнике 60 °. Тогда  внешний угол АОВ равен сумме двух внутренних ∠ В и ∠С, с ним не смежными, т.е. он равен 60°+60°=120°, а тогда в равнобедренном треуг. АОВ ∠ А =∠ В= 30 °,

(180°-120°)/2=30°,  как углы при основании равнобедренного ΔАОВ, т.к. АО и ВО радиусы одной окружности и ∠DАС = 90°, т.к. радиус, проведенный в точку касания перпендикулярен касательной АD, значит, искомый  ∠ DАВ =90°-30°=60°

ответ 60 °

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Равные отрезкт ab и сd точкой пересечения о делятся в отношении ао: ов=со: do=2: 1 а) докажите равенство треугольников bad и dcb б) найдите угол obc, если угол oda=40
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