Question

如图，点a.b.c.d在圆心o上，o在角d内部，四边形oabc为平行四边形，则角oad+角ocd=

补充图片

Answer 1

解：连接OB、OD
∵平行四边形OABC，OA＝OC
∴菱形OABC
∴OA＝AB，OC＝BC
∵OB＝OA＝OC
∴等边△OAB、等边△OBC
∴∠AOB＝∠COB＝60
∴∠AOC＝∠AOB+∠COB＝120
∴∠ADC＝∠AOC/2＝60
∵OD＝OA，OD＝OC
∴∠OAD＝∠ADO，∠OCD＝∠CDO
∴∠OAD+∠OCD＝∠ADO+∠CDO＝∠ADC＝60°

Answer 2

解：连接OB、OD
∵平行四边形OABC，OA＝OC
∴菱形OABC
∴OA＝AB，OC＝BC
∵OB＝OA＝OC
∴等边△OAB、等边△OBC
∴∠AOB＝∠COB＝60
∴∠AOC＝∠AOB+∠COB＝120
∴∠ADC＝∠AOC/2＝60
∵OD＝OA，OD＝OC
∴∠OAD＝∠ADO，∠OCD＝∠CDO
∴∠OAD+∠OCD＝∠ADO+∠CDO＝∠ADC＝60°