Question

如图，直线AB和CD相交于点O∠COE=90°，OF平分∠AOE,∠COF=34°，求∠BOD的度数

Answer 1

解：
∵∠COE＝90, ∠COF＝34
∴∠EOF＝∠COE-∠COF＝90-34＝56
∵OF平分∠AOE
∴∠EOF＝∠AOE/2
∴∠AOE/2＝56
∴∠AOE＝112
∴∠AOC＝∠AOE-∠COE＝112-90＝22
∵直线AB和CD相交于点O
∴∠BOD＝∠AOC＝22

Answer 2

cof=34,then eof=56.
aoc=aof-cof=eof-cof=56-34=22.
eob=180-aoe=180-2eof=68.
bod=180-cob=180-coe-eob=180-90-68=22.