题目内容：

已知四边形ABCD中,DE,CE分别平分为∠ADC,∠BCD,∠A=100°,∠B=92°,求∠DEC的度数

优质解答

DE平分∠ADC
∠ADE = ∠CDE = 1/2 ∠ADC
CE平分∠BCD
∠BCE = ∠DCE = 1/2 ∠BCD
延长DE交BC于F
∠FEC =∠CDE + ∠DCE = 1/2 (∠ADC + ∠BCD) = 1/2 (360° -∠A - ∠B)
= 1/2 (360° -100 °- 92°) = 84°
∠DEC = 180° - 84° = 96°