Question

如图DC平分∠ADB,EC平分∠AEB,∠DAE=50°,∠DBE=150°,求∠DCE的度数

Answer 1

∠DCE=100°

连接DE,则∠BDE+∠BED=180°-∠DBE=30°,

则∠ADB+∠AEB=180°-∠DAE-(∠BDE+∠BED)=180°-50°-30°=100°,

则∠DCE=180°-∠CDE-∠CED

=180°-(∠CDB+∠BDE)-（∠CEB+∠BED)

=180°-(∠BDE+∠BED)-(∠ADB/2+∠AEB/2)

=180°-(∠BDE+∠BED)-(∠ADB+∠AEB)/2

=180°-30°-100°/2

=100°