Question

在△ABC中，∠ABC=∠C，点D是AC边上一点，∠A=∠ADB,∠DBC=30°。∠BDC的度数。

Answer 1

∠A+∠ABC+∠C=180° ∠ABC=∠C ∴∠A+2∠ABC=180° (1)
∠A+∠ABD+∠ADB=180° ∠A=∠ADB ∴2∠A+∠ABD=180° ∠ABD=∠ABC+∠DBC=∠ABC+30° ∴2∠A+∠ABC+30°=180° (2)
解（1）（2）组成方程组，得： ∠A=30° ∠ABC=75°

∠BDC=∠A=30°

Answer 2

∠BDC=∠A+∠ABD=∠ADB+∠ABD=∠C+∠DBC+∠ABD=∠C+∠ABC=2∠C

∵∠BDC+∠C+∠DBC=180°

∴∠BDC+∠C=150°

故∠BDC=100°

Answer 3

∠BDC=180-30-∠C=150-(180-∠A)/2=60+1/2∠A
∵∠A=∠ADB ∴60+1/2∠A+∠A=180 ∠A=80 。∠BDC=100

Answer 4

∠BDC=100°

Answer 5

∠BDC=80

Answer 6

50度