Question

如图,三角形ABC中角A=角ABD,角C=角BDC=角ABC求角BDC的度数

Answer 1

∠BDC=∠A+∠ABD

∵∠A=∠ABD,∠C=∠BDC

∴∠C=2∠A

又∵∠C=∠ABC,∴∠ABC+∠C+∠A=∠C+∠C+1/2*∠C=2.5∠C=180°

∴∠C=72°=∠BDC