Question

已知在三角形ABC中,AB=AC,角BAC和角ACB的平分线,角ADC=130度,求角CAB

Answer 1

∵AD ,CD是角平分线
∠ADC=180°-1/2(∠ACB +∠瞎饥陆BAC )=130°
∴1/2(∠ACB +∠BAC )=50°
∴肢型∠ACB +∠BAC=100°
∴∠ABC=80°
∵AB=AC
∴∠ACB=80°
∴∠磨顷BAC=20°