AB=AC,BAC=70,D为三角形ABC外一点,AD=AC,角BCD=120,求角BDC(全部分都送上~~帮下忙吧~)

∠ACB=1/2(180-70)=55
∠ADC=∠ACD=120-55=65
∠CAD=180-2×65=50
∠BAD=70+50=120
∠BDA=1/2(180-120)=30
∠BDC=65-30=35(度)

AB=AC,角BAC=70,角ACB=55,角BCD=120,AD=AC,角ACD=65,角CAD=50,角BAD=BAC+CAD=120,AB=AC,AD=AC,AB=AC,角BDA=30,角BDC=角CDA-角BDA=65-30=35

因为:AB=AC且角BAC=70
所以:角ABC=角ACB=55
又因为角BCD=120
所以角ACD=角ADC=65 且角CAD=50
又因为AB=AC=AD
所以三角形ABD为等腰三角形,角ADB=[180-(70+50)]/2=30
所以角BDC=角ADC-角ADB=65-30=35
回答完毕