如图,△ABC中,AC=BC,∠BAC的外角平分线交BC的延长线于点D,若∠ADC=1/2∠CAD,则∠ABC=_度.

问题描述:

如图,△ABC中,AC=BC,∠BAC的外角平分线交BC的延长线于点D,若∠ADC=

1
2
∠CAD,则∠ABC=______度.

设∠CDA=α,∵∠ADC=12∠CAD,∴∠CAD=2α,而AD平分∠CAE,∴∠CAD=∠DAE=2α,而∠EAD=∠B+∠ADC,∴∠B=2α-α=α,又∵AC=BC,∴∠BAC=∠B=α,在△ABD中,∴∠B+∠CAB+∠CAD+∠ADC=180°,即α+α+2α+α=180...