如图所示,在△ABC中,∠B=1/2∠BAC,△ABC的外角平分线交BC的延长线于点D,若∠ADC=1/3∠CAD,求∠B的度数.

∠B=1/2∠BAC∠B=1/2(180°-∠BAC)+∠ADC=1/2(180°-∠BAC)+1/3∠CAD=1/2(180°-∠BAC)+1/3(180°-1/2(180°-∠BAC))=60°+(1-1/3)*1/2(180°-∠BAC)=60°+60°-2/3*(1/2∠BAC)=120°-2/3*∠B故而∠B=120°/(1+2/3)=...我想要“∵ ∴”形式的麻烦了谢谢……这有什么区别么？……好吧：∵△ABC的外角平分线交BC的延长线于点D，∠ADC=1/3∠CAD ∴∠B=1/2(180°-∠BAC)+∠ADC=1/2(180°-∠BAC)+1/3∠CAD=1/2(180°-∠BAC)+1/3(180°-1/2(180°-∠BAC))=60°+(1-1/3)*1/2(180°-∠BAC)=60°+60°-2/3*(1/2∠BAC)∵ ∠B=1/2∠BAC ∴∠B=60°+60°-2/3*(1/2∠BAC)=120°-2/3*∠B ∴∠B=120°/(1+2/3)=72°