已知,△ABC中,AB=AC,∠BAC=108°,BD平分∠ABC,求证:BC=AB+CD

问题描述:

已知,△ABC中,AB=AC,∠BAC=108°,BD平分∠ABC,求证:BC=AB+CD

在BC上截取BE=AB,连接DE∵AB=AC∴∠ABC=∠C=(180°-108°)/2=36°∵BD平分∠ABC即∠ABD=∠DBEBD=BD,AB=BE∴△ABD≌△BDE∴∠BED=∠BAC=108°∴∠DEC=180°-∠BED=180°-108°=72°∴∠EDC=180°-∠DEC-∠C=180°-72...