Question

如图,在△ABC中,(AB>AC),若直线AD平分∠BAC外角,且与△ABC的外接圆相交于点E,与BC边相交于D.

.求证:ABXAC=ADXAE
.

Answer 1

AD平分∠BAC==>∠清棚枣答拆BAD=∠EAC,
∠ABC=∠AEC(弧相等，圆周角相等）

==>△ABD 和△和慎AEC 相似==>AD/AC=AB/AE==>AD*AE=AB*AC

Answer 2

由AD平分∠BAC外角，得∠灶者粗DAB=∠EAC 又由圆隐镇的内嫌唤接四边形的性质得：∠ABD=∠AEC
得△ABD相似△AEC 得 AB:AE=AD:AC 从而得到ABXAC=ADXAE

Answer 3

AD平分∠BAC==>∠清棚枣答拆BAD=∠EAC,
∠ABC=∠AEC(弧相等，圆周角相等）

==>△ABD 和△和慎AEC 相似==>AD/AC=AB/AE==>AD*AE=AB*AC

Answer 4

AD平分∠BAC==>∠清棚枣答拆BAD=∠EAC,
∠ABC=∠AEC(弧相等，圆周角相等）

==>△ABD 和△和慎AEC 相似==>AD/AC=AB/AE==>AD*AE=AB*AC