Question

(1/2)在三角形ABC中,AB=AC,AD垂直于BC,垂足为点D,AN是三角形ABC外角角CAM的平分线,CE垂直于A

Answer 1

证明:AB=AC,AD⊥BC,则BD=CD;AD平分∠BAC.

又AN平分∠CAM,则∠CAD+∠CAE=90度,又∠ADC=∠AEC=90°.

则四边形ADCE为矩形,得AF=FC.故DF∥AB,且DF=AB/2.

当∠BAC=90°时:∠CAD=45°=∠ACD,AD=CD.故矩形ADCE为正方形.