Question

几何,已知在三角形ABC中,角ABC=3角C,AD是角BAC的角平分线,BE垂直AD于点E,求证：BE=1/2(AC-A

Answer 1

延长BE交AC于F

AD平分BAC,BF丄AD

易知：AB=AF,BE=EF

∠ABF=∠AFB

=>

AC-AB = FC

2BE=BF

则只需证：BF=CF

只需证：∠C=∠CBF

设∠C=X,∠ABF=∠AFB=Y

=>

∠ABC=3X

=>

∠CBF=3X-Y

而

∠CBF+∠C=∠AFB

=>

3X-Y + X=Y

=>

Y=2X

=>

∠C=∠CBF

得证