Question

在三角形ABC中,AB=AC,角BAC=90度,BF平分角ABC,CD垂直BF,交BF的延长线于点D,求证:BF=2CD

Answer 1

tanπ/4=(2tanπ/8)/[1-(tanπ/8)^2]=1

设tanπ/8=x,则2x/(1-x^2)=1

x^2+2x-1=0

x=-1±√2

因为tanπ/8>0

所以tanπ/8=(√2)-1

首先tan(π/8) = √2 - 1

CD = （√2 - 1）BD 即 BD = （√2 +1）CD FD = （√2 - 1）CD

BF = BD - FD = 2CD