△ABC中,AB=AC,∠A=36°,CD是角平分线,则△DBC的面积与△ABC的面积比是?

Question

△ABC中,AB=AC,∠A=36°,CD是角平分线,则△DBC的面积与△ABC的面积比是?

Answer 1

∠A＝36
∠B=∠ACB=(180-∠A)/2=(180-36)/2=72
∠BCD=∠ACD=∠ACB/2=72/2=36
∠ADC=180-∠B-∠ACD=180-72-36=72
设BC=x,BD=y
则:CD=BC=x,AD=CD=x
AB=AD+BD=x+y,AC=AB=x+y
易证,△ABC~△CBD
所以,AB/BC=CB/BD
(x+y)/x=x/y
1+y/x=x/y
设y/x=z
则:1+z=1/z
z^2+z-1=0
z=(-1±√5)/2
把不合题意的负数去掉
则:z=(√5-1)/2
因为△ABC~△CBD
所以:
△DBC的面积与△ABC的面积的比值
=(BD/BC)^2
=(y/x)^2
=z^2
=((√5-1)/2)^2
=(6-2√5)/4
=(3-√5)/2