(2014•大庆)如图,等腰△ABC中,AB=AC,∠BAC=36°,BC=1,点D在边AC上且BD平分∠ABC,设CD
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![](http://img.wesiedu.com/upload/4/70/470300311dd632a2002d1e72d1e77d82.jpg)
(1)求证:△ABC∽△BCD;
(2)求x的值;
(3)求cos36°-cos72°的值.
![(2014•大庆)如图,等腰△ABC中,AB=AC,∠BAC=36°,BC=1,点D在边AC上且BD平分∠ABC,设CD](/uploads/image/z/17528770-10-0.jpg?t=%EF%BC%882014%E2%80%A2%E5%A4%A7%E5%BA%86%EF%BC%89%E5%A6%82%E5%9B%BE%EF%BC%8C%E7%AD%89%E8%85%B0%E2%96%B3ABC%E4%B8%AD%EF%BC%8CAB%3DAC%EF%BC%8C%E2%88%A0BAC%3D36%C2%B0%EF%BC%8CBC%3D1%EF%BC%8C%E7%82%B9D%E5%9C%A8%E8%BE%B9AC%E4%B8%8A%E4%B8%94BD%E5%B9%B3%E5%88%86%E2%88%A0ABC%EF%BC%8C%E8%AE%BECD)
(1)∵等腰△ABC中,AB=AC,∠BAC=36°,
∴∠ABC=∠C=72°,
∵BD平分∠ABC,
∴∠ABD=∠CBD=36°,
∵∠CBD=∠A=36°,∠C=∠C,
∴△ABC∽△BCD;
(2)∵∠A=∠ABD=36°,
∴AD=BD,
∵BD=BC,
∴AD=BD=BC=1,
设CD=x,则有AB=AC=x+1,
∵△ABC∽△BCD,![](http://img.wesiedu.com/upload/a/72/a72ccc79f8e498e40bb28851fa9e9d60.jpg)
∴
AB
BD=
BC
CD,即
x+1
1=
1
x,
整理得:x2+x-1=0,
解得:x1=
−1+
5
2,x2=
−1−
5
2(负值,舍去),
则x=
−1+
5
2;
(3)过B作BE⊥AC,交AC于点E,
∵BD=BC,
∴E为CD中点,即DE=CE=
−1+
5
4,
在Rt△ABE中,cosA=cos36°=
AE
AB=
1+
−1+
5
4
−1+
∴∠ABC=∠C=72°,
∵BD平分∠ABC,
∴∠ABD=∠CBD=36°,
∵∠CBD=∠A=36°,∠C=∠C,
∴△ABC∽△BCD;
(2)∵∠A=∠ABD=36°,
∴AD=BD,
∵BD=BC,
∴AD=BD=BC=1,
设CD=x,则有AB=AC=x+1,
∵△ABC∽△BCD,
![](http://img.wesiedu.com/upload/a/72/a72ccc79f8e498e40bb28851fa9e9d60.jpg)
∴
AB
BD=
BC
CD,即
x+1
1=
1
x,
整理得:x2+x-1=0,
解得:x1=
−1+
5
2,x2=
−1−
5
2(负值,舍去),
则x=
−1+
5
2;
(3)过B作BE⊥AC,交AC于点E,
∵BD=BC,
∴E为CD中点,即DE=CE=
−1+
5
4,
在Rt△ABE中,cosA=cos36°=
AE
AB=
1+
−1+
5
4
−1+
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