(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=x.(1)求
(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=x.(1)求证:△ABC∽△BCD;(2)求x的...
(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=x.(1)求证:△ABC∽△BCD;(2)求x的值;(3)求cos36°-cos72°的值.
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(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,
∴
=
,即
=
,
整理得:x2+x-1=0,
解得:x1=
,x2=
(负值,舍去),
则x=
;
(3)过B作BE⊥AC,交AC于点E,
∵BD=BC,
∴E为CD中点,即DE=CE=
,
在Rt△ABE中,cosA=cos36°=
=
∴∠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,
∴
AB |
BD |
BC |
CD |
x+1 |
1 |
1 |
x |
整理得:x2+x-1=0,
解得:x1=
?1+
| ||
2 |
?1?
| ||
2 |
则x=
?1+
| ||
2 |
(3)过B作BE⊥AC,交AC于点E,
∵BD=BC,
∴E为CD中点,即DE=CE=
?1+
| ||
4 |
在Rt△ABE中,cosA=cos36°=
AE |
AB |
1+
| ||||
|