如图四,在△ABC中,AB=AC,∠A=36°,BD平分角ABC交AC于点D,DE平分角BDC交BC于点E,则EC/AD=
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![如图四,在△ABC中,AB=AC,∠A=36°,BD平分角ABC交AC于点D,DE平分角BDC交BC于点E,则EC/AD=](/uploads/image/z/13858654-22-4.jpg?t=%E5%A6%82%E5%9B%BE%E5%9B%9B%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CAB%3DAC%2C%E2%88%A0A%3D36%C2%B0%2CBD%E5%B9%B3%E5%88%86%E8%A7%92ABC%E4%BA%A4AC%E4%BA%8E%E7%82%B9D%2CDE%E5%B9%B3%E5%88%86%E8%A7%92BDC%E4%BA%A4BC%E4%BA%8E%E7%82%B9E%2C%E5%88%99EC%2FAD%3D)
如图四,在△ABC中,AB=AC,∠A=36°,BD平分角ABC交AC于点D,DE平分角BDC交BC于点E,则EC/AD=
如图四,在△ABC中,AB=AC,∠A=36°,BD平分角ABC交AC于点D,DE平分角BDC交BC于点E,则EC/AD=
如图四,在△ABC中,AB=AC,∠A=36°,BD平分角ABC交AC于点D,DE平分角BDC交BC于点E,则EC/AD=
∵在△ABC中,AB=AC,∠A=36°,
∴∠ABC=∠C=72°,
∵BD平分∠ABC交AC于点D,
∴∠ABD=∠CBD=36°,
∴∠ABD=∠A,
∴AD=BD,
∴∠BDC=180°-∠CBD-∠C=72°,
∴∠BDC=∠C,
∴BD=BC,
∴AD=BD=BC,
∵DE平分∠BDC交BC于点E,
∴∠CDE=∠BDE=36°,
∴∠BDE=∠CBD,
∴BE=DE,
∴∠DEC=180°-∠C-∠CDE=72°,
∴DE=CD=BE,
∴△DEC∽△BDC,
∴
CD
BC
=
EC
CD
,
设CD=x,则EC=BC-BE=AD-CD=AD-x,BC=BD=AD,
∴
x
AD
=
AD−x
x
,
解得:x=
−1+
5
2
AD,
∴EC=
3−
5
2
AD,
∴
EC
AD
=
3−
5
2
.
故答案为:
3−
5
2
.