△ABC中,AB=AC=10,∠A=36°,BD是角平分线交AC于D,则DC=______.
人气:286 ℃ 时间:2020-03-27 04:21:45
解答
∵AB=AC,∠A=36°,
∴∠C=∠ABC=
(180°-∠A)=72°,
∵BD平分∠ABC,
∴∠ABD=∠DBC=36°,
∴∠BDC=∠A+∠ABD=72°,
∴AD=BD,BD=BC,
∴AD=BD=BC,
设CD=x,则AD=BD=BC=10-x,
∵∠A=∠DBC=36°,∠C=∠C,
∴△BDC∽△ABC,
∴
=
,
∴
=
,
x
1=15+5
,x
2=15-5
,
∵CD<AC,AC=10,
∴x
1=15+5
舍去,
故答案为:15-5
.
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