题目内容
| EC |
| EB |
| 1 |
| 3 |
| ED |
| EA |
| 1 |
| 2 |
| DC |
| AB |
考点:与圆有关的比例线段
专题:立体几何
分析:由已知得△ECD∽△EAB,
=
=
=
,从而BE=
DE,由此能求出
=
=
=
.
| DC |
| AB |
| EC |
| AE |
| DE |
| BE |
| ||
| 2DE |
| 6 |
| DC |
| AB |
| DE |
| BE |
| DE | ||
|
| ||
| 6 |
解答:
解:∵A,B,C,D四点共圆
∴∠EDC=∠EBF,
又∵∠DEC=∠AEC
∴△ECD∽△EAB,
又∵
=
,
=
,
∴
=
=
=
,
∴BE2=6DE2,即BE=
DE,
∴
=
=
=
.
故答案为:
.
∴∠EDC=∠EBF,
又∵∠DEC=∠AEC
∴△ECD∽△EAB,
又∵
| EC |
| EB |
| 1 |
| 3 |
| ED |
| EA |
| 1 |
| 2 |
∴
| DC |
| AB |
| EC |
| AE |
| DE |
| BE |
| ||
| 2DE |
∴BE2=6DE2,即BE=
| 6 |
∴
| DC |
| AB |
| DE |
| BE |
| DE | ||
|
| ||
| 6 |
故答案为:
| ||
| 6 |
点评:本题考查两线段长的比值的求法,是中档题,解题时要注意圆的性质的合理运用.
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