题目内容
如图,过点B作DB⊥AB于点B,使BD=
AB,在AD上截取DE=BD,在AB上截取AC=AE,则
=______.

1 |
2 |
BC |
AB |

DB=x,
∵BD=
AB,
∴AB=2x,
∴由勾股定理得:AD=
x,
∵DE=BD,AC=AE,
∴DE=DB=x,AC=AE=AD-DE=(
-1)x,
BC=AB-AC=2x-(
-1)x=(3-
)x,
∴
=
=
,
故答案为:
.
∵BD=
1 |
2 |
∴AB=2x,
∴由勾股定理得:AD=
5 |
∵DE=BD,AC=AE,
∴DE=DB=x,AC=AE=AD-DE=(
5 |
BC=AB-AC=2x-(
5 |
5 |
∴
BC |
AB |
(3-
| ||
2x |
3-
| ||
2 |
故答案为:
3-
| ||
2 |

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