如圖,在△
ABC中,
AD是
BC上的高,
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,
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(1) 求證:
AC=BD;
(2)若
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,
BC=12,求
AD的長.
(1)∵
AD是
BC上的高,∴
AD⊥
BC.
∴∠
ADB=90°,∠
ADC=90°. …………………………………………1分
在Rt△
ABD和Rt△
ADC中,
∵
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=
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,
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=

…………………………………………3分
又已知

∴
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=

.∴
AC=BD. ………………………………4分
(2)在Rt△
ADC中,

,故可設(shè)
AD=12
k,
AC=13
k.
∴
CD=

=5
k. ………………………………5分
∵
BC=BD+CD,又
AC=BD,∴
BC=13
k+5
k=18
k ………………………………6分
由已知
BC=12, ∴18
k=12.∴
k=
. ………………………………7分
∴
AD=12
k=12

=8. ……………………………8分
(1)在直角三角形中,表示
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,根據(jù)它們相等,即可得出結(jié)論
(2)利用

和勾股定理表示出線段長,根據(jù)
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,求出
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長
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