如圖,△ABC中,AB=AC,∠BAC=90°,AE=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226720327.png)
AC,BD=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226720327.png)
AB,點(diǎn)F在BC上,且CF=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226720327.png)
BC.求證:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240532267988331.png)
(1)EF⊥BC;
(2)∠ADE=∠EBC.
證明:設(shè)AB=AC=3a,
則AE=BD=a,CF=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226814344.png)
a.
(1)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226829546.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226845560.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226860447.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226892547.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226907545.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226860447.png)
.
又∠C為公共角,故△BAC∽△EFC,
由∠BAC=90°.∴∠EFC=90°,∴EF⊥BC.
(2)由(1)得EF=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226814344.png)
a,
故
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226970551.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226985471.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053227001413.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053227032577.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053227048542.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053227001413.png)
,
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053226970551.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824053227094578.png)
.∵∠DAE=∠BFE=90°,
∴△ADE∽△FBE,∴∠ADE=∠EBC.
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