Rt△ABC中,∠BAC=90°,AB=AC=2.以AC為一邊,在△ABC外部作等腰直角三角形ACD,則線段BD的長為 .
【答案】
分析:分情況討論,①以A為直角頂點,向外作等腰直角三角形DAC;②以C為直角頂點,向外作等腰直角三角形ACD;③以AC為斜邊,向外作等腰直角三角形ADC.分別畫圖,并求出BD.
解答:解:①以A為直角頂點,向外作等腰直角三角形DAC,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/images0.png)
∵∠DAC=90°,且AD=AC,
∴BD=BA+AD=2+2=4;
②以C為直角頂點,向外作等腰直角三角形ACD,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/images1.png)
連接BD,過點D作DE⊥BC,交BC的延長線于E.
∵△ABC是等腰直角三角形,∠ACD=90°,
∴∠DCE=45°,
又∵DE⊥CE,
∴∠DEC=90°,
∴∠CDE=45°,
∴CE=DE=2×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/1.png)
,
在Rt△BAC中,BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/2.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/3.png)
,
∴BD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/5.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/6.png)
;
③以AC為斜邊,向外作等腰直角三角形ADC,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/images9.png)
∵∠ADC=90°,AD=DC,且AC=2,
∴AD=DC=ACsin45°=2×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/8.png)
,
又∵△ABC、△ADC是等腰直角三角形,
∴∠ACB=∠ACD=45°,
∴∠BCD=90°,
又∵在Rt△ABC中,BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/9.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/10.png)
,
∴BD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/13.png)
.
故BD的長等于4或2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/14.png)
或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202637212267047/SYS201311032026372122670011_DA/15.png)
.
點評:分情況考慮問題,主要利用了等腰直角三角形的性質(zhì)、勾股定理等知識.