题目内容
As in right figure,in a quadrilateral ABCD,we have its diagonal AC bisects∠DAB,and AB=21,AD=9,BC=DC=10,then the distance from point C to line AB is ,and the length of AC is . (英汉词典:quadrilateral 四边形;bisect 平分)![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163028550493504/SYS201310221630285504935022_ST/images0.png)
【答案】分析:作辅助线构建直角三角形,求证△CED≌△CFB,即可得DE=FB,即可求得BF,根据BF求CF,根据CF、AF求AC.
解答:
解:过C作CE⊥AD,CF⊥AB,
∴△CED≌△CFB,
∴DE=FB,
∴9+DE=21-BF,
∴BF=6,
根据勾股定理,可知CF=CE=8,即点C到AB的距离为8.
∴AC=
=
=17.
故答案为:8,17.
点评:本题考查了角平分线的性质及全等三角形的证明,勾股定理在直角三角形中的应用,解答本题构建直角△CFD是关键.
解答:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163028550493504/SYS201310221630285504935022_DA/images0.png)
∴△CED≌△CFB,
∴DE=FB,
∴9+DE=21-BF,
∴BF=6,
根据勾股定理,可知CF=CE=8,即点C到AB的距离为8.
∴AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163028550493504/SYS201310221630285504935022_DA/0.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163028550493504/SYS201310221630285504935022_DA/1.png)
故答案为:8,17.
点评:本题考查了角平分线的性质及全等三角形的证明,勾股定理在直角三角形中的应用,解答本题构建直角△CFD是关键.
![](http://thumb.zyjl.cn/images/loading.gif)
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