证明直角三角形
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/08/19 18:10:49
点E位正方形ABCD的边AD上一点,连接be,过点A作AH⊥BE垂足为H,延长AH交CD于点F.求证:DE=CF![](http://img.wesiedu.com/upload/f/7d/f7d5551713b397923dfeadbf8b504d52.png)
![](http://img.wesiedu.com/upload/f/7d/f7d5551713b397923dfeadbf8b504d52.png)
![证明直角三角形](/uploads/image/z/20136499-43-9.jpg?t=%E8%AF%81%E6%98%8E%E7%9B%B4%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2)
解题思路: 主要是证明△ABE≌△DAF即可
解题过程:
点E位正方形ABCD的边AD上一点,连接be,过点A作AH⊥BE垂足为H,延长AH交CD于点F.求证:DE=C F E![](http://img.wesiedu.com/upload/4/a5/4a57e2f051f5fe24b129a5c64d480240.png)
![](http://img.wesiedu.com/upload/3/e1/3e18f9833f4fe1a7dee4ecedfc30a37f.png)
A D F B C 解:∵ABCD是正方形,∴∠BAD=∠ADF=90,∴∠ABE+∠AEB=90;又AH⊥BE, ∴∠FAD+∠AEB=90,∴∠ABE=∠FAD,又AB=AD, ∴△ABE≌△DAF,所以AE=DF,又AD=DC,∴DE=C F
最终答案:略
解题过程:
点E位正方形ABCD的边AD上一点,连接be,过点A作AH⊥BE垂足为H,延长AH交CD于点F.求证:DE=C F E
![](http://img.wesiedu.com/upload/4/a5/4a57e2f051f5fe24b129a5c64d480240.png)
![](http://img.wesiedu.com/upload/3/e1/3e18f9833f4fe1a7dee4ecedfc30a37f.png)
![](http://img.wesiedu.com/upload/4/64/464ccfc6a73db2bd63f38c8bf8e8f2bb.png)
最终答案:略