急用拜托了,
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/06/30 23:51:19
急用拜托了,
![](http://img.wesiedu.com/upload/d/62/d62b30094bebc492be1b76cd79a54065.jpg)
![](http://img.wesiedu.com/upload/d/62/d62b30094bebc492be1b76cd79a54065.jpg)
![急用拜托了,](/uploads/image/z/15105704-32-4.jpg?t=%E6%80%A5%E7%94%A8%E6%8B%9C%E6%89%98%E4%BA%86%2C%26nbsp%3B)
证明:
作AH//CB交CD延长线于H
则AD/BD=AH/BC
∵AC=BC
∴AD/BD=AH/AC
∵AC=BC,∠ACB=90°
∴∠CAB=∠CBA=45°
∵AH//CB
∴∠HAB=∠CBA=45°
∴∠CAH=∠CAB+∠HAB=90°
∵EF⊥CD
∴∠CEF+∠ACH=90°
∵∠H+∠ACH=90°
∴∠CEF=∠H
又∵∠ECF=∠HAC=90°
∵△ECF∽△HAC(AA)
∴AH/AC=CE/CF
∴AD/BD=CE/CF
![](http://img.wesiedu.com/upload/6/b5/6b5bb89b6f11a5868bc06be1371c0f56.jpg)
作AH//CB交CD延长线于H
则AD/BD=AH/BC
∵AC=BC
∴AD/BD=AH/AC
∵AC=BC,∠ACB=90°
∴∠CAB=∠CBA=45°
∵AH//CB
∴∠HAB=∠CBA=45°
∴∠CAH=∠CAB+∠HAB=90°
∵EF⊥CD
∴∠CEF+∠ACH=90°
∵∠H+∠ACH=90°
∴∠CEF=∠H
又∵∠ECF=∠HAC=90°
∵△ECF∽△HAC(AA)
∴AH/AC=CE/CF
∴AD/BD=CE/CF
![](http://img.wesiedu.com/upload/6/b5/6b5bb89b6f11a5868bc06be1371c0f56.jpg)