如图,△ABC内接于⊙O,AB是⊙O的直径,C是弧AD的中点,弦CE⊥AB于点H,连接AD,分别交CE、BC于点P、Q,
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如图,△ABC内接于⊙O,AB是⊙O的直径,C是弧AD的中点,弦CE⊥AB于点H,连接AD,分别交CE、BC于点P、Q,连接BD.
![](http://img.wesiedu.com/upload/9/72/972b634a0dc1999999d6b4877a100ec2.jpg)
(1)求证:P是线段AQ的中点;
(2)若⊙O的半径为5,AQ=
![](http://img.wesiedu.com/upload/9/72/972b634a0dc1999999d6b4877a100ec2.jpg)
(1)求证:P是线段AQ的中点;
(2)若⊙O的半径为5,AQ=
15 |
2 |
![如图,△ABC内接于⊙O,AB是⊙O的直径,C是弧AD的中点,弦CE⊥AB于点H,连接AD,分别交CE、BC于点P、Q,](/uploads/image/z/15309342-54-2.jpg?t=%E5%A6%82%E5%9B%BE%EF%BC%8C%E2%96%B3ABC%E5%86%85%E6%8E%A5%E4%BA%8E%E2%8A%99O%EF%BC%8CAB%E6%98%AF%E2%8A%99O%E7%9A%84%E7%9B%B4%E5%BE%84%EF%BC%8CC%E6%98%AF%E5%BC%A7AD%E7%9A%84%E4%B8%AD%E7%82%B9%EF%BC%8C%E5%BC%A6CE%E2%8A%A5AB%E4%BA%8E%E7%82%B9H%EF%BC%8C%E8%BF%9E%E6%8E%A5AD%EF%BC%8C%E5%88%86%E5%88%AB%E4%BA%A4CE%E3%80%81BC%E4%BA%8E%E7%82%B9P%E3%80%81Q%EF%BC%8C)
(1)证明:∵AB是⊙O的直径,弦CE⊥AB,
∴
![](http://img.wesiedu.com/upload/6/a3/6a3bc38fee8e9057a31dd1025b0ce6f1.jpg)
AC=
![](http://img.wesiedu.com/upload/3/b0/3b0f3290352b51c6925a899ebaa1d177.jpg)
AE.
又∵C是
![](http://img.wesiedu.com/upload/1/6d/16de0551a831f52f89b3ac30da5317a3.jpg)
AD的中点,
∴
![](http://img.wesiedu.com/upload/6/a3/6a3bc38fee8e9057a31dd1025b0ce6f1.jpg)
AC=
![](http://img.wesiedu.com/upload/f/f7/ff72cbf5bd08076fc228899df1210579.jpg)
CD,
∴
![](http://img.wesiedu.com/upload/3/b0/3b0f3290352b51c6925a899ebaa1d177.jpg)
AE=
![](http://img.wesiedu.com/upload/f/f7/ff72cbf5bd08076fc228899df1210579.jpg)
CD.
∴∠ACP=∠CAP.
∴PA=PC,
∵AB是直径.
∴∠ACB=90°.
∴∠PCQ=90°-∠ACP,∠CQP=90°-∠CAP,
∴∠PCQ=∠CQP.
∴PC=PQ.
∴PA=PQ,即P是AQ的中点;
(2)∵
![](http://img.wesiedu.com/upload/6/a3/6a3bc38fee8e9057a31dd1025b0ce6f1.jpg)
AC=
![](http://img.wesiedu.com/upload/f/f7/ff72cbf5bd08076fc228899df1210579.jpg)
CD,
∴∠CAQ=∠ABC.
又∵∠ACQ=∠BCA,
∴△CAQ∽△CBA.
∴
AC
BC=
AQ
AB=
15
2
10=
3
4.
又∵AB=10,
∴AC=6,BC=8.
根据直角三角形的面积公式,得:AC•BC=AB•CH,
∴6×8=10CH.
∴CH=
24
5.
又∵CH=HE,
∴CE=2CH=
48
5.
∴
![](http://img.wesiedu.com/upload/6/a3/6a3bc38fee8e9057a31dd1025b0ce6f1.jpg)
AC=
![](http://img.wesiedu.com/upload/3/b0/3b0f3290352b51c6925a899ebaa1d177.jpg)
AE.
又∵C是
![](http://img.wesiedu.com/upload/1/6d/16de0551a831f52f89b3ac30da5317a3.jpg)
AD的中点,
∴
![](http://img.wesiedu.com/upload/6/a3/6a3bc38fee8e9057a31dd1025b0ce6f1.jpg)
AC=
![](http://img.wesiedu.com/upload/f/f7/ff72cbf5bd08076fc228899df1210579.jpg)
CD,
∴
![](http://img.wesiedu.com/upload/3/b0/3b0f3290352b51c6925a899ebaa1d177.jpg)
AE=
![](http://img.wesiedu.com/upload/f/f7/ff72cbf5bd08076fc228899df1210579.jpg)
CD.
∴∠ACP=∠CAP.
∴PA=PC,
∵AB是直径.
∴∠ACB=90°.
∴∠PCQ=90°-∠ACP,∠CQP=90°-∠CAP,
∴∠PCQ=∠CQP.
∴PC=PQ.
∴PA=PQ,即P是AQ的中点;
(2)∵
![](http://img.wesiedu.com/upload/6/a3/6a3bc38fee8e9057a31dd1025b0ce6f1.jpg)
AC=
![](http://img.wesiedu.com/upload/f/f7/ff72cbf5bd08076fc228899df1210579.jpg)
CD,
∴∠CAQ=∠ABC.
又∵∠ACQ=∠BCA,
∴△CAQ∽△CBA.
∴
AC
BC=
AQ
AB=
15
2
10=
3
4.
又∵AB=10,
∴AC=6,BC=8.
根据直角三角形的面积公式,得:AC•BC=AB•CH,
∴6×8=10CH.
∴CH=
24
5.
又∵CH=HE,
∴CE=2CH=
48
5.
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如图,AB是圆O的直径,BC是弦,延长BC到D,使CD=BC,CE切圆O于点C,交AD于E,求证:CE⊥AD.
如图,△ABC是⊙O的内接三角形且AB=AC,BD是⊙O的直径.过点A做AP‖BC交DB的延长线于点P,连接AD.
如图,AB是⊙O的直径,C是BD的中点,CE⊥AB于E,BD交CE于点F.
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