20.(1) 连结CB.∵AB是⊙O的直径.∴∠ACB=90°. ························· 1分 而∠CAH=∠BAC.∴△CAH∽△BAC . ················································· 2分 ∴. 即AHAB=AC2 . ·························································· 3分 (2) 连结FB.易证△AHE∽△AFB. ······················································ 4分 ∴ AEAF=AHAB. ············································································· 5分 ∴ AEAF=AC2 . ···················································································· 6分 (也可连结CF.证△AEC∽△ACF) (3) 结论APAQ=AC2成立 . ··································································· 7分 查看更多

 

题目列表(包括答案和解析)

(2013•遂宁)如图,在⊙O中,直径AB⊥CD,垂足为E,点M在OC上,AM的延长线交⊙O于点G,交过C的直线于F,∠1=∠2,连结CB与DG交于点N.
(1)求证:CF是⊙O的切线;
(2)求证:△ACM∽△DCN;
(3)若点M是CO的中点,⊙O的半径为4,cos∠BOC=
14
,求BN的长.

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(2013•江东区模拟)如图,抛物线y=
1
4
x2-m2(m>0)与x轴相交于点A、C,与y轴相交于点P,连结PA、PC,过点A画PC的平行线分别交y轴和抛物线于点B、C1,连结CB并延长交抛物线于点A1,在过点A1画AC1的平行线分别交y轴和抛物线于点B1、C2,连结C1B1并延长交抛物线于点A2,…,依次得到四边形,记四边形AnBnCnBn-1的面积为Sn
(1)求证:四边形ABCP是菱形.
(2)设∠A1B1C1=a,且90°<a<120°,求m的取值范围.
(3)当m=1时,
①填表:
序号 S1 S2 S3 Sn
四边形的面积
②是否存在2个四边形,他们的面积Sp、Sq满足:Sp×Sq=214(p<q)?若存在,求p、q的值;若不存在,请说明理由.

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如图,直线y=x+3与坐标轴分别交于A、B两点,抛物线y=ax2+bx-3a经过点A、B,顶点为C,连结CB并延长交x轴于点E,点D与点B关于抛物线的对称轴MN对称。
(1)求抛物线的解析式及顶点C的坐标;
(2)求证:四边形ABCD是直角梯形。

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(2011内蒙古赤峰,24,12分)如图,直线y=x+3与坐标轴分别交于A、B两点,抛物线经过点A、B,顶点为C,连结CB并延长交x轴于点E,点D与点B关于抛物线的对称轴MN对称。

(1)求抛物线的解析式及顶点C的坐标;

(2)求证:四边形ABCD是直角梯形。

 

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如图,在⊙O中,直径AB⊥CD,垂足为E,点M在OC上,AM的延长线交⊙O于点G,交过C的直线于F,∠1=∠2,连结CB与DG交于点N.

(1)求证:CF是⊙O的切线;
(2)求证:△ACM∽△DCN;
(3)若点M是CO的中点,⊙O的半径为4,cos∠BOC=,求BN的长.

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