2003年辽宁省中考数学试题答案

七、 27．

（1）证明：

B ①连结BD F

∵AB是⊙O的直径

·O ∴∠ADB＝90°

A ∴∠AGC＝∠ADB＝90°

l 又∵ACDB是⊙O内接四边形

C G E D ∴∠ACG＝∠B

图(a) ∴∠BAD＝∠CAG ······································ 3分

②连结CF

∵∠BAD＝∠CAG ∠EAG＝∠FAB ∴∠DAE＝∠FAC 又∵∠ADC＝∠F ∴△ADE∽△AFC ·············································································· 5分

ADAE =AFAC∴AC·AD＝AE·AF ·············································································· 6分 （其他方法相应给分）

（2）①图形正确 ················································ 8分

C(D) G E ②两个结论都成立，证明如下：

B ①连结BC

F ∵AB是直径

O ∴∠ACB＝90° · ∴∠ACB＝∠AGC＝90° A ∵GC切⊙O于C ∴∠GCA＝∠ABC ∴∠BAC＝∠CAG（即∠BAD＝∠CAG） ······· 10分 图(b)

②连结CF

∵∠CAG＝∠BAC，∠GCF＝∠GAC

∴∠GCF＝∠CAE，∠ACF＝∠ACG－∠GFC，∠E＝∠ACG－∠CAE ∴∠ACF＝∠E ∴△ACF∽△AEC ∴

ACAF =AEAC∴AC2＝AE·AF（即AC·AD＝AE·AF） ········································· 12分 （注：其他方法证明，按相应步骤给分） ∴

八、

28．解：（1）直线y=?22x?8与x轴、y轴分别交于点C、P

y Q B ∴C（?22，0），P(0,-8)

∴cot∠OCD=22 cot∠OPC=22 ·D(0,1) C O F · A x ∴∠OCD=∠OPC

∵∠OPC+∠PCO=90° ∴∠OCD+∠PCO=90° ∴PC是⊙D的切线 ······································ 5分 （2）设直线PC上存在一点E(x,y)，使S△EOP＝4S△CDO

11?8?x?4??1?22 22P

由y??22x?8可知：

解得 x=±2 当x=2时，y=-12， 当x=-2时，y=-4·············································· 9分 ∴在直线PC上存在点E（2，-12）或（-2，-4）使S△EOP＝4S△CDO ······ 10分

（注：只求出一个点，扣2分）

（3）解法一：

⌒ 于F，交⊙D于Q，连结DQ 作直线PF交劣弧AC

由切割线定理得：PC2＝PF·PQ ····················· ① 在△CPD和△OPC中

∵∠PCD＝∠POC＝90° ∠CPD＝∠OPC

PCPD ＝POPC即PC2＝PO·PD ········································· ② 由①、②得：PO·PD＝PF·PQ，又∵∠FPO＝∠DPQ

PFPDm9=∴△FPO∽△DPQ ，即= FODQn3∴△CPD∽△OPC ∴

∴m=3n ··························································································· 13分 （2

y B ⌒ 于F 解法二：作直线PF交劣弧AC

设F(x,y)，作FM⊥y轴，M为垂足，连结DF， ∵m2-(8+y)2=x2 n2-y2=x2

∴m2-64-16y-y2=n2-y2 即 m2-64-16y=n2 ·········① 又∵32-(1-y)2=x2 x ∴32-(1-y)2=n2-y2

·D(0,1) C O M F · A n2?8解得y= ··············②

2将②代入①，解得：m=3n，m=-3n(舍) ∴m=3n ·················································· 13分 (2

P