复习圆

（3）Sn

1π （n为正整数）． ··················· 10分 2n 12n 1

3 (1) 证明： 如图，

∵ 点I是△ABC的内心， ∴ ∠BAD=∠CAD，∠ABI=∠CBI． ………………2分 ∵ ∠CBD=∠CAD， ∴ ∠BAD=∠CBD． ……………………………3分 ∴ ∠BID=∠ABI+∠BAD =∠CBI+∠CBD=∠IBD． ∴ ID=BD． ………………………5分 (2)解：如图， ∵∠BAD=∠CBD=∠EBD， ∠D=∠D， ∴ △ABD∽△BED． …………………………7分

BDAD∴ ． ∴ AD DE BD2 ID2． …………………8分

DEBD∵ ID=6，AD=x，DE=y，∴ xy=36． ………………9分 又∵ x=AD>ID=6， AD不大于圆的直径10， ∴ 6<x≤10．

36

∴ y与x的函数关系式是y ．（6 x≤10） …………………………10分

x

说明：只要求对xy=36与6<x≤10，不写最后一步，不扣分．

的中点， 4 （1）证明：∵C是劣弧BD

∴ DAC CDB． ·············································· 1分 而 ACD公共，

∴△DEC∽△ADC． ··········································· 3分

DCEC

， ACDC

∵CE 1.AC AE EC 2 1 3，

2

∴DC AC EC 3 1 3 ．

∴DC ． ··················································································································· 4分

由已知BC DC AB是⊙O的直径， ∴ ACB 90 ，

（2）证明：连结OD，由⑴得∴AB AC CB 3

2

2

2

2

2

12．

∴AB

∴OD OB BC DC ， ∴四边形OBCD是菱形． ∴DC∥AB，DC AB， ∴四边形ABCD是梯形． ···················································· 5分 法一：

过C作CF垂直AB于F，连结OC

，则OB BC OC ∴ OBC 60 ． ··············································································································· 6分

CF3

，CF BC sin60 ， BC2113∴S梯形ABCD＝CF

AB＋DC ＝ ·········································· 7分

222∴sin60

法二：（接上证得四边形ABCD是梯形）

又DC∥AB ∴AD BC，连结OC，则△AOD，△DOC和△

OBC 6分

∴△AOD≌△DOC≌△OBC，

京翰教育1对1家教 /