中考数学压轴题详解—圆1(6)

（3）证明：连结OC交BD于G由（2）得四边形OBCD是菱形，

∴OC BD且OG GC． ····················································································· 8分 又已知OB＝BH ， ∴BG∥CH． ··································································· 9分 ∴ OCH OGB 90 ， ∴CH是⊙O的切线． ············································· 10分

5 解： （1）添加 AB=BD ························································································· 2分

∴∵AB=BD ∴ ∠BDE =∠BCD···································································· 3分 AB=BD又∵∠DBE =∠DBC ∴△BDE∽△BCD ∴

BDBC

BEBD

················································································································· 4分

的中点 ·（2）若AB∥DO，点D所在的位置是BC························································ 5分

∵AB∥DO ∴∠ADO =∠BAD ············································································ 6分

C · =D∵∠ADO =∠OAD ∴∠OAD =∠BAD ∴DB··················································· 7分

（3）在（1）和（2）的条件下，．

C ∴ =D∵ ∠BDA =∠DAC ∴ BD∥OA AB=BD

又∵AB∥DO ∴四边形AODB是平行四边形 ············································· 9分 ∵OA=OD ∴平行四边形AODB是菱形 ·················································· 10分

6 解：（1）FB＝FE ，PE＝PA ············································································· 2分

（2）四边形CDPF的周长为

FC＋CD＋DP＋PE＋EF＝FC＋CD＋DP＋PA＋BF ·················&