中考数学压轴题详解—圆(6)

复习圆

（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

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

BDBE

································································································································ 4分 BCBD

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

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

=DC ·∵∠ADO =∠OAD ∴∠OAD =∠BAD ∴DB······················································· 7分

（3）在（1）和（2）的条件下，．

=DC ∴∠∵ 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 ·········································· 3分

＝BF＋FC＋CD＋DP＋PA ········································ 4分 ＝BC＋CD＋DA ··············································· 5

分 ＝

3＝ ················································· 6分

（3）存在．

7分

OF,则 若BF FG CF

BFCF

OFFG

BFCF

∵ cos∠OFB＝ ，cos∠GFC＝

OFFG

∴ ∠OFB＝∠GFC

又 ∵ ∠OFB＝∠OFE

∴ ∠OFE＝∠OFB＝∠GFC=60 ······································································· 8分 ∴ 在Rt△OFB中 FE＝FB＝∴ 在Rt△GFC中

CG

＝CF

tan GFC CF tan60 1tan60 6∴ DG CG CD

6 ∴ DP DG ··········································· 9

分 tan PGD DG tan30 3 ∴

AP AD DP 3 3 ··············································· 10分

京翰教育1对1家教 /

OB

＝1

tan60