中考证明题答案
C
22、(1)证明:如图示,∵OA ⊥OB ,∴∠1与∠2互余,
又∵四边形ABCD 是矩形,∴∠BAD =90o , B
∴∠2与∠3互余,∴∠1=∠3,·········1分
∵OA ⊥OB ,DE ⊥OA ,∴∠BOA =∠DEA =90o ··2分 O ∴△OAB ∽△EDA .···························3分
(2) 解:在Rt △OAB 中,AB =32+42=5,········· 4分 1D C 3D
B F 由(1)可知∠1=∠3,∠BOA =∠DEA =90o , 1∴当a =AD=AB=5时,△OAB 与△EDA 全等.···5分
32当a =AD=AB=5时,可知矩形ABCD 为正方形, O A E ∴BC =AB ,如图,过点C 作CH ⊥OE 交OE 于点H ,
则CH 就是点C 到OE 的距离,过点B 作BF ⊥CH 交CH 于点F ,
则∠4与∠5互余,∠1与∠5互余,∴∠1=∠4,·························6分 又∵∠BFC =∠BOA ,BC =AB ,∴△OAB ≌△FCB (AAS ),···············7分 ∴CF =OA =4,BO =BF ,∴四边形OHFB 为正方形,
∴HF =OB =3,∴点C 到OE 的距离CH =CF+HF=4+3=7.················8分 4
22、(1)证明:如图,∵△ABC 是等腰三角形,∴AC=BC , ∴∠BAD =∠ABE ,··1分
又∵AB=BA、∠2=∠1, ∴△ABD ≌△BAE (ASA ),·············2分
∴BD=AE,又∵∠1=∠2,∴OA=OB,
∴BD-OB=AE-OA,即:OD=OE.································3分
(2) 证明:由(1)知:OD=OE,∴∠OED =∠ODE ,
1(180 -∠DOE ),···4分 2
1 同理:∠1=(180-∠AOB ), 2∴∠OED=
又∵∠DOE =∠AOB ,∴∠1=∠OED ,∴D E ∥AB ,··············5分
∵AD 、BE 是等腰三角形两腰所在的线段,∴AD 与BE 不平行,
∴四边形ABED 是梯形, 又由(1)知∴△ABD ≌△BAE ,∴AD=BE
∴梯形ABED 是等腰梯形.·····································6分
24. (1)证明:∵OA=OB,
∴∠OAB=∠OBA ,
∵OA ⊥CD ,
∴∠OAB+∠AGC=90°,
又∵∠FGB=∠FBG ,∠FGB=∠AGC ,
∴∠FBG+∠OBA=90°,
即∠OBF=90°,
∴OB ⊥FB ,
∵AB 是⊙O 的弦,
∴点B 在⊙O 上,
∴BF 是⊙O 的切线;
(2)解:∵AC ∥BF ,
∴∠ACF=∠F ,
∵CD=a,OA ⊥CD ,
∴CE=CD=a ,
∵tan ∠F=,
∴tan ∠ACF=
即=, =,
解得AE=a ,
连接OC ,设圆的半径为r ,则OE=r﹣a , 在Rt △OCE 中,CE +OE=OC,
即(a )+(r ﹣a )=r,
解得r=a ; 222222
(3)证明:连接BD ,
∵∠DBG=∠ACF ,∠ACF=∠F (已证),
∴∠DBG=∠F ,
又∵∠F=∠F ,
∴△BDG ∽△FBG ,
∴=
2, 即GB =DG•GF ,
222∴GF ﹣GB =GF﹣DG •GF=GF(GF ﹣DG )=GF•DF , 22即GF ﹣GB =DF•GF .