摘要: 证明:(1)连结OD. ························································································· 1分 由O.E分别是BC.AC中点得OE∥AB. ∴∠1=∠2.∠B=∠3.又OB=OD. ∴∠2=∠3. 而OD=OC.OE=OE ∴△OCE≌△ODE. ∴∠OCE=∠ODE. 又∠C=90°.故∠ODE =90°. ································ 2分 ∴DE是⊙O的切线. ·········································· 3分 (2)在Rt△ODE中.由.DE=2 得 ····························································· 5分 又∵O.E分别是CB.CA的中点 ∴AB=2· ∴所求AB的长是5cm. ····················································································· 7分
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(本题满分8分)如图1,已知反比例函数y=
过点P, P点的坐标为(3-m,
2m),m是分式方程
的解,PA⊥x轴于点A,PB⊥y轴于点B.


(1)求m值
(2)试判断四边形PAOB的形状,并说明理由.
(2)如图2,连结AB,E为AB上的一点,EF⊥BP于点F,G为AE的中点,连结OG、FG,试问FG和OG有何数量关系?请写出你的结论并证明. 查看习题详情和答案>>
2m),m是分式方程
(1)求m值
(2)试判断四边形PAOB的形状,并说明理由.
(2)如图2,连结AB,E为AB上的一点,EF⊥BP于点F,G为AE的中点,连结OG、FG,试问FG和OG有何数量关系?请写出你的结论并证明. 查看习题详情和答案>>
(本题满分8分)
如图,E、F分别是平行四边形ABCD对角线BD所在直线上两点,BE=DF,请你以F为一个端点,和图中己标明字母的某一点连成一条新的线段,猜想并证明它和图中已有的某一条线段相等(只需研究一组线段相等即可)

(1)连结_________
(2)猜想:_________
(3)证明: 查看习题详情和答案>>
如图,E、F分别是平行四边形ABCD对角线BD所在直线上两点,BE=DF,请你以F为一个端点,和图中己标明字母的某一点连成一条新的线段,猜想并证明它和图中已有的某一条线段相等(只需研究一组线段相等即可)
(1)连结_________
(2)猜想:_________
(3)证明: 查看习题详情和答案>>
(本题满分10分)如图,⊙O是△ABC的外接圆,FH是⊙O 的切线,切点为F,
FH∥BC,连结AF交BC于E,∠ABC的平分线BD交AF于D,连结BF.
(1)证明:AF平分∠BAC;
(2)证明:BF=FD;
(3)若EF=4,DE=3,求AD的长.
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