《2025年点拨训练九年级数学上册人教版》

1. 从圆上任意一点出发的两条弦所组成的折线,称为该圆的一条折弦。如图,PA,PB组成⊙O的一条折弦。C是劣弧AB的中点,直线CD⊥PA于点E,则AE= PE+PB。证明该结论是否成立,请写出证明过程。
(提示：可连接AD,BD,延长DB,AP相交于点F)

答案：
证明：如图，连接 AD,DB,延长 DB,AP 相交于点 F.
∵ 四边形 ADBP 是$\odot O$的内接四边形，$\therefore ∠A + ∠PBD = 180^{\circ }$.
∵$∠PBD + ∠PBF = 180^{\circ },$$\therefore ∠PBF = ∠A.$
∵ C 是劣弧 AB 的中点，$\therefore \widehat {AC} = \widehat {CB}.$

$\therefore ∠CDA = ∠CDF.$ 易得$\triangle AED\cong \triangle FED,$$\therefore AE = EF, ∠F = ∠A,$$\therefore ∠PBF = ∠F.\therefore PB = PF.$ $\because EF = PE + PF,\therefore AE = PE + PB.$
∴ 该结论成立.

2. 如图,M是等边三角形ABC的外接圆$\overset{\frown}{BC}$上的一点,判断MA,MB,MC之间的数量关系,并给出证明过程。

答案：
解：$MA = MB + MC$. 证明: 延长 BM 到点 N,使$MN = MC$,连接 CN,如图.
∵$\triangle ABC$是等边三角形，$\therefore AC = BC, ∠BAC = ∠ABC = 60^{\circ }.$
∴ 易得$∠CMN = 60^{\circ }.$
∴$\triangle CMN$是等边三角形.$\therefore ∠BNC = 60^{\circ }.$

$\because ∠AMC = ∠ABC = 60^{\circ },$$\therefore ∠AMC = ∠BNC.$ $\because ∠AMC = ∠BNC, ∠MAC = ∠NBC, AC = BC,$$\therefore \triangle MAC\cong \triangle NBC.\therefore MA = NB.$ $\because BN = BM + MN = BM + MC,$$\therefore MA = MB + MC.$

3. 如图,在等腰三角形ABC中,AC= BC,⊙O为△ABC的外接圆,D为⊙O上一点,CE⊥AD于点E,求证：AE= BD+DE。

答案：
证明：如图,在 AE 上截取$AF = BD$,连接 CF,CD. 在$\triangle ACF$和$\triangle BCD$中,
$\begin{cases}AC = BC,\\∠CAF = ∠CBD,\\AF = BD,\end{cases}$

$\therefore \triangle ACF\cong \triangle BCD.\therefore CF = CD.$ $\because CE⊥AD,\therefore EF = DE.$ $\therefore AE = AF + EF = BD + DE.$

4. 如图,△ABC内接于⊙O,AC= BC,CD是⊙O的一条弦,且$\overset{\frown}{BC}= \overset{\frown}{BD}$,过点A作AP⊥CD,分别交CD,⊙O于点E,P,连接BP,若CD= 6,△ABP的周长为13,求AE的长。

答案：
解：如图,在 AE 上截取$AF = BP$,连接 CF,PC. 又$\because AC = BC, ∠CAF = ∠CBP,$$\therefore \triangle CAF\cong \triangle CBP.\therefore CF = CP.$ $\because CD⊥AP,\therefore EF = PE.$ $\therefore AE = AF + FE = PB + PE.$ $\because AC = BC,\therefore \widehat {AC} = \widehat {BC}.$ $\because \widehat {BC} = \widehat {BD},\therefore \widehat {AB} = \widehat {CD}.$

$\therefore AB = CD = 6.$
∵$\triangle ABP$的周长是 13,$\therefore AP + PB = 13 - 6 = 7.$ $\because AE = PE + PB,$$\therefore 2AE = AP + PB = 7.\therefore AE = \frac{7}{2}.$