《2025年大联考单元期末测试卷九年级数学全一册沪科版》

23. 【原题呈现】如图 1,在等边三角形$ABC$中,点$D$,$E$是$AB$,$AC$上的点,且$CD = AE$,求$\angle CFD$度数.
解答过程:在等边三角形$ABC$中,$\angle A = \angle DCB = 60^{\circ}$,$BC = AC$.
又$\because CD = AE$,$\therefore\triangle CDB\cong\triangle AEC(SAS)$,
$\therefore\angle CFD = \angle FCB + \angle DBC = \angle FCB + \angle ACE = \angle ACB = 60^{\circ}$.
【操作探究】如图 2,将$CB$绕点$C$逆时针旋转 $60^{\circ}$到$CQ$,连接$BQ$,连接$FQ$交$BC$于点$O$,求证:$\angle CFQ = 60^{\circ}$.
【深入思考】如图 3,延长$QF$交$AC$于点$P$.若点$P$恰好是$AC$的中点.
① 请直接写出$\frac{OC}{OB}=$$\frac{1}{2}$;
② 若$AC = 6$,求$FO$的长.

答案：
23.解:【操作探究】证明:如图2，延长$FD$至点$G$，使得$FG = FC$，连接$CG$。
$\because FG = FC$，且$\angle GFC = 60^{\circ}$，
$\therefore \triangle GFC$是等边三角形，
$\therefore CG = FG = FC$，$\angle GCF = 60^{\circ}$。(2分)
又$\because \angle BCQ = 60^{\circ}$，
$\therefore \angle GCF + \angle FCB = \angle BCQ + \angle FCB$，
$\therefore \angle GCB = \angle QCF$。
在$\triangle GCB$与$\triangle FCQ$中，$\because \begin{cases}CG = CF\\\angle GCB = \angle FCQ\\CB = CQ\end{cases}$
$\therefore \triangle GCB \cong \triangle FCQ(SAS)$，
$\therefore \angle CFQ = \angle G = 60^{\circ}$。(5分)
【深入思考】①$\frac{1}{2}$。(8分)
由【操作探究】知，$\triangle GCB \cong \triangle FCQ$，
$\therefore \angle CGB = \angle CFO = 60^{\circ}$。
$\because \angle QCB = \angle ABC = 60^{\circ}$，
$\therefore CQ // AB$。$\because CQ = CB = BA$，
$\therefore$四边形$ABQC$是平行四边形。
又$\because AC = AB$，$\therefore$四边形$ABQC$是菱形，
$\therefore CP // BQ$，
$\therefore \angle OCP = \angle OBQ$，$\angle OPC = \angle OQB$，
$\therefore \triangle OCP \sim \triangle OBQ$，$\therefore \frac{CO}{OB}=\frac{CP}{BQ}=\frac{CP}{AC}=\frac{1}{2}$。
②如图3，连接$BP$。
$\because \triangle ABC$是等边三角形，
$\therefore AB = AC = BC = 6$，$\angle A = \angle ABC = \angle ACB = 60^{\circ}$。由①知$\frac{CO}{OB}=\frac{1}{2}$，$\therefore CO = 2$，$OB = 4$。
由【操作探究】知，$\triangle GCB \cong \triangle FCQ$，
$\therefore \angle CQO = \angle CBD$。
又$\because CQ = CB$，$\angle OCQ = \angle DCB = 60^{\circ}$，
$\therefore \triangle COQ \cong \triangle CDB(ASA)$，
$\therefore CO = CD = 2 = AE$，$\therefore BE = AB - AE = 4$。(10分)
$\because$点$P$是$AC$的中点，$\therefore BP \perp AC$，$\angle ABP = \frac{1}{2}\angle ABC = 30^{\circ}$，$DP = PC - CD = \frac{1}{2}AC - CD = 3 - 2 = 1$。
$\therefore$在$Rt\triangle PCB$中，$BP = \sqrt{BC^{2} - PC^{2}} = \sqrt{6^{2} - 3^{2}} = \sqrt{27} = 3\sqrt{3}$。
$\therefore$在$Rt\triangle BPD$中，$BD = \sqrt{(3\sqrt{3})^{2} + 1^{2}} = 2\sqrt{7} = CE$。(12分)
由①知，$\angle CFO = 60^{\circ}$，
$\therefore \angle CFO = \angle CBE = 60^{\circ}$，$\angle FCO = \angle BCE$，
$\therefore \triangle FCO \sim \triangle BCE$，$\therefore \frac{CO}{CE}=\frac{FO}{EB}$，$\therefore \frac{2}{2\sqrt{7}}=\frac{FO}{4}$，
$\therefore FO = \frac{4\sqrt{7}}{7}$。(14分)