《2025年学霸提优大试卷九年级数学全一册苏科版》

14. (14分)如图,$AB是\odot O$的直径,圆内接四边形$ACDE的边CD与直径AB交于点F$,点$G在DE$的延长线上,$EA平分\angle CEG$.
(1)求证:$AB\perp CD$;
(2)若$AC = CE$,$AF = 9$,$BF = 1$,求$\triangle ACE$的面积.

答案：
14.
(1) $\because$四边形ACDE是$\odot O$的内接四边形，$\therefore$易证$\angle ACD=\angle AEG$。$\because EA$平分$\angle CEG$，$\therefore \angle AEG=\angle AEC$，$\therefore \angle ACD=\angle AEC$，$\therefore \overset{\frown }{AD}=\overset{\frown }{AC}$。$\because AB$为直径，$\therefore AB\perp CD$。
(2)如图，连接AD、OD，$\because AC=CE$，$\therefore \overset{\frown }{AC}=\overset{\frown }{CE}$。$\because \overset{\frown }{AC}=\overset{\frown }{AD}$，$\therefore \overset{\frown }{CE}=\overset{\frown }{AD}$，$\therefore CE=AD$，$\angle EAC=\angle DCA$，$\overset{\frown }{CE}-\overset{\frown }{DE}=\overset{\frown }{AD}-\overset{\frown }{DE}$，$\therefore \overset{\frown }{AE}=\overset{\frown }{DC}$，$\therefore AE=DC$。在$\triangle ACE$与$\triangle CAD$中，$\begin{cases}AC=CA\\CE=AD\\AE=CD\end{cases}$，$\therefore \triangle ACE\cong \triangle CAD(SSS)$，$\therefore S_{\triangle ACE}=S_{\triangle CAD}$。$\because AF=9$，$BF=1$，$\therefore AB=10$，$\therefore OD=OB=5$，$\therefore OF=4$，$\therefore DF=\sqrt{5^{2}-4^{2}}=3$，$\therefore DC=3× 2=6$，$\therefore S_{\triangle CAD}=\frac{1}{2}CD\cdot AF=\frac{1}{2}× 6× 9=27$，$\therefore S_{\triangle ACE}=27$。
　　　　　　　　　　　　　　　　　　

15. (12分)如图,$\triangle ABC是\odot O$的内接三角形,直径$HF交AC于点D$,$HF和BC的延长线交于点E$.
(1)若$HF\perp AB$,求证:$\angle OAD = \angle E$.
(2)若点$A$在下半圆上运动,则当点$A$运动到什么位置时,$\triangle CDE的外心在\triangle CDE$的一边上?请说明理由.

答案： 15.
(1)连接OB，$\because HF\perp AB$，$\therefore \overset{\frown }{BH}=\overset{\frown }{AH}$，$\therefore \angle AOH=\angle BOH=\frac{1}{2}\angle AOB$。又$\angle ACB=\frac{1}{2}\angle AOB$，$\therefore \angle AOH=\angle ACB$。$\because \angle AOD+\angle AOH=180^{\circ}$，$\angle ECD+\angle ACB=180^{\circ}$，$\therefore \angle AOD=\angle ECD$。又$\angle ODA=\angle CDE$，$\therefore \angle OAD=\angle E$。
(2)当AB是$\odot O$的直径或$AC\perp HF$时，$\triangle CDE$的外心在$\triangle CDE$的一边上。理由：易知$\angle DEC$不可能为$90^{\circ}$，分两种情况讨论：①当$\angle DCE=90^{\circ}$时，$\angle BCA=90^{\circ}$，$\therefore AB$为$\odot O$的直径。此时$\triangle CDE$的外心在$\triangle CDE$的边DE上；②当$\angle CDE=90^{\circ}$时，$\triangle CDE$是直角三角形，$\therefore AC\perp HF$。此时$\triangle CDE$的外心在$\triangle CDE$的边CE上。综上所述，当点A运动到使AB是$\odot O$的直径或$AC\perp HF$时，$\triangle CDE$的外心在$\triangle CDE$的一边上。

16. (14分)(2024·包头模拟)如图,四边形$ABCD是\odot O$的内接四边形,$BD是\odot O$的直径,点$F是CD$延长线上的一点,且$DA平分\angle BDF$,$AE\perp CD于点E$.
(1)求证:$AB = AC$;
(2)若$BD = 12$,$DE = 2$,求$BC$的长.

答案：
16.
(1) $\because DA$平分$\angle BDF$，$\therefore \angle ADF=\angle ADB$。$\because \angle ABC+\angle ADC=\angle ADC+\angle ADF=180^{\circ}$，$\therefore \angle ADF=\angle ABC$。$\because \angle ACB=\angle ADB$，$\therefore \angle ABC=\angle ACB$，$\therefore AB=AC$。
(2)如图，过点A作$AG\perp BD$，垂足为点G.$\because DA$平分$\angle BDF$，$AE\perp CF$，$AG\perp BD$，$\therefore AG=AE$，$\angle AGB=\angle AEC=90^{\circ}$。在$Rt\triangle AED$和$Rt\triangle AGD$中，$\begin{cases}AD=AD\\AE=AG\end{cases}$，$\therefore Rt\triangle AED\cong Rt\triangle AGD(HL)$，$\therefore GD=ED=2$。在$Rt\triangle AEC$和$Rt\triangle AGB$中，$\begin{cases}AC=AB\\AE=AG\end{cases}$，$\therefore Rt\triangle AEC\cong Rt\triangle AGB(HL)$，$\therefore BG=CE$。$\because BD=12$，$\therefore BG=BD - GD=12 - 2=10$，$\therefore CE=BG=10$，$\therefore CD=CE - DE=10 - 2=8$。$\because BD$是$\odot O$的直径，$\therefore \angle BCD=90^{\circ}$，$\therefore$在$Rt\triangle BCD$中，$BC=\sqrt{DB^{2}-DC^{2}}=\sqrt{12^{2}-8^{2}}=4\sqrt{5}$