《2025年名校调研跟踪测试卷九年级数学下册人教版》

25. 如图，在四边形$ABCD$中，$AC$平分$\angle BAD$，$\angle ADC = \angle ACB = 90^{\circ}$，$E$为$AB$的中点，$AC$与$DE$交于点$F$.
(1)求证：$CE// AD$；
(2)求证：$AC^2 = AB\cdot AD$；
(3)若$AC = 2\sqrt{3}$，$AB = 4$，求$\frac{DF}{EF}$的值.
　　

答案：
(1)证明：$\because E$为$AB$中点，$\angle ACB = 90^{\circ}$，$\therefore CE=\frac{1}{2}AB = AE$，$\therefore\angle EAC=\angle ECA$. $\because AC$平分$\angle BAD$，$\therefore\angle DAC=\angle CAB$，$\therefore\angle DAC=\angle ECA$，$\therefore CE// AD$.
(2)证明：$\because AC$平分$\angle DAB$，$\therefore\angle DAC=\angle CAB$. $\because\angle ADC=\angle ACB = 90^{\circ}$，$\therefore\triangle ADC\sim\triangle ACB$，$\therefore\frac{AD}{AC}=\frac{AC}{AB}$，$\therefore AC^{2}=AB\cdot AD$.
(3)解：由
(2)知，$AC^{2}=AB\cdot AD$. $\because AC = 2\sqrt{3}$，$AB = 4$，$\therefore AD = 3$. $\because\angle ACB = 90^{\circ}$，$E$为$AB$的中点，$\therefore CE=\frac{1}{2}AB = 2$. $\because CE// AD$，$\therefore\triangle AFD\sim\triangle CFE$，$\therefore\frac{DF}{EF}=\frac{AD}{CE}=\frac{3}{2}$.