《2025年大联考单元期末测试卷九年级数学全一册沪科版》

22. 如图,在四边形$ABCD$中,$AC$平分$\angle DAB$,$AC^2 = AB· AD$,$\angle ADC = 90^{\circ}$,点$E$为$AB$的中点.
(1) 求证:$\triangle ADC\sim\triangle ACB$.
(2) 求证:$CE// AD$.
(3) 若$AD = 4$,$AB = 6$,求$AF$的长.

答案： 22.解:
(1)证明:$\because AC$平分$\angle DAB$，
$\therefore \angle DAC = \angle CAB$。
$\because AC^{2} = AB· AD$，$\therefore \frac{AC}{AB}=\frac{AD}{AC}$，
$\therefore \triangle ADC \sim \triangle ACB$。(4分)
(2)证明:$\because \triangle ADC \sim \triangle ACB$，
$\therefore \angle ACB = \angle ADC = 90^{\circ}$。(5分)
$\because$点$E$为$AB$的中点，$\therefore CE = AE = \frac{1}{2}AB$，
$\therefore \angle EAC = \angle ECA$。
$\because \angle DAC = \angle EAC$，
$\therefore \angle DAC = \angle ECA$，$\therefore CE // AD$。(8分)
(3)由
(2)，得$CE = \frac{1}{2}AB = 3$。
$\because CE // AD$，$\therefore \triangle AFD \sim \triangle CFE$，
$\therefore \frac{CF}{AF}=\frac{CE}{AD}=\frac{3}{4}$，$\therefore \frac{AF}{AC}=\frac{4}{7}$。(10分)
$\because AC^{2} = AB· AD$，$AD = 4$，$AB = 6$，
$\therefore AC = 2\sqrt{6}$，$\therefore AF = \frac{4}{7}×2\sqrt{6} = \frac{8\sqrt{6}}{7}$。(12分)