《2025年领智优选浙江期末复习卷九年级数学全一册浙教版》

23. (9 分)如图,等腰$\triangle ABC$内接于$\odot O$,$AB = AC$,D 为$\overset{\frown}{AC}$上一点,连结 BD 交 AC 于点 E,连结 AD 并延长交 BC 的延长线于点 F.
(1)求证:$\triangle CDF\backsim\triangle ABF$;
(2)若$BD\perp AC$.
①求证:$\angle BAC = 2\angle CAF$;
②当$\frac{AB}{BC}=\frac{\sqrt{10}}{2}$时,求$\frac{S_{\triangle CDF}}{S_{\triangle ABF}}$的值.

答案：
23.
(1)证明：$\because$等腰$\triangle ABC$内接于$\odot O$，$AB = AC$，D为$\overset{\frown}{AC}$上一点，$\therefore \angle BAD+\angle BCD = 180^{\circ}$.
$\because \angle BCD+\angle FCD = 180^{\circ}$，$\therefore \angle BAD = \angle FCD$，
$\because \angle F = \angle F$，$\therefore \triangle CDF\backsim \triangle ABF$.
(2)①证明：等腰$\triangle ABC$内接于$\odot O$，如图1，过点A作$AG\perp BC$于点G，则$\angle AGC = 90^{\circ}$.

$\therefore \angle BAC = 2\angle CAG$，$\angle ACG+\angle CAG = 90^{\circ}$.
$\because BD\perp AC$，$\therefore \angle BEC = 90^{\circ}$，
$\therefore \angle BCE+\angle CBE = 90^{\circ}$，$\therefore \angle CAG = \angle CBE$.
$\because \angle CAF = \angle CBD$，$\therefore \angle BAC = 2\angle CAF$.
②解：延长AG交$\odot O$于点H，连结BH，如图2，

由①知，$AG\perp BC$，$BG = CG$，
$\therefore AG$过点O，
$\therefore \angle ABH = \angle AGB = 90^{\circ}$.
$\because \angle BAH = \angle GAB$，
$\therefore \triangle AHB\backsim \triangle ABG$，
$\therefore \frac{BH}{BG}=\frac{AB}{AG}$，
$\because \frac{AB}{BC}=\frac{\sqrt{10}}{2}$，$\therefore \frac{AB}{BG}=\sqrt{10}$，$AB = \sqrt{10}BG$，
在直角三角形ABG中，由勾股定理，得$AG = \sqrt{AB^{2}-BG^{2}} = 3BG$，$\therefore BH=\frac{\sqrt{10}}{3}BG$，
$\because \angle BAH = \angle CAH = \angle CBD = \angle CAD$，
$\therefore CD = BH=\frac{\sqrt{10}}{3}BG$，
由
(1)知，$\triangle CDF\backsim \triangle ABF$，
$\therefore \frac{S_{\triangle CDF}}{S_{\triangle ABF}}=(\frac{CD}{AB})^{2}=(\frac{\frac{\sqrt{10}}{3}BG}{\sqrt{10}BG})^{2}=\frac{1}{9}$.