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在平面四边形 ${\displaystyle ABCD}$ 中, ${\displaystyle AB\perp BC}$, ${\displaystyle BC\perp CD}$, ${\displaystyle \mathrm{\angle }CAD={45}^{\circ }}$; ${\displaystyle BC=1}$, ${\displaystyle \mathrm{\Delta }ACD}$ 和 ${\displaystyle \mathrm{\Delta }ABC}$ 的面积分别为 ${\displaystyle S_1,S_2}$, 则 ${\displaystyle \frac{S_1}{S_2}}$ 的最小值为_____.

按图示建系. 设 ${\displaystyle k_{CA}=1}$, ${\displaystyle k_{AD}=k_2}$. 根据倒角公式, ${\displaystyle \mathrm{\angle }CAD={45}^{\circ }}$ 意味着 $$\frac{k_1-k_2}{1+k_1{k}_2}=1\Rightarrow k_1=\frac{1+k_2}{1-k_2}.$$ 这样 $$\begin{array}{rl}\frac{S_1}{S_2}& =1-\frac{S_{\mathrm{\Delta }ADE}}{S_2}=1-\frac{DE}{AB}=1-\frac{k_1}{k_2}=1-(\frac{1}{k_2}-\frac{2}{k_2-1}).\end{array}$$ 根据反向柯西不等式, $$(\frac{1}{k_2}-\frac{2}{k_2-1})(k_2-(k_2-1))\le (1-\sqrt{2}{)}^2,$$ 所以 ${\displaystyle \frac{S_1}{S_2}\ge 1-(1-\sqrt{2}{)}^2=2\sqrt{2}-2}$.

作 ${\displaystyle AF\perp AC}$, 与 ${\displaystyle CD}$ 的延长线交于 ${\displaystyle F}$. 作 ${\displaystyle AE\perp CF}$ 于 ${\displaystyle E}$. 设 ${\displaystyle \mathrm{\angle }CAB=\mathrm{\angle }ACF=\mathrm{\angle }EAF=\alpha }$. 这样 ${\displaystyle AD}$ 是直角 ${\displaystyle \mathrm{\angle }CAF}$ 的角平分线. 根据角平分线定理, $$\frac{CD}{CD+DF}=\frac{AC}{AC+AF}\Rightarrow CD=CF\cdot \frac{AC}{AC+AF}=\frac{1}{\sin \alpha \cos \alpha }\cdot \frac{\frac{1}{\sin \alpha }}{\frac{1}{\sin \alpha }+\frac{1}{\cos \alpha }}.$$ 而 ${\displaystyle AB=\frac{1}{\tan \alpha }}$, 所以 $$\frac{S_1}{S_2}=\frac{CD}{AB}=\frac{1}{\cos \alpha (\sin \alpha +\cos \alpha )}=\frac{1}{\frac{1}{2}\sin 2\alpha +\frac{1}{2}\cos 2\alpha +\frac{1}{2}}\ge \frac{1}{\frac{\sqrt{2}}{2}+\frac{1}{2}}=2\sqrt{2}-2.$$