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在 ${\displaystyle \mathrm{\Delta }ABC}$ 中，${\displaystyle AB=1,BC=3}$, 在 ${\displaystyle AC}$ 的右侧取点 ${\displaystyle D}$, 构成平面四边形 ${\displaystyle ABCD}$.

1. 若 ${\displaystyle \cos B+\cos D=0}$ 且 ${\displaystyle B={120}^{\circ }}$，求 ${\displaystyle \mathrm{\Delta }ACD}$ 面积的最大值;
2. 若 ${\displaystyle AD=CD=2}$, 当四边形 ${\displaystyle ABCD}$ 的面积最大时，求对角线 ${\displaystyle BD}$ 的长.

1. 由条件，${\displaystyle D={60}^{\circ }}$.在 ${\displaystyle \mathrm{\Delta }ABC}$ 中由余弦定理得$$\begin{array}{r}AC=\sqrt{1^2+3^2-2\times \frac{1}{2}\times 3}=\sqrt{13},\end{array}$$
   故在${\displaystyle \mathrm{\Delta }ADC}$中$$\begin{array}{r}\frac{1}{2}=\frac{x^2+y^2-13}{2xy}\ge \frac{2xy-13}{2xy}\Rightarrow xy\le 13,\end{array}$$
   从而 ${\displaystyle S_{\mathrm{\Delta }ACD}={\displaystyle \frac{1}{2}xy\sin D\le \frac{13\sqrt{3}}{4}}}$.
2. 由题意，用两次余弦定理得$$\begin{array}{r}A{C}^2=10-6\cos B=8-8\cos D\Rightarrow 3\cos B-4\cos D=1.\end{array}$$
   这样$$\begin{array}{rl}{S}_{ABCD}=& \frac{1}{2}(3\sin B+4\sin D)=\frac{3\sin \alpha +4\sin \beta +\alpha (3\cos \alpha -4\cos \beta )-\alpha }{2}\\ =& \frac{3(\sin B+\alpha \cos B)+4(\sin D-\alpha \cos D)-\alpha }{2}\le \frac{7\sqrt{1+{\alpha }^2}-\alpha }{2},\end{array}$$
   由待定系数法可得 ${\displaystyle {\displaystyle \alpha =\frac{1}{4\sqrt{3}}}}$，取等时 ${\displaystyle \cos B={\displaystyle \frac{\alpha }{\sqrt{1+{\alpha }^2}}=\frac{1}{7}}}$, ${\displaystyle \cos D=\frac{-\alpha }{\sqrt{1+{\alpha }^2}}=-\frac{1}{7}}$，从而 ${\displaystyle AC={\displaystyle \frac{8\sqrt{7}}{7}}}$. 由 ${\displaystyle B,D}$ 互补，得 ${\displaystyle A,B,C,D}$ 共圆，因此由托勒密定理得 ${\displaystyle {\displaystyle BD=\frac{AB\cdot CD+AD\cdot BC}{AC}=\sqrt{7}}}$.