浙江 2017 · 数学 q14

题面

14．（6分）已知\triangle A B C$$A B = A C = 4，BC=2，点D 为AB延长线上一点，$B D = 2$，连结CD，则$\triangle{D C}$的面积是$c o m \_B D C =$

答案

${\frac{\sqrt{1 5}}{2}} , \ {\frac{\sqrt{1 0}}{4}}$

解析

分析

如图，取 BC 得中点 E，根据勾股定理求出 AE，再求出$S_{\Delta A B C}$，再根据$S_{\triangle B D C} = \frac{1}{2} S_{\triangle A B C}$即可求出，根据等腰三角形的性质和二倍角公式即可求出

解答

解：如图，取BC 得中点E，\because A B = A C = 4 , B C = 2 ,$$\therefore B E = \frac{1}{2} B C = 1 , A E \bot B C ,$$\therefore A E = \sqrt{A B^{2} - B E^{2}} = \sqrt{1 5} ,$$\therefore S_{\Delta A B C} = \frac{1}{2} \mathrm{B C} \bullet \mathrm{A E} = \frac{1}{2} \times 2 \times \sqrt{1 5} = \sqrt{1 5} ,$$\because B D = 2 ,$$\therefore S_{\Delta \mathrm{B D C}} = \frac{1}{2} S_{\Delta \mathrm{A B C}} = \frac{\sqrt{1 5}}{2} ,$$\because B C = B D = 2 ,$$\therefore \angle B D C = \angle B C D ,$$\therefore \angle A B E = 2 \angle B D C在$R t \triangle A B E^{\mathrm{,}}$中，\because \cos \angle A B E = \frac{B E}{A B} = \frac{1}{4} ,$$\therefore \cos \angle A B E = 2 \cos^{2} \angle B D C - 1 = \frac{1}{4} ,$$\therefore \cos \angle B D C = \frac{\sqrt{1 0}}{4} ,故答案为：${\frac{\sqrt{1 5}}{2}} , \ {\frac{\sqrt{1 0}}{4}}$