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### 将军饮马
#### 模型3
![](http://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/2023/02/4be9532200d21449d7715028d96c3571.png)
**本题由12升级而来不再是两静一定而是两动一定了**
但是很显然,它符合将军饮马问题的基本条件,就是 <font color='red' size=4><b>有公共端点的,两个线段和的最小值</b></font>
那该怎么做呢?
答:先不管什么$N$是不是动点,我们直接按模型$2$来做,视$N$为定点。则有关于河$BD$做出对称点$N'$,连接$C \sim N'$,则$CN'$就是答案。
![](http://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/2023/02/881cbfec64e9a0ca208640813d361609.png)
但问题是$N$是动点,$N'$也是动点,没法求啊~
我们再来看一下条件,有一个$BD$是角平分线的条件,所以$N$关于角平分线的对称点,**一定** 是在$BA$上!
问题就进一步转化为 **求$C$到$AB$上某一个点的距离最短**,那不就是垂线段最短吗?
所以,从$C$引垂线到$AB$,垂足为$H$,求出$HC$就是答案。
因为知道三角形面积$S_{\triangle ABC}=1/2 \times AB \times CH$
$18=1/2 \times 12 \times CH$
$\therefore CH=3$