@[da_ke](/user/766675) 注意:MN不平行CB
by da_ke @ 2024-04-10 21:47:28
@[da_ke](/user/766675) 建系（猜的）
by Dream_Creator @ 2024-04-10 21:58:46
@[Dream_Creator](/user/529722) 很明显不对
by da_ke @ 2024-04-10 21:59:37
@[da_ke](/user/766675) 确定不是求 AMN 面积吗
by Dream_Creator @ 2024-04-10 22:06:15
@[Dream_Creator](/user/529722) c,搞错了
by da_ke @ 2024-04-10 22:08:19
求$\triangle$ AMN 面积
by da_ke @ 2024-04-10 22:08:35
2.5
by Zi_Gao @ 2024-04-11 15:23:09
答案肯定是2.5，但我忘了怎么证了。
by 聊机 @ 2024-04-11 17:51:48
解：
$$
令 BD,CE 交于 O
\\
连接 DN,BN
\\
\because BM=DM
\\
\therefore S_{\triangle ADM}=\frac{1}{2}S_{\triangle ABD},S_{\triangle DMN}=\frac{1}{2}S_{\triangle BDN}
\\
同理 S_{\triangle ANC}=\frac{1}{2}S_{\triangle AEC}
\\
由图可知：
\\
\begin{aligned}
&S_{\triangle AMN}
\\
=&S_{\triangle ADM}+S_{\triangle DMN}-S_{\triangle ADN}
\\
=&\frac{1}{2}S_{\triangle ABD}+\frac{1}{2}S_{\triangle DBN}-(S_{\triangle ANC}-S_{\triangle CDN})
\\
=&\frac{1}{2}(S_{\triangle ABD}+S_{\triangle DBN}-S_{\triangle AEC})+S_{\triangle CDN}
\\
=&\frac{1}{2}(S_{\triangle BEN}-S_{\triangle CDN})+S_{\triangle CDN}
\\
=&\frac{1}{2}(S_{\triangle BEN}+S_{\triangle CDN})
\end{aligned}
\\
设 OD=m 则 OB=4-m
\\
\because BD \perp CE
\\
\therefore S_{\triangle BEN}=5-\frac{5m}{4},S_{\triangle CDN}=\frac{5m}{4}
\\
\therefore S_{\triangle AMN}=\frac{5}{2}
$$
by ALnAYuLvM @ 2024-04-11 19:27:36
@[da_ke](/user/766675) 问下楼主，这是几年级的题目
by ALnAYuLvM @ 2024-04-11 19:30:06