{ "steps": [ { "title": "将军饮马问题" }, { "subtitle": "如图，∠ABC=30°，D是它内部一点，BD=2，E、F分别是，BABC上的两个动点，则△DEF周长的最小值为多少" }, { "draw": "创建角ABC，其中∠ABC=30°", "voice_text": "画一个角ABC，角度为30度。" }, { "draw": "创建点D，在角ABC内部，BD=2", "voice_text": "在角内取点D，使BD=2。" }, { "draw": "在边BA上创建动点E，在边BC上创建动点F", "voice_text": "在BA、BC上分别取动点E、F。" }, { "draw": "连接DE、DF、EF，形成三角形DEF", "voice_text": "连接DE、DF、EF" }, { "draw": "展示点E和点F运动动效", "voice_text": "观察三角形DEF周长的变化。" }, { "voice_text": "首先，我们看下题目中的条件，D、E、F三个都是动点，所以DE、DF、EF三条线段都是不固定的，我们要求的是这三条线段的最小值" }, { "voice_text": "我们通过对称的方式，不改变线段长度的前提下，改变线段的方向。" }, { "draw": "作点D关于直线AB的对称点G", "voice_text": "过D点，做AB的对称点G" }, { "draw": "连接GE", "voice_text": "连接GE。" }, { "draw": "高亮DE、GE", "process_text": "DE=GE", "voice_text": "DE的长度就是GE的长度" }, { "draw": "作点D关于直线BC的对称点H", "voice_text": "再过D点,做关于BC的对称点H" }, { "draw": "连接HF", "process_text": "HF=DF", "voice_text": "连接HF" }, { "voice_text": "这个时候，这三条线段变为同向的折线。" }, { "voice_text": "什么时候最短呢？" }, { "draw": "两点G、H之间最短路径，其中动点E在AB上，动点F在BC上", "voice_text": "那就是当G、E、F、H共线的时候，三角形DEF的周长最小。" }, { "voice_text": "我们怎么去求周长的最小值呢？" }, { "draw": "连接GH，连接BG，BH", "voice_text": "连接GH，连接BG，BH" }, { "draw":"标记：角ABC=30°", "voice_text": "题目中已知角ABC等于三十度" }, { "draw":"连接BD", "voice_text": "连接BD之后" }, { "draw":"标记：角ABD=1", "voice_text": "我们假设角ABD是角一" }, { "draw":"标记：角DBC=2", "voice_text": "角DBC是角二" }, { "draw":"标记：角ABH=2，角CBH=2", "voice_text": "那么角ABH就是角一，CBH就是角二" }, { "voice_text": "角一加角二等于三十度", }, { "process_text": "∴∠GBH=2∠ABC ∵∠ABC=30° ∴∠GBH=60°", "voice_text": "那么角GBH就是六十度" }, { "draw": "标记：BD=2", "voice_text": "同时BD的长度是二" }, { "draw": "标记：BG=2，BH=2", "process_text": "BD=BG=BH=2", "voice_text": "那么BG的长度也是二，BH的长度也是二" }, { "process_text": "∴△GBH是等边三角形", "voice_text": "那么三角形GBH是等边三角形" }, { "process_text": "∴GH=BD ∵BD=2", "voice_text": "GH的长度就是BDE的长度就等于二" }, { "process_text": "∴△DEF周长的最小值为2。", "voice_text": "所以三角形DEF周长的最小值就是二。这道题答案就是2" } ] } 把上面的步骤变成讲题视频