如图,将▱ABCD(纸片)沿过对角线交点O的直线EF折叠,点A落在点A1处,点B落在点B1处,设FB1交CD于点G,A1B1分别交CD,DE于点H,I.求证:EI=FG
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![如图,将▱ABCD(纸片)沿过对角线交点O的直线EF折叠,点A落在点A1处,点B落在点B1处,设FB1交CD于点G,A1B1分别交CD,DE于点H,I.求证:EI=FG](/uploads/image/z/3181816-64-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B0%86%26%239649%3BABCD%EF%BC%88%E7%BA%B8%E7%89%87%EF%BC%89%E6%B2%BF%E8%BF%87%E5%AF%B9%E8%A7%92%E7%BA%BF%E4%BA%A4%E7%82%B9O%E7%9A%84%E7%9B%B4%E7%BA%BFEF%E6%8A%98%E5%8F%A0%2C%E7%82%B9A%E8%90%BD%E5%9C%A8%E7%82%B9A1%E5%A4%84%2C%E7%82%B9B%E8%90%BD%E5%9C%A8%E7%82%B9B1%E5%A4%84%2C%E8%AE%BEFB1%E4%BA%A4CD%E4%BA%8E%E7%82%B9G%2CA1B1%E5%88%86%E5%88%AB%E4%BA%A4CD%2CDE%E4%BA%8E%E7%82%B9H%2CI.%E6%B1%82%E8%AF%81%EF%BC%9AEI%3DFG)
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如图,将▱ABCD(纸片)沿过对角线交点O的直线EF折叠,点A落在点A1处,点B落在点B1处,设FB1交CD于点G,A1B1分别交CD,DE于点H,I.求证:EI=FG
如图,将▱ABCD(纸片)沿过对角线交点O的直线EF折叠,点A落在点A1处,点B落在点B1处,设FB1交CD于点G,A1B1分别交CD,DE于点H,I.求证:EI=FG如图,将▱ABCD(纸片)沿过对角线交点O的直线EF折叠,点A落在点A1处,点B落在点B1处,设FB1交CD于点G,A1B1分别交CD,DE于点H,I.求证:EI=FG
简易证法:
▱ABCD(纸片)沿过对角线交点O的直线EF折叠,▱ABCD沿折线分成的两部分为全等的四边形,
∠A1=∠A=∠C,A1E=CF,∠A1EI=∠CFG(∵∠A1EI的两边分别平行于∠CFG的两边)
∴⊿A1EI≌⊿CFG,即得EI=FG.