已知:△ABC是等边三角行.在边BC外有一点D ;△BCD的顶角∠BDC=120° 且BD=CD,过顶点D做一60° 角∠EDF .与边AB相交于E、与边AC相交于F DG为△DEF的垂线垂足为G.求证:△BDE≌△DEG应为三角形BDE也是直
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![已知:△ABC是等边三角行.在边BC外有一点D ;△BCD的顶角∠BDC=120° 且BD=CD,过顶点D做一60° 角∠EDF .与边AB相交于E、与边AC相交于F DG为△DEF的垂线垂足为G.求证:△BDE≌△DEG应为三角形BDE也是直](/uploads/image/z/8143831-55-1.jpg?t=%E5%B7%B2%E7%9F%A5%3A%E2%96%B3ABC%E6%98%AF%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E8%A1%8C.%E5%9C%A8%E8%BE%B9BC%E5%A4%96%E6%9C%89%E4%B8%80%E7%82%B9D+%EF%BC%9B%E2%96%B3BCD%E7%9A%84%E9%A1%B6%E8%A7%92%E2%88%A0BDC%3D120%C2%B0+%E4%B8%94BD%3DCD%2C%E8%BF%87%E9%A1%B6%E7%82%B9D%E5%81%9A%E4%B8%8060%C2%B0+%E8%A7%92%E2%88%A0EDF+.%E4%B8%8E%E8%BE%B9AB%E7%9B%B8%E4%BA%A4%E4%BA%8EE%E3%80%81%E4%B8%8E%E8%BE%B9AC%E7%9B%B8%E4%BA%A4%E4%BA%8EF+DG%E4%B8%BA%E2%96%B3DEF%E7%9A%84%E5%9E%82%E7%BA%BF%E5%9E%82%E8%B6%B3%E4%B8%BAG.%E6%B1%82%E8%AF%81%EF%BC%9A%E2%96%B3BDE%E2%89%8C%E2%96%B3DEG%E5%BA%94%E4%B8%BA%E4%B8%89%E8%A7%92%E5%BD%A2BDE%E4%B9%9F%E6%98%AF%E7%9B%B4)
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