在三角形ABC中,AB=AC,角A=30°P为BC上一点,PE⊥AB,PF⊥AC,垂足分别为E,F,则PE+PF为?1/2AB.但我不知道如何做辅助线大侠帮帮吧.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 00:10:14
![在三角形ABC中,AB=AC,角A=30°P为BC上一点,PE⊥AB,PF⊥AC,垂足分别为E,F,则PE+PF为?1/2AB.但我不知道如何做辅助线大侠帮帮吧.](/uploads/image/z/8577660-12-0.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CAB%3DAC%2C%E8%A7%92A%3D30%C2%B0P%E4%B8%BABC%E4%B8%8A%E4%B8%80%E7%82%B9%2CPE%E2%8A%A5AB%2CPF%E2%8A%A5AC%2C%E5%9E%82%E8%B6%B3%E5%88%86%E5%88%AB%E4%B8%BAE%2CF%2C%E5%88%99PE%2BPF%E4%B8%BA%3F1%2F2AB.%E4%BD%86%E6%88%91%E4%B8%8D%E7%9F%A5%E9%81%93%E5%A6%82%E4%BD%95%E5%81%9A%E8%BE%85%E5%8A%A9%E7%BA%BF%E5%A4%A7%E4%BE%A0%E5%B8%AE%E5%B8%AE%E5%90%A7.)
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