设⊙O为正三角形ABC的内切圆,E F是AB AC上的切点,劣弧EF上任一点P到BC CA AB的距离分别为d1,d2,d3.求证:根号d1=根号d2+根号d3.答案是这样的:作PP1垂直BC于P1,作PP2垂直AC于P2,PP3垂直AB于P3,连结EF交PP1
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![设⊙O为正三角形ABC的内切圆,E F是AB AC上的切点,劣弧EF上任一点P到BC CA AB的距离分别为d1,d2,d3.求证:根号d1=根号d2+根号d3.答案是这样的:作PP1垂直BC于P1,作PP2垂直AC于P2,PP3垂直AB于P3,连结EF交PP1](/uploads/image/z/3028546-10-6.jpg?t=%E8%AE%BE%E2%8A%99O%E4%B8%BA%E6%AD%A3%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%86%85%E5%88%87%E5%9C%86%2CE+F%E6%98%AFAB+AC%E4%B8%8A%E7%9A%84%E5%88%87%E7%82%B9%2C%E5%8A%A3%E5%BC%A7EF%E4%B8%8A%E4%BB%BB%E4%B8%80%E7%82%B9P%E5%88%B0BC+CA+AB%E7%9A%84%E8%B7%9D%E7%A6%BB%E5%88%86%E5%88%AB%E4%B8%BAd1%2Cd2%2Cd3.%E6%B1%82%E8%AF%81%EF%BC%9A%E6%A0%B9%E5%8F%B7d1%3D%E6%A0%B9%E5%8F%B7d2%2B%E6%A0%B9%E5%8F%B7d3.%E7%AD%94%E6%A1%88%E6%98%AF%E8%BF%99%E6%A0%B7%E7%9A%84%EF%BC%9A%E4%BD%9CPP1%E5%9E%82%E7%9B%B4BC%E4%BA%8EP1%2C%E4%BD%9CPP2%E5%9E%82%E7%9B%B4AC%E4%BA%8EP2%2CPP3%E5%9E%82%E7%9B%B4AB%E4%BA%8EP3%2C%E8%BF%9E%E7%BB%93EF%E4%BA%A4PP1)
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