Question

已知三角形abc,p是任意一点p到三边距离分别为H1,H2,H3三角形abc的高为h试求h1,h2,h3,h之间的关系

Answer 1

结论是 h=h1+h2+h3

证明：连接AP,Bp,CP,将原来的△ABC,分割成△APB、△BPC、△APC.

则△APB面积 = (1/2)AB・DP = (1/2)ah1 (a为等边△ABC的边长)

△BPC面积 = (1/2)BC・FP = (1/2)ah3

△APC面积 = (1/2)AC・EP = (1/2)ah2

三个三角形的面积总和 = (1/2)ah1+(1/2)ah2+(1/2)ah3 = (1/2)a(h1+h2+h3)

三个三角形面积总和,就是原△ABC的面积,即等于(1/2)ah

所以,可得 h=h1+h2+h3