Question

已知等边三角形ABC和点P 设点P到三角形三边AB,AC,BC的距离分别为h1 h2 h3 △ABC的高为h

Answer 1

结论是 h=h1+h2+h3

证明：连接AP,BO,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