Question

证明三角形的重心与其三个顶点的连线的向量之和为零向量

Answer 1

设,三顶点A(x1,y1),B(x2,y2),C(x3,y3)

则重心O[(x1+x2+x3)/3,(y1+y2+y3)/3]

向量OA=[x1-(x1+x2+x3)/3,y1-(y1+y2+y3)/3]

向量OB=[x2-(x1+x2+x3)/3,y2-(y1+y2+y3)/3]

向量OC=[x3-(x1+x2+x3)/3,y3-(y1+y2+y3)/3]

则,向量OA+向量OB+向量OC

=[x1-(x1+x2+x3)/3,y1-(y1+y2+y3)/3]+[x2-(x1+x2+x3)/3,y2-(y1+y2+y3)/3]+[x3-(x1+x2+x3)/3,y3-(y1+y2+y3)/3]

=(0,0)

即得证