4·2三角形的重心和几个著名的“难题”三角形的重心用向量方法来处理是最简单的了.对于△ABC,我们仍然任意定一个原点O,并作出有关各点的位置向量(图12中的虚线),于是BC的中点L对应于OL=21(OB+OC),先不问三条中线是否共点,先看一条中线AL,其上的点都可用λOA+(1-λ)OL=λOA+12 The center of gravity of a triangle and the center of gravity of several famous “puzzle” triangles are most straightforward to deal with by vector methods. For △ ABC, we still arbitrarily set an origin O and make a position vector for each point 12 in the dotted line), so the BC middle point L corresponds to OL = 21 (OB + OC), first do not ask whether the three midline point, first look at a center AL, the point can be used λOA + (1-λ) OL = λOA + 12