52 Statics. 



Fi. 31. ^* Hence, (1.) In order to have the centre of gravity of the 

 perimeter of a polygon, it is necessary, from the middle of each 

 of the sides, to let fall perpendiculars upon two fixed lines AE r 

 AC, taken in the plane of this polygon ; and, considering the 

 gg. weight of each side as united in the middle of this side, to seek 

 the common centre of gravity of these weights in the manner 

 already explained. 



92. (2.) The centre of gravity of the surface of a parallelogram is 



the middle point of the line which joins the middle of two opposite 



Fig. 32. sides. For, by considering the parallelogram as composed of 



material lines, parallel to these two sides, each will have its cen- 



tre of gravity in the line which passes through the middle of 



these same sides. The common centre of gravity, therefore, 



of all the lines will be in the bisecting line. Jt will, moreover, 



be in its middle point, since this line, considered as sustaining the 



90. weights of all the other lines, is uniformly heavy. 



Fig. 33. 93. (3.) To find the centre of gravity of a triangle ABC ; we 

 draw from the vertex A to the middle D of the opposite side J5C, 

 the straight line DA, and from the point D we take DG = 



Indeed, the straight line DA, which divides BC into two 

 equal parts at the point D, divides also into two equal parts 

 every other line ZJV, parallel to BC-, accordingly, if we consider 

 the surface of the triangle as an assemblage of material lines 

 parallel to BC, the line DA, which passes through the particular 

 centres of gravity of all these lines, will also pass through their 

 common centre of gravity, that is, through the centre of gravity 

 of the triangle. For the same reason, the line CE, which passes 

 through the middle of AB, will in like manner pass through the 

 centre of gravity of the triangle. This centre is consequently at 

 the point of intersection G of the two lines CE and DA. Now, if 

 we join ED, it will be parallel to AC, since it divides into two 



Geom. equal parts the sides AB, BC. The two triangles EGD, AGC, 

 are accordingly similar, as well as the triangles ABC, EBD ; we 



Geom. have, therefore, 



DG : AG : : DE : AC : : BD : BC : : 1 : 2 ; 

 that is, DG is half of AG, and therefore one third of AD. 



