122 CENTER OF GRAVITY 



the center of gravity must lie in the line BE, where E is the 

 middle point of AC. 



These two results fully fix the position of the center of gravity ; 

 it must be at the intersection of the lines AD, BE. 



Join DE. Then the triangles DCE, BCA are two similar 

 triangles, the former being just half the size of the latter. Thus 

 DE must be parallel to AB, and of half the length of AB. 



It now follows that DGE, AGB are similar triangles, of which 



the former is half the size of the latter. 

 Hence GD is half of AG. 



Thus G divides AD in the ratio 2 : 1. 

 If we join C to F, the middle point of 

 AB, we can in the same way show that 

 CF must divide AD in the ratio 2:1. 

 Thus CF must also pass through G. 

 FIG 67 The three lines AD, BE, CF, which 



join the vertices of the triangle to the 



middle points of the opposite sides, are called the medians of the 

 triangle. We have shown that the three medians meet in the same 

 point G, and that this point is the center of gravity of the triangle. 

 We have also shown that the center of gravity divides any median 

 in the ratio 2:1, i.e. that it is one third of the way up the median, 

 starting from the base. 



93. Center of gravity of any polygon. The center of gravity 

 of any rectilinear polygon can be found by dividing it up into 

 triangles and replacing each triangle by a particle at its center of 

 gravity. 



EXAMPLES 



1. Show that the center of gravity of a triangle coincides with that of three 

 equal particles placed at its angular points. 



2. Prove that if the center of gravity of a triangle coincides with its ortho- 

 center the triangle is equilateral. 



3. A cardboard square is bent along a diagonal until the two parts are at 

 right angles. Find the position of its center of gravity. 



4. A quarter of a triangular lamina is cut off by a line parallel to its base. 

 Where is the center of gravity of the remainder? 



