<s id="A18-2.38.01">38 If <abg> is an evenly thick and heavy triangle and under the points <abg> there are supports in the same position, we want to show how to find the amount of the weight of the triangle <abg> that each of them bears.</s>
<s id="A18-2.38.02">Let us bisect <bg> at point <d> and connect the two points <a> and <d>, divide the line <ad> at point <e> so that the part <ae> is twice the amount of <ed>, then the point <e> is the point of the entire weight of the triangle.</s>
<s id="A18-2.38.03">Now we have to distribute it on the supports.</s>
<s id="A18-2.38.04">But if we imagine the line <ad> in equilibrium when it is suspended at point <e>, then the load at <d> is twice the amount of that at <a>, because the line <ae> is twice the amount of line <de>.</s>
<s id="A18-2.38.05">And if we imagine the weight at <d> distributed to the two points <b>, <g> and the line <bg> is in equilibrium, then at each of the two points <b>, <g> rests half of the weight that is at <d>, because the two lines <bd> and <dg> are equal to one another.</s>
<s id="A18-2.38.06">The weight at <d> was, however, twice the amount of the weight at <a>; consequently, the loads at points <a>, <b>, <g> are equal to one another and thus the supports bear equal weights.</s>
