<s id="A18-2.39.01">39 Let further the triangle <abg> be evenly heavy and thick, on supports in the same position, and let any weight be put on or suspended at point <e>, in fact, let point <e> have any random position, then we want to find out, how much of the weight at <e> each of the supports bears.</s>
<s id="A18-2.39.02">Let us draw <ae> and extend it towards <d>, divide the weight at <e> so that, if the triangle lies on line <ab> in equilibrium, the load at <d> relates to the load at <a> like the line <ae> to the line <ed>. Let us further divide the weight at <d> so that <bg>, if it is suspended, is in equilibrium, then the weight of <g> relates to the weight of <b> like the line <bd> to the line <gd>. </s>
<s id="A18-2.39.03">The weight at <d> has been determined; consequently the two weights <g>, <b> are determined.</s>
<s id="A18-2.39.04">But the weight at <a> has also been determined; consequently the weights that rest on the supports, are determined.</s>
