<s id="A18-2.35.01">35 Now we still have to explain some things that we require for pull and pressure, but not of the kind mentioned in the last book, rather, of greater importance than those, things that Archimedes and others have already clarified.</s>
<s id="A18-2.35.02">First now we want to show how one finds the center of gravity of an evenly thick and heavy triangle.</s>
<s id="A18-2.35.03">Let the known triangle be the triangle <abg> and let us divide the line <bg> at point <d> into two halves and connect the two points <a>, <d>.</s>
<s id="A18-2.35.04">If we now put the triangle onto the line <ad>, then it does not incline to any side, because the triangles <abd> and <adg> are equal.</s>
<s id="A18-2.35.05">If we further divide the line <ag> at point <e>, and connect the two points <b>, <e>, then put the triangle onto line <be>, then it does not incline to any side.</s>
<s id="A18-2.35.06">Since now the triangle, when put onto each of the two lines <ad> and <be>, is in balance in all its parts and does not incline to any side, then the common point of intersection is the center of this weight, namely the point <z>.</s>
<s id="A18-2.35.07">We have, however, to imagine the point <z> in the middle of the thickness of the triangle <abg>.</s>
<s id="A18-2.35.08">Now it turns out that if we connect the two points <a>, <d> and divide the line <ad> at point <z> into two parts in a manner that one of them, namely <az>, is twice the amount of <dz>, that point <z> is the center of gravity; for if we connect the two points <d>, <e>, then the line <ab> is parallel to line <de>, since the two lines <ag> and <bg> were bisected at the points <d> and <e>.</s>
<s id="A18-2.35.09">Then <ag> relates to <ge> like <ab> to <ed>; <ag> is, however, twice the amount of <ge>; consequently, <ab> is twice the amount of <ed>.</s>
<s id="A18-2.35.10">Furthermore, <ab> relates to <ed> like <az> to <dz>; consequently, <az> is twice the amount of <zd>, because the two figures <abz> and <dze> equal one another in their angles.</s>
