<s id="A18-1.08.01">[8] As for the case that a point that is moving in two motions of constant speed can follow unequal lines, we are going to prove that now.</s>
<s id="A18-1.08.02">Let us assume a rectangle, namely <abgd>, and let the line <ad> be a diagonal; further let the point <a> run in steady motion on the line <ab> and let the line move in steady motion on the two lines <ag>, <bd>, so that it is always parallel to line <gd>; also let the time in which point <a> runs to <b> be equal to the time in which line <ab> reaches <gd>; thus I say that point <a> in a certain time covers two unequal lines.</s>
<s id="A18-1.08.03">Proof for this: If line <ab> moves for a certain time and is placed on line <ez>, then the point moving on line <ab> in the same time comes to rest on line <ez> and a constant relation sets in.</s>
<s id="A18-1.08.04">The relation between line <ag> and line <ab>, i.e., with line <gd>, is, namely, equal to the relation of line <ae> with the line that lies between point <e> and the point moving on it.</s>
<s id="A18-1.08.05">Line <ag>, however, is to line <gd> as <ae> is to <eh>.Thus the point moving on line <ab> after <h> falls on the line <ad>, which is the diagonal.</s>
<s id="A18-1.08.06">Similarly we prove that the point running on line <ab> constantly moves forward on line <ad> and is at the same time moving on lines <ad> and <ab>.The two lines <ad> and <ab> are, however, different; thus, the point moving forward in steady motion covers unequal lines in the same time.</s>
<s id="A18-1.08.07">But, as I said, the motion of the point on line <ab> is simple, its motion on the diagonal <ad> however is derived from the motion of <ab> on the two lines <ag> and <bd> and the motion of <a> on line <ab>.Thus the one point <a> covers in steady motion two unequal lines. q.e.d.</s>
