In other words, given a straight line (call it BC), and some point A not on the given line, construct a line through A that is parallel to BC.

1) Let A be the given point, [br]2) and BC the given straight line.[br][br]Note: We must construct a straight line, through point A, such that the line is parallel to line BC.[br][br]3) Let a point D be taken at random on BC,[br]4) and let AD be joined.[br][br]5) On the straight line DA, and at the point A on it, let the angle DAE be constructed equal to the angle ADC [br][I. 23].[br][br]6) And let the straight line AF be produced in a straight line with EA.[br][br]Explanation: Then, since the straight line AD falling on the two straight lines BC, EF has made the alternate angles EAD, ADC equal to one another; therefore, EAF is parallel to BC.[br][br]Therefore, through the given point A, the straight line EAF has been drawn parallel to the given straight line BC. █