Theorem 6.1: If two parallel lines are transected by a third, the alternate interior angles are the same size.
Proof: Let A and B denote the points where the transverse line meets the parallel lines. Let C be the midpoint of the line segment between A and B. Let DE be the line through C which is perpendicular to one, and hence, by Theorem 1.14, both of the parallel lines.
Let D be the point where the perpendicular line meets the line which contains the point A and let E be the point where the perpendicular line meets the line that contains the point B. If we rotate the figure 180o about point C, then point A is moved to point B and vice versa. Let D' be the point to which D is moved. Then D' and B are points on the line to which the line which is determined by AD is moved, So the line determined by D' and B is the line to which the line determined by AD is moved . Since rotations preserve the size of angles, by Theorem 5.6, / CD'B is a right angle. Thus, the line determined by D' and B is a line through B which is parallel to AD by Theorem 1.15. But, there is only one line through B, which is parallel to AD, so BE = BD'. So / CAD is moved to / CBE. Since a rotation is an invertible isometry by Theorem 5.12, it preserves the size of the angles by Theorem 5.6. We can conclude that the alternate interior angles are the same size.