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.

next theorem (6.2)