Let A and B be two points. These points exist by Euclid's first postulate as does the line AB. We can use his 3rd postulate to produce a point C on line AB that has the same distance from A that B does. We can use the 2nd betweenness axioms to find points midway between A and B and A and C. We use again the third posutlate to construct two circles centered a B and C and their intersections of which there are only two form another line, l, by the first postulate this line will be perpendicular with AB and have A as a point. We can do the same thing to the point B producing the line m. We still have not used the fifth postulate. [br][br]The flawed proof asserts that there is a line through a point on the line m, not equal to B that is perpendicular to l. The fifth postulate is not used in the construction of this line as the applet below demonstrates. Where the proof is flawed is now where it is suggested that this line is also perpendicular to m. The definition of a rectangle is that all four angles are of 90 degrees. and just because m is perpendicular to AB and there is a line perpendicular to l that intersects m at a point on m not equal to B does not mean that this line and m are perpendicular.