document.write( "Question 1075375: Find the flaw in the following argument purporting to construct a rectangle. Let A and B be any two points. There is a line l through A perpendicular to line AB (proposition 3.16) and, similarly, there is a line m through B perpendicular to line AB. Take any point C on m other than B. There is a line through C perpendicular to l- let it intersect l at D. Then ABCD is a rectangle.\r
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document.write( "Proposition 3.16: For every line l and every point P there exists a line P perpendicular to l. \n" );
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Algebra.Com's Answer #690092 by KMST(5328)![]() ![]() You can put this solution on YOUR website! Let AB be a line along the length of the (horizontal) base of a shoe box. \n" ); document.write( "Let l be a line through A along the width of the base of the shoe box. \n" ); document.write( "You are going to construct line m perpendicular to AB. \n" ); document.write( "There is no requirement for that line to be parallel to l. \n" ); document.write( "It does not need to be on the same plane determined by AB and l. \n" ); document.write( "Let m be a vertical line through B. \n" ); document.write( "lines l and m are both perpendicular to line AB. \n" ); document.write( "They do not intersect, but they are not parallel \n" ); document.write( "(or even in the same plane). \n" ); document.write( "Take any point C on m, other than B. \n" ); document.write( "Points A, B and C determine a plane perpendicular to line l. \n" ); document.write( "All lines on that plane that go through A are perpendicular to line l. \n" ); document.write( "Line AC is one of them. It intersects line l at point D, which happens to be point A. \n" ); document.write( "You have just constructed triangle ABC. \n" ); document.write( "You have other options for a line m that is not coplanar with l, \n" ); document.write( "and may give you a point D different from A, \n" ); document.write( "but ABCD will not be planar, so it will not be a rectangle. \n" ); document.write( " |