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.
Proposition 3.16: For every line l and every point P there exists a line P perpendicular to l.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Let AB be a line along the length of the (horizontal) base of a shoe box.
Let l be a line through A along the width of the base of the shoe box.
You are going to construct line m perpendicular to AB.
There is no requirement for that line to be parallel to l.
It does not need to be on the same plane determined by AB and l.
Let m be a vertical line through B.
lines l and m are both perpendicular to line AB.
They do not intersect, but they are not parallel
(or even in the same plane).
Take any point C on m, other than B.
Points A, B and C determine a plane perpendicular to line l.
All lines on that plane that go through A are perpendicular to line l.
Line AC is one of them. It intersects line l at point D, which happens to be point A.
You have just constructed triangle ABC.
You have other options for a line m that is not coplanar with l,
and may give you a point D different from A,
but ABCD will not be planar, so it will not be a rectangle.
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