Tuesday, October 19, 2004

Find the Fallacy

How to prove a right angle congruent to an obtuse angle?

Please click the following link for the geometrical figure:

Consider the above picture: ABCD is a rectangle and E is a point near D slightly outside the rectangle so that AE is equal to AD. H is the middle of CD and K is the middle of CE. The perpendicular to CD going through H and the perpendicular to CE going though K intersect at a certain point J.
Now consider the sides of the triangles BCJ and AEJ: First, BC=AE (since both of these are equal to AD). Second, JB=JA (J is on the perpendicular bisector of CD, which is also that of AB). Third, CJ=EJ (J is on the perpendicular bisector of CE, by construction). The inescapable conclusion (I swear it's true!) is that BCJ and AEJ are congruent triangles (as their 3 sides are congruent), therefore the angles CBJ and EAJ are equal...
The angles JBA and JAB are equal (since JAB is an isosceles triangle), the picture clearly tells you that if you subtract JBA from CBJ and and JAB from JAE you obtain ABC and BAE. Yet one of these (ABC) is a right angle, whereas the other (BAE) is obtuse by construction. How can this be? Find the flaw or the fallacy here?


The geometrical figure is not possible to draw and hence misleading. Hence the argument depending upon the assumption of the possibility of the drawing leads us to the wrong conclusion. The possible right geometrical figure could be seen here:

The point J is so far out that the line JE is outside the angle AJD, not inside it, as the bad picture led you to believe...


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