Here is a false proof. I will prove that
an angle cuts two parallel lines into equal (congruent) segments. On the left,
I have drawn an arbitrary angle E which intersects the two arbitrary parallel
lines AB and CD. I will prove that AB=CD. Return to my Mathematics pages
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© Copyright 2002, Jim Loy
Here is a false proof. I will prove that
an angle cuts two parallel lines into equal (congruent) segments. On the left,
I have drawn an arbitrary angle E which intersects the two arbitrary parallel
lines AB and CD. I will prove that AB=CD.
Proof: The lengths AE/CE=BE/DE [This is a theorem of Euclid's: If a line intersects two sides of a triangle, and is parallel to the third side, then it cuts the first two sides into proportional segments (I will try to find out Euclid's wording here)]. AExDE=BExCE (Here x stands for multiplication). AExDExAB-AExDExCD=BExCExAB-BExCExCD [multiplying both sides by AB-CD]. AExDExAB-BExCExAB=AExDExCD-BExCExCD [adding and subtracting equals to and from equals]. AB(AExDE-BExCE)=CD(AExDE-BExCE). AB=CD [dividing both sides by AExDE-BExCE]. And that is what I was trying to prove.
Flaw in the proof: This is just a case of division by zero, which is illegal (see Two Equals One?). Since AE/CE=BE/DE, then AExDE-BExCE is zero. I tracked down this proof in order to show you that division by zero can be found in geometry. You have to stay alert, in order to avoid the trap.