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Proof that 90=100



We construct a 4-sided figure ABED as follows:
|∠ABE| =  90 degrees
|∠BED| = 100 degrees
|AB|   = |ED|

Once we have that, we'll now tinker a bit to show that 90=100:
Draw the perpendicular bisectors to BE and AD; they'll meet
at some point we'll name C.
Note that while the figure is indeed not necessarily to scale,
this argument stays the same whether C is below 
BD, or above it, on BD.
(You should verify this later.)

(Lemma: we must prove that those two lines indeed meet (i.e. that the 
point C even exists); wouldn't be too hard to do this, since it's 
pretty clear that AD and BE aren't parallel.)


 1.   |AB|  = |ED|    by construction
 2.   |BC|  = |EC|    C on perp.bisector of AE   (thus triangle BEC is isosceles)
 3.   ∠CBE  ≈  ∠BEC  base angles of isosocles triangle BEC are congruent
 4.  |∠CBE| = |∠BEC|  congruent angles have equal measures, and line3
 5.   |AC|  = |DC|    C on perp.bisector of AE
 6.    ABC  ≈ DEC     (!!) by SSS, and lines 1,2,5.

From here, it's just routine steps to conclude 90=100:
 7.   ∠ABC  ≈  ∠DEC      corresponding parts of corresponding triangles; 6
 8.  |∠ABC| = |∠DEC|   congruent angles have equal measures; line 7
 9.  |∠ABC| = |∠ABE| + |∠CBE|   by construction
10.  |∠DEC| = |∠DEB| + |∠BEC|   by construction
11.  |∠DEC| = |∠DEB| + |∠CBE|   subst. equals for equals; lines 10 and 4
12.  |∠ABC| = |∠DEB| + |∠CBE|   subst. equals for equals; lines 11 and 8
13.  |∠ABE| + |∠CBE| = |∠DEB| + |∠CBE|   subst. equals for equals; lines 12 and 9
14.  |∠ABE| = |∠CBE|   subtract equals from equals remains equal
15.      90 = |∠CBE|   by construction, and replace-equals-by-equals
16.      90 = 100      by construction, and replace-equals-by-equals






A useful corollary: 0=1.
 1. 90 = 100    Previous theorem.
 2.  0 = 10     Subtract 90 from both sides remains equal
 3.  0 = 1      Divide by 10 (non-zero) on both sides remains equal.





Note that were C to occur above BE,
then the proof is essentially the same;
we just change lines 9 and 10 to have a - instead of a +,
and on line 13 we add (rather than subtract) the same to both sides.

(If C occured exactly on BE, then either proof would hold,
as we'd be adding or subtracting zero.)




Challenge:
Identify exactly, which is the first line in error.


(If you can't, you are welcome to give me $100,
and i'll return you an equal amount. :-)









This result suggests we need to be good at proofs.