90 = 100: A Proof

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Construct a four-sided figure ABED as follows:

|∠ABE| = 90 ◦
|∠DEB| = 100 ◦
|AB| = |ED|

Using that as a starting point, we now tinker a bit to show that 90 = 100:

Draw the perpendicular bisectors to BE and AD; call the point where they meet "C".

Note: Actually, we must prove that those two perpendicular bisectors really do meet at all (i.e., that the point C even exists). In this case, it turns out to be pretty clear it's not hard to argue that lines AD and BE aren't parallel, and therefore their perpendicular bisectors aren't parallel, and so they must intersect (in Euclidean geometry). Still, be alert for people making glib assertions in proofs.

Figure 1.1 A construction to help prove that 90=100

Looking at this figure, some warning flags should be going up: How do we know C lies below BD? Might it lie above BD? Or exactly on BD? It turns out that the argument below is the same in all of these cases, though you'll certainly want to verify this to yourself later.

Table 1.1 A useful corollary: 0 = 1.
1		By construction.
2		C is on the perpendicular bisector of BE (thus ,BEC is isosceles).
3	$\angle CBE\cong \angle BEC$	Base angles of isosceles triangle BEC are congruent.
4	$\|\angle CBE\|=\|\angle BEC\|$	Congruent angles have equal measures; line 3.
5		C is on the perpendicular bisector of AD (thus ,ADC is isosceles).
6	$\triangle ABC \cong \triangle DEC (!!)$	Triangles with three congruent sides are congruent (Euclid's Side-Side-Side congruence theorem); lines 1,2,5.
7	(From here, it's just routine steps to conclude 90 = 100:)
8	$\angle ABC \cong \angle DEC$	Corresponding parts of congruent triangles are congruent; line 6.
9	$\|\angle ABC\| = \|\angle DEC\|$	Congruent angles have equal measures; line 8.
10	$\|\angle ABC\|=\|\angle ABE\|+\|\angle CBE\|$	By construction.
11	$\|\angle DEC\|=\|\angle DEB\|+\|\angle BEC\|$	By construction.
12	$\|\angle DEC\|=\|\angle DEB\|+\|\angle CBE\|$	Substituting equals with equals; lines 11 and 4.
13	$\|\angle ABC\|=\|\angle DEB\|+\|\angle CBE\|$	Substituting equals with equals; lines 12 and 9.
14	$\|\angle ABC\|+\|\angle CBE=\|\angle DEB\|+\|\angle CBE\|$	Substituting equals with equals; lines 13 and 10.
15	$\|\angle ABE\|=\|\angle DEB\|$	Subtracting equals from equals remains equal.
16	$90=\|\angle DEB\|$	By construction, and substituting equals with equals; line 15.
17		By construction, and substituting equals with equals; line 16.

1		Previous theorem.
2		Subtracting equals (90) from equals remains equal.
3		Dividing equals by non-zero equals (10) remains equal.

Exercise 1.1.1

If you feel this result is incorrect, then the challenge for you is to find the first line which is false. You may have noticed that the proof given here has some very minuscule steps e.g. "Congruent angles have equal measure." Usually such simple steps can be omitted, since they are obvious to any reader. We include them for a few reasons:

As a careful thinker, you should recognize that such small steps really are part of the complete reasoning, even if they're not worth mentioning continually.
If a computer is checking a proof, it needs to actually include those steps.
Programmers do need to be concerned with distinctions about (abstract) types the difference between angles and their measures, in this case.
Sometimes a line's justification is glibly given as "by construction", when that may not even be correct !-).

In this course, we'll spend a few weeks working with proofs which do include all the small, pedantic steps, to instill a mental framework for what a rigorous proof is. But after that, you can relax your proofs to leave out such low-level steps, once you appreciate that they are being omitted.

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90 = 100: A Proof