Postulate 1: Law of Trichotomy

For any two values x,y, there are only three possible states:

(a) x = y

(b) x is less than y

(c) x is greater than y

This is one of the postulates of real numbers. See here for details on constructing real numbers.

Lemma 1: Similar polygons inscribed in circles are to one another as the squares on their diameters

Proof:

(1) By assumption, polygon ABCDE is similar to polygon FGHKL with BM and GN being the diameters of circles.

(2) From the property of similar polygon (see definition above), we have:

∠ BAE ≅ ∠ GFL

BA/AE = GF/GL

(3) triangle ABE is equiangular to triangle FGL [See Lemma 1 here for details]

(4) Since they are equiangular, we know that:

∠ AEB ≅ ∠ FLG

(5) Since both angles open on the same length of the circumference (see here):

∠ AEB ≅ ∠ AMB

∠ FLG ≅ ∠ FNG

(6) From (4) and (5), we can conclude that:

∠ AMB ≅ ∠ FNG

(7) Since BM and GN are both diameters of the circle (see here),

we can conclude that both ∠ BAM and ∠ GFN are right angles.

(8) Since the angles of a triangle add up to 180 degrees (see Lemma 4 here), we can conclude from step #6 and step #7 that triangle ABM is equiangular to triangle FGN.

(9) From the properties of equiangular triangles (see Lemma 3 here), we know that:

BM/GN = BA/GF

(10) From (9), we can conclude that:

(BM/GN)

^{2}= (BA/GF)

^{2}= BM

^{2}/GN

^{2}= BA

^{2}/GF

^{2}

(11) From similar polygons (See Theorem here), we know that:

(Area of ABCDE)/(Area of FGHKL) = BA

^{2}/GF

^{2}

(12) Putting this all together, gives us:

BM

^{2}/GN

^{2}= (Area of ABCDE)/(Area of FGHKL)

QED

Lemma 2: if A/B = C/D with A greater than C, then D is less than B.

Proof:

(1) Let A/B = C/D with A greater than C.

(2) So that AD = BC

(3) Now, D ≠ B since if D = B, then AD is greater than BC which contradicts step #2.

(4) Now, D cannot be greater than B since then AD is greater than BC which contradicts step #2.

(5) So, by the Law of Trichotomy (see Postulate above), we can conclude that D is less than B.

QED

Theorem: Two circles are to each other as the squares of their diameters.

Proof:

(1) Let C

_{1}be the circle formed with diameter BD and area A

_{1}.

(2) Let C

_{2}be the circle formed with diameter FH and area A

_{2}.

(3) Assume that A

_{1}/A

_{2}≠ (BD)

^{2}/(FH)

^{2}

(4) There exists an area S such that: (BD)

^{2}/(FH)

^{2}= A

_{1}/S

(5) Assume that S is less than A

_{2}

(6) Then, there exists a polygon EKFLGMHN such that the area of this polygon is greater than the area of S. [From the Method of Exhaustion, see Lemma 2.]

(7) We can inscribe a similar polygon into circle C

_{1}[See here for details on this construction]

(8) From Lemma 1 above, we can conclude:

BD

^{2}/FH

^{2}= (Area polygon AOBPCQDR)/(Area polygon EKFLGMHN)

(9) But then, from step #4:

BD

^{2}/FH

^{2}= A

_{1}/S

(10) So we can conclude that:

(Area polygon AOBPCQDR)/(Area polygon EKFLGMNH) = A

_{1}/S

(11) Since A

_{1}is greater than Area polygon AOBPCQDR, we can conclude from step #15 that S is greater than Area polygon EKFLGMN from Lemma 3 above.

(12) But this is impossible since in step #6 we showed that S is less than this same regular polygon so we have a contradiction and we reject our assumption in step #5.

(13) Now, let's assume that S is greater than A

_{2}

(14) So this means that (FH)

^{2}/(BD)

^{2}= S/A

_{1}

(15) Let T be the area such that S/A

_{1}= A

_{2}/T

(16) We can see that T is less than A

_{1}from Lemma 2 above.

(17) There exists a regular polygon that is greater in area than T but smaller than the area of the circle by the Method of Exhaustion (see Lemma 2)

(18) We can inscribe a similar polygon in A

_{2}.

(19) From Lemma 1 above, we can conclude:

FH

^{2}/BD

^{2}= (Area polygon EKFLGMHN)/(Area polygon AOBPCQDR)

(20) Likewise from step #14, we have:

(FH)

^{2}/(BD)

^{2}= S/A

_{1}

(21) And from step #15, this means that:

(Area polygon EKFLGMNH)/(Area polygon AOBPCQDR) = A

_{2}/T

(22) Now A

_{2}is greater in area than polygon EKFLGMNH so that T must be greater than the area of polygon AOBPCQDR

(23) But this is impossible from step #17 so we have a contradiction and we reject step #13.

(29) We now apply the Law of Trichotomy (see Postulate above) and we are done.

QED

References

- Phil Schultz, Eudoxus's Proof that the area of a circle is a constant times diameter squared
- David Joyce, Euclid's Elements

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