## Sunday, April 23, 2006

### Similar Polygons

In today's blog, I present a proof from Euclid on similar polygons. I use this theorem later in my proof for the existence of pi.

Definition: Similar Polygons

Two polygons are similar if corresponding angles are congruent and corresponding sides are proportional.

Lemma 1: A/B = C/D, then (A+C)/(B+D) = A/B = C/D

Proof:

(1) Let A/B = C/D

(2) Then AD=BC and A = BC/D

(3) So,

(A+C)/(B+D) = (BC/D + C)/(B + D) = (BC/D + CD/D)/(B+D) = [(BC+CD)/D][1/(B+D)] =
= [C(B+D)]/[D(B+D)] =C/D

QED

Corollary 1.1: A/B = C/D = E/F → (A+C+E)/(B+D+F)=A/B=C/D=E/F

Proof:

(1) (A+C)/(B+D) = A/B = C/D = E/F [See Lemma 1 above]

(2) Let U = A+C, V = B+D

(3) U/V = A/B = C/D [From step #1]

(4) U+E/V+F = A/B = C/D=E/F [From Lemma 1 above]

(5) But then:

(A + C + E)/(B + D + F)

QED

Lemma 2: if A/B = C/D, then A/C = B/D

Proof:

(1) A/B = C/D

(2) A*D = B*C

(3) Dividing both sides by C*D gives us:

A/C = B/D

QED

Lemma 3: if BE/AB = GL/GF and AB/BC = GF/GH,
then: BE/BC = GL/GH

Proof:

(1) BE/AB = GL/GF → BE*GF = GL*AB

(2) AB/BC = GF/GH → AB*GH = BC*GF

(3) So that:
BE*GF*AB*GH = GL*AB*BC*GF

(4) Dividing both sides by GF*AB*BC*GH gives us:
BE/BC = GL/GH

QED

Therem 1: Similar polygons have a ratio equal to the square of the ratio of two corresponding sides.

Proof:

(1) Let ABCDE and FGHKL be similar polygons.

(2) From the properties of similar polygons (see definition above), we know that:

(a)
∠ ABC ≅ ∠ FGH
AB/BC = FG/GH

(b)
∠ BCD ≅ ∠ GHK
BC/CD = GH/HK

(c)
∠ CDE ≅ ∠ HKL
CD/DE = HK/KL

(d)
∠ DEA ≅ ∠ KLF
DE/EA = KL/LF

(e)
∠ EAB ≅ ∠ LFG
EA/AB = LF/FG

(3) From congruent angles and corresponding sides (see here), we know that:

(a) triangle ABC is equiangular with triangle FGH (from #2a)

(b) triangle BCD is equiangular with triangle GHK (from #2b)

(c) triangle ECD is equiangular with triangle LHK (from #2c)

(d) triangle ABE is equiangular with triangle FGL. (from #2e)

(4) BE/BC = GL/GH (from Lemma 3 above) since:

(a) BE/AB = GL/GF (from #3d)

(b) AB/BC = GF/GH (from #3a)

(5) ∠ EBC ≅ ∠ LGH since:

(a) ∠ ABE ≅ ∠ FGL (#3d)

(b) ∠ ABC ≅ ∠ FGH (#3a)

(c) ∠ EBC = ∠ ABC - ∠ ABE

(d) ∠ LGH = ∠ FGH - ∠ FGL = ∠ ABC - ∠ ABE

(6) triangle EBC ≅ triangle LGH (from here) since:

(a) BE/BC = GC/GH (#4)

(b) ∠ EBC ≅ ∠ LGH (#5)

(7) triangle BOC is equiangular with triangle GPH since:

(a) ∠ OBC ≅ ∠ PGH (from #3b)

(b) ∠ BCO ≅ ∠ GHP (from #6)

(8) From the properties equiangular triangles we know that:

(a) ∠ BAM ≅ ∠ GFN (#3a)

(b) ∠ ABM ≅ ∠ FGN (#3d)

(c) ∠ MBC ≅ ∠ NGH (#6)

(d) ∠ BCM ≅ ∠ GHN (#3a)

(e) ∠ ODC ≅ ∠ PKH (#3b)

(f) ∠ OCD ≅ ∠ PHK (#3c)

(9) From congruent angles and corresponding sides again (see here), we know that:

(a) triangle AMB equiangular with triangle FNG from ∠ BAM ≅ ∠ GFN (#7a) and ∠ ABM ≅ ∠ FGN (#7b)

(b) triangle BMC equiangular with triangle GNH from ∠ MBC ≅ ∠ NGH (#7c) and ∠ BCM ≅ ∠ GHN (#7d)

(c) triangle COD equiangular with triangle HPK from ∠ ODC ≅ ∠ PKH (#7e) and ∠ OCD ≅ ∠ PHK (#7f).

(9) From properties of equiangular triangles, we know that:

(a) CO/OD = HP/PK (from #8c)

(b) BO/OC = GP/PH (from #7)

(c) AM/MB = FN/NG (from #8a)

(d) BM/MC = GN/NH (from #8b)

(10) Using Lemma 3 above gives us:

(a) BO/OD = GP/PK (from #9b and #9a)

(b) AM/MC = FN/NH (from #9c and #9d)

(11) Since triangles with the same height have their areas proportional to their bases (see here), we know that:

(a) BO/OD = BOC/COD

(b) BO/OD = BOE/OED

(c) GP/PK = GPH/HPK

(d) GP/PK = LGP/LPK

(e) AM/MC = ABM/MBC

(f) AM/MC = AME/EMC

(g) FN/NH = FGN/NGH

(h) FN/NH = FNL/LNH

(12) From step #11, we find that:

(a) BOC/COD = BOE/EOD

(b) GPH/HPK = LGP/LPK

(c) ABM/MBC = AME/MEC

(d) FGN/NGH = FNL/LNH

(13) From Lemma 1 above, we get:

(a) ABM/MBC = ABE/CBE since [ABE/CBE = (ABM + AME)/(MBC + MEC)]

(b) FGN/NGH = FGL/HGL since [FGL/HGL = (FGN + FNL)/(NGH +LNH)]

(c) BOC/COD = CBE/CED since [CBE/CED = (BOC + BOE)/(COD + OED)]

(d) GPH/HPK = HGL/HLK since [HGL/HLK = (GPH + LGP)/(HPK + LPK)]

(14) Combining step #11 with step #13 gives us:

(a) BO/OD = CBE/CED (see #11a and #13c)

(b) GP/PK = HGL/HLK (see #11c and #13d)

(c) AM/MC = ABE/CBE (see #11e and #13a)

(d) FN/NH = FGL/HGL (see #11g and #13b)

(15) Combining step #10 with step #14 gives us:

(a) CBE/CED = HGL/HLK (see #14a, #14b and #10a)

(b) ABE/CBE = FGL/HGL (see #14c, #14d and #10b)

(16) Using step #15 and Lemma 2, we get:

(a) CBE/HGL = CED/HLK (from #15a)

(b) ABE/FGL = CBE/HGL (from #15b)

(c) Thus, CBE/HGL = CED/HLK = ABE/FGL

(17) Now, polygon ABCDE and polygon FGHKL divide up into three triangles where:

polygon ABCDE = triangle ABE + triangle CBE + triangle CED

polygon FGHKL = triangle FGL + triangle HGL + triangle HLK

(18) Using step #16, we can apply Corollary 1.1 above to get:

triangle ABE/triangle FGL = polygon ABCDE/polygon FGHKL

(19) Since similar triangles are one to another in square ratio of the corresponding sides (see here), we have (see step #3d):

triangle ABE/triangle FGL = (AB)2/(FG)2

(20) Combining step #18 with step #19 gives us:

polygon ABCDE/polygon FGHKL = (AB)2/(FG)2

QED

References