## 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

### Isoceles Triangles

In today's blog, I show a very elementary property of isoceles triangles. This is one of the many proofs that I use to prove the existence of pi. Today's proof is taken straight from Euclid.

Definition 1: Isoceles Triangle

An isoceles triangle that has two sides of equal length.

Theorem 1: In an isoceles triangle, the base angles are congruent.

Proof:

(1) Let triangle ABC be an isoceles triangle with AB ≅ AC

(2) Let D be a point that extends AB such that A,B,D are on the same line.

(3) Let E be a point that extends AC such that A,C,E are on the same line and AE ≅ AD.

(4) triangle DAC ≅ triangle EAB by Side-Angle-Side (see here) since:

(a) AB ≅ AC (step #1)

(b) ∠ DAC is a common angle.

(5) We also know that triangle DBC triangle ECB by Side-Side-Side (see here if needed) since:

(a) BD ≅ CE since AD ≅ AG (step #3) and AB ≅ AC (step #1)

(b) BE ≅ CD (step #4) [By properties of congruent triangles, see here if needed]

(c) BC is a common side.

(6) So, now it follows that ∠ ABC ≅ ∠ ACB since:

(a) ∠ ABE ≅ ∠ ACD [By Properties of congruent triangles, see here if needed]

(b) ∠ EBC ≅ ∠ DCB [By Properties of congruent triangles and step #5]

(c) And finally, we know that:

∠ ABC = ∠ ABE - ∠ EBC

∠ ACB = ∠ ACD - ∠ DCB

QED

Corollary: If base angles are congruent, then sides are congruent

Proof:

(1) Let ABC be a triangle such that ∠ ABC ≅ ∠ ACB

(2) Assume that AB does not equal AC.

(3) Then one of them is greater. Let's assume AB. (Otherwise, we can make the same argument for side AC)

(4) Then there exists a point D such that BD is less than AB and BE ≅ AC

(5) From Theorem 1 above, we know that ∠ ABC ≅ ∠ DCB

(6) But this implies ∠ DCB ≅ ∠ ACB which is impossible.

(7) So we have a contradiction and we reject our assumption in #2.

QED

Theorem 2: The angle bisector of an isoceles triangle, is perpendicular to the base and divides up the base into two congruent segments.

Proof:

(1) Let AD be the angle bisector of ∠ BAC

(2) From this we, see that triangle BAD ≅ triangle CAD by S-A-S (see here) since:

(b) AB ≅ AC since ABC is an isoceles triangle.

(c) AD is a shared side between the two triangles.

(3) From congruent triangles (see here), we know that:

BD ≅ DC

(4) Now, since ∠ ADB, ∠ ADC are congruent and add up to 180 degrees (see here), we can conclude that they are both right angles.

QED

References

### Some properties of circles

In today's blog, I go over properties of a circle that I use later to prove the existence of pi. Today's proof is taken straight from Euclid.

Theorem 1: In a circle, if an angle that opens on the diameter, then it is a right angle.

Proof:

(1) Let E be the center of the circle.

(2) BE ≅ BA ≅ CE since they are all radii.

(3) Since triangle AEB and triangle AEC are isoceles triangles (see here if needed), we can conclude (see here) that:

∠ ABE ≅ ∠ BAE

∠ ACE ≅ ∠ CAE

(4) And step #3 gives us that ∠ BAC = ∠ ABC + ∠ ACB.

(5) But we also know that ∠ FAC = ∠ ABC + ∠ ACB since:

(a) ∠ FAC = 180 degrees - BAC [Angles of a straight line add up to 180 degrees, see here if needed]

(b) ∠ ABC + ∠ ACB = 180 degrees - BAC [Angles in a triangle add up to 180 degrees, see here if needed]

(6) And since ∠ FAC ≅ ∠ BAC, both must be right angles [since 2*x = 180 degrees → x = 90 degrees]

QED

Postulate 1: Similar segments of circles on equal straight lines equal one another.

Euclid originally presented this postulate as a theorem using the principle of superposition (see here for details). I am presenting it as a postulate in order to avoid the superposition.

Lemma 1: In equal circles, angles stand on equal circumferences whether they stand at the centers or the circumferences.

Proof:

(1) Let ABC and DEF be congruent circles with ∠ G ≅ ∠ H and ∠ A ≅ ∠ D.

(2) Since they are congruent, all radii are congruent so that:

BG ≅ CG ≅ EH ≅ FH

(3) So we have triangle BGC ≅ triangle EHF by side-angle-side (see here if needed)

(4) From step #3, we know that BC ≅ EF

(5) So that segment BAC ≅ segment EDF. [See Postulate I above]

(6) And this implies that segment BKC ≅ segment ELF.

QED

Lemma 2: In a circle, the angle at the center is double the angle at the circumference when the angles have the same circumference as base.

Proof:

(1) Let ABC be a circle with center E.

(2) EA ≅ EB since both are radii.

(3) ∠ EAB ≅ ∠ EBA since the base angles of an isoceles triangle are congruent (see here for details if needed).

(4) ∠ BEF = ∠ EAB + ∠ EBA (since angles of a triangle add up to 180 degrees and since two angles of a straight line add up to 180 degrees) so ∠ BEF is double ∠ EAB.

(5) We can use the same line of reasoning to establish that ∠ FEC is double ∠ EAC.

(6) Putting step #4 and step #5 together gives us that ∠ BEC is double ∠ BAC

(7) We can use this same reasoning to prove that ∠ GEC is double ∠ EDC.

(8) We can also prove that ∠ GEB is double ∠ EDB.

(9) Therefore the remaining ∠ BEC is double ∠ BDC.

QED

Theorem 2: In equal circles, angles standing on equal circumferences are equal to one another, whether they stand at the center or at the circumference.

Proof:

(1) So, we can assume that circumference BC circumference EC

(2) Assume ∠ BGC ≠ ∠ EHF

(3) Then one of them is greater; let's assume ∠ BGC is greater (if ∠ EHF is greater, we can make the same argument in terms of EHF)

(4) Construct ∠ BGK equal to ∠ EHF on the straight BG and at the point G on it (see here for details on the construction).

(5) Now equal angles stand on equal circumferences when they are at the centers, therefore circumference BK equals circumference EF (see Lemma 1 above)

(6) But EF equals BC so therefore BK equals BC which is a contradiction since BK is smaller than BC.

(7) So, we reject our assumption and conclude that ∠ BGC ≅ ∠ EHF

(8) The angle at A is half of the angle BGC. [See Lemma 2 above]

(9) The angle at D is half of the angle EHF [See Lemma 2 above]

(10) Therefore, the angle at A also equals the angle at D.

QED

Lemma 3: A line tangent to a point on a circle forms a right angle with a line drawn from that point to the center of the circle.

Proof:

(1) Assume that ∠ FCD is not a right angle

(2) Let FG be a line that is perpendicular to DE

(3) So triangle FGC is a right angle with the hypotenuse at FC.

(4) So FC is greater than FG by the Pythaogrean Theorem [See the Corollary here for details]

(5) But FC = FB since radii are congruent.

(6) So FC is less than FG which contradicts step #4.

(7) So we have a contradiction and we reject our assumption.

QED

References: