Division Algorithm for Polynomials

In today's blog, I will go over a result that I use in the proof for the Fundamental Theorem of Algebra.

Today's proof is taken from Joseph A. Gallian's Contemporary Abstract Algebra.

Theorem: Division Algorithm for Polynomials

Let F be a field, f(x), g(x) ∈ F[x] with g(x) ≠ 0.

Then there exists unique polynomials q(x), r(x) in F[x] such that f(x) = g(x)q(x) + r(x) and r(x)=0 or deg r(x) is less than deg g(x).

Proof:

(1) Let f(x) = g(x)q(x) + r(x) with g(x) ≠ 0

(2) If f(x) = 0 or deg f(x) is less than g(x), then q(x)=0, r(x)=f(x)

(3) So, we can assume that f(x) ≠ 0 and deg f(x) ≥ deg g(x)

(4) Let f(x) = a_nxⁿ + ... + a₀

(5) Let g(x) = b_mx^m + ... + b₀

(6) Let f₁(x) = f(x) - a_nb_m^-1x^n-mg(x)

(7) We note that deg f₁(x) is less than deg f(x) since:

f(x) - a_nb_m^-1x^n-mg(x) =

= a_nxⁿ + ... + a₀ - a_nb_m^-1x^n-m(b_mx^m + ... + b₀) =

= a_nⁿ - a_nxⁿ + a_n-1x^n-1 + ... + a₀ - a_nb_m^-1b_m-1x^n-1 - ... - a_nb_m^-1b₀x^n-m =

= a_n-1x^n-1 + ... + a₀ - a_nb_m^-1b_m-1x^n-1 - ... - a_nb_m^-1b₀x^n-m

So that f₁(x) has a degree of n-1 while f(x) has a degree of n.

(8) Now, we are ready to prove this theorem by induction.

(9) The assumption is true for deg f(x) = 0

deg f(x) is 0 → f(x)=C where C is a constant.

If deg g(x) is 0, then g(x) = D where D is a nonzero constant and q(x) = C/D and r(x)=0.

If deg g(x) is greater than 0, then q(x)=0 and r(x)=C.

(10) We can now assume that the assumption holds for all polynomials up to degree n-1.

(9) We see that:

f(x) = a_nb_m^-1x^n-mg(x) + f₁(x)

where the degree of f₁(x) is n-1 [See step #7]

(10) But by the induction hypothesis (step #10), we can assume that there exists q₁(x) and r₁(x) where r₁(x) has a degree lower than g(x).

(11) Therefore, we have:

a_nb_m^-1x^n-mg(x) + f₁(x) = a_nb_m^-1x^n-mg(x) + q₁(x)g(x) + r₁(x) =

= [a_nb_m^-1x^n-m + q₁(x)]g(x) + r₁(x)

(12) Which proves that degree r(x) is less than degree g(x) by principle of induction.

(13) Now, we still need to prove uniqueness of q(x),r(x)

(14) Suppose that:

f(x) = g(x)q(x) + r(x) = g(x)q'(x) + r'(x) where r(x),r'(x)=0 or deg r(x),r'(x) is less than deg g(x)

(15) Now, if we substract both equations, we get:

0 = g(x)[q(x) - q'(x)] + [r(x) - r'(x)]

which is the same as:

r'(x) - r(x) = g(x)[q(x) - q'(x)]

(16) Now since r'(x) and r(x) have degree less than g(x), the only way that this can be true is if r'(x) - r(x) = 0

(17) But then r'(x) = r(x) and q(x) = q'(x)

QED

References

• Joseph A. Gallian, Contemporary Abstract Algebra.
  

Properties of cos θ + i sin θ

In today's blog, I will go over some basic trigonometric properties that I use in my proof for the Fundamental Theorem of Algebra.

Lemma 1: [(cos α + i sin α) ]^a = cos (a*α) + i sin (a*α)

Proof:

(1) From Euler's Formula, we know that:

e^iα = cos α + i sin (α)

(2) So, we see that:

(e^iα)^a = e^i*a*α

(3) Finally,

e^i(a*α) = cos (a*α) + i sin (a*α)

QED

Lemma 2: (cos α + i sin α)(cos β + i sin β) = cos(α + β) + i sin(α + β)

Proof:

(1) Again, using Euler's Formula, we have:

e^iα = cos α + i sin α

e^iβ = cos β + i sin β

(2) So multiplying these two values together gives us [see here if a review of exponents are needed]:

(e^iα)(e^iβ) = e^{iα + iβ} = e^{i(α + β)}

(3) Finally,

e^{i(α + β)} = cos(α + β) + i sin(α + β)

QED

Argand Diagram

An Argand Diagram is a way of visualizing complex numbers. It involves graphing a complex number to the Cartesian coordinates.

There are two ways to represent a complex number:

x + iy

r (cos φ + i sin φ)

Both of these ways are equivalent since:



Here is the Argand Diagram:

One important point to remember is that in the r (cos φ + i sin φ) form, all values can be positive since r is an absolute magnitude and φ is a value between 0 and 2π using radians (or in degrees, between 0 and 360).

References

• Complex Number, Wikipedia
  

Using Maclaurin Series to define exponents

It is easy to define exponents in terms of positive integers.

xⁿ = x₁ * x₂ * ... * x_n

It is also straight forward to define exponents for 0, negative integers, and rational numbers. I wrote more about this in a previous blog.

But what about the complex numbers? What does it mean to have a value such as 5ⁱ. In fact, a very well known equation is Euler's Identity (see here):

e^iπ + 1 = 0

To handle this, we need a new definition for exponents that is consistent with their use for rational numbers but which can also handle the complex domain.

One way is to use the Maclaurin Series. In a previous blog, I showed that the Maclaurin series is:

f(x) = f(0) + (x/1!)f'(0) + (x²/2!)f''(0) + (x³/3!)f'''(0) + ... + (xⁿ/n!)fⁿ(0)


Now, if we define an exponent as a function so that e^x becomes:

f(x) = e^x

Applying derivatives, we get:

f(0) = e⁰ = 1

And likewise, all fⁿ(0) = 1

Substituting these results into the above equation gives us:

e^x = 1 + (x/1!) + (x²/2!) + ... + xⁿ/n!

Now a^y = e^y*ln(a) since:

(a) e^ln(a) = a [See here for details on ln if needed]

(b) ln(a^y) = y*ln(a) [See here for detalis on ln if needed]

So, we can use the equation for e^x to define a^y

a^y = e^y*ln(a) = 1 + (y*ln(a))/1! + (y*ln(a))²/2! + ... + (y*ln(a))ⁿ/n!

So for example,

1³ = 1 + (3*ln(1))/1! + (3*ln(2))²/2! + ... + =

= 1 + (0)/1! + (0)/2! + ...

2¹ = 1 + (1*ln(2))/1! + (1*ln(2))²/2! + ... + =

1 + ln(2) + ln(2)*ln(2)/2 + ln(2)*ln(2)*ln(3)/6 + ...

= 1 + 0.693... + 0.240... + 0.0555... + ... = 2

Finally, let's ask the question about aⁱ, which equals:

aⁱ = 1 + (i*ln(a)) + (i*ln(a))²/2! + ... + (i*ln(a))ⁿ/n!

= 1 + i*ln(a) - [ln(a)]²/2 -i*[ln(a)]³/3! + [ln(a)]⁴/4! + ...

= (1 - [ln(a)]²/2 + [ln(a)⁴/4! + ...) + i(ln(a) - [ln(a)³/3! + ln(a)⁵/5! + ...)

radians

The idea of radians is to define degrees in terms of pi. In a previous blog, I showed how Archimedes' proof for the area of a circle can be used to establish pi.

Definition Radian: There are 2π radians in a circle.

In other words 2π radians = 360 degrees.

A straight line has π radians and a right angle has π/2 radians.

This is useful because it enables trigonometric functions to be expressed in terms of taking π as a value.

Here are some examples:

sin (π/2) = 1
sin (π) = 0
sin(2π) = 0

cos (π/2) = 0
cos(π) = -1
cos(2π) = 1

Lemma 1: sin(nπ) = 0 where n is any integer

Proof:

(1) sin(0) = 0 and sin(π)=0 [See Property 1, here]

(2) sin(x + &2pi;) = x [See Property 5, here]

(3) So, we can see that:

if n is even, then n = 2x and sin (nπ) = sin(x*2π) = sin(0 + x*2π) = sin(0) = 0

if n is odd, then n = 2x+1 and sin(nπ) = sin(x*2π + π) = sin(π) = 0

QED

Lemma 2: The area of a sector = (1/2)θr²

Proof:

(1) We know that the area of a circle = πr² (see Corollary 2, here)

(2) An angle of a circle = (θ/2π) of a circle.

(3) So therefore, the area of a sector = (θ/2π)(πr²) = (1/2)θr²

QED

Euclid's Proof for the Pythagorean Theorem

In today's blog, I will go over Euclid's proof for the Pythagorean Theorem as presented in Euclid's elements. There is a very large number of different proofs for this result. The simplest involve tiles that show how squares of the two sides can be refitted to form a square on the hypotenuse. One proof was discovered by American president, James Garfield.

Theorem: In a right triangle, the hypotenuse squared is equal to the sum of the other sides squared.





























Proof:

(1) Since BAC is a right angle, we can extend AC to make a square GFBA and we can extend AB to make a square ACKH. (See here for details on the construction)

(2) We can also build a square based on BC. (See here for details on the construction)

(3) Let AL be a line that is parallel to BD and CE. (See here for details on the construction)

(4) ∠ DBA ≅ ∠ FBC since ∠ FBA and ∠ DBC are right angles and we get this result if we add ∠ ABC to both.

(5) triangle DBA ≅ triangle FBC by Side-Angle-Side (see here) since:

(a) ∠ DBA ≅ ∠ FBC (#4)

(b) FB ≅ AB since they are sides of the same square.

(c) BC ≅ BD since they are sides of the same square.

(6) The parallelogram BL is double the triangle ABD since they have the same base (BD) and since they are in the same parallel. (see Corollary 3.1 here for details)

(7) Likewise, the parallelogram GFBA is double the triangle FBC for the same reason as step #6.

(8) From step #5, we can conclude that the parallelogram BL is congruent to the parallelogram GFBA.

(9) We can follow the same steps to show that parallelogram CL is congruent to parallelogram ACKH since:

(a) ∠ KCB ≅ ∠ ACE since ∠ BCE, ∠ ACK are right angles and we can add ∠ ACB to each.

(b) AC ≅ CK and BC ≅ CE since the sides of the same square are congruent.

(c) So that triangle KCB ≅ triangle ACE by Side-Angle-Side.

(d) parallelogram CL is double the area of triangle ACE

(e) parallelogram CA is double the aera of triangle KCB

(f) So therefore parallelogram CL ≅ parallelogram CA

(10) So that we have the square of BC is congruent to the square of AB added to the square of AC since:

(a) the square BDEC = parallelogram BL + parallelogram CL

(b) parallelogram BL ≅ square of AB

(c) parallelogram CL ≅ square of AC

QED

Corollary 1: In a right triangle, the hypotenuse is longest of the three sides of a triangle.

Proof:

(1) Assume that the hypotenuse h is equal or less than one side s₁

(2) Then, h² is less than s₁² + s₂²

(3) But this contradicts the Pythagorean Theorem so we can reject our assumption.

QED

Corollary 2: sin²(θ) + cos²(θ) = 1.

Proof:

(1) From the main theorem, for any right triangle with base sides x and y and with hypotenuse z, we have:

x² + y² = z²

(2) If θ is the angle between z and x, then we have (see here):

sin θ = y/z

cos θ = x/z

So that:

sin²(θ) = y²/z²

cos²(θ) = x²/z²

(3) Now if we divide the equation in step #1, by z² on both sides we get:

x²/z² + y²/z² = 1.

(4) Now inserting the values in step #2 gives us:

sin²(θ) + cos²(θ) = 1.

QED

Archimedes and the area of a circle

In today's blog, I will go over Archimede's proof that the area of a circle is (1/2)rc. When this proof is combined with Euclid's proof (most likely from Eudoxus), it is possible to show that for all circles, the ratio of C/D is constant.

Postulate 1: For a given chord of a circle, the segment of the circumference is longer than the chord.













In the example above, BDC is longer than BC.

Postulate 2: For a given segment of the circumference, lines connected above it combined are longer.





























In the example above, AG + GE is greater than the segment of the circumference AE.

Lemma 1: The area of a regular polygon is (1/2)h*Q where Q is the sum of the perimeter.





















Proof:

(1) A regular polygon can be divided up into a sum of triangles.

(2) The area of each triangle is equal to (1/2)*(GH)*(DE) [See here for details if needed]

(3) So, the area of the polygon = (# of sides)*(1/2)*(GH)*(DE)

(4) Let Q = the sum of the bases (for example, for the hexagon above, Q = DE + EF + FA + AB + BC + CD)

(5) Then, the area of the polygon = (1/2)*(GH)*Q

QED


Theorem: The area of a circle is (1/2)circumference * radius

Proof:

(1) Let K = (1/2)*C*R where C = circumference and R = radius.

(2) Let A = the area of the circle.

(3) Assume that A is greater than K

(4) We can inscribe a polygon inside A that is greater than K and less than A. [See Lemma 2 and the Method of Exhaustion here for details]

(5) So, the area of the polygon = (1/2)*Q*h where h is the distance from the center to the base and where Q is the perimeter of the polygon. [See Lemma 1 above]

(6) But Q is less than C (see Postulate 1 above) and h is less than R.

(7) So we have Area Polygon = (1/2)Q*h which is less than (1/2)*C*R

(8) But this contradicts step #4 so we reject step #3.

(9) Now, let's assume that K is greater than A.

(10) We can circumscribe a polygon P around A such that P is greater than A but less than K. [See Lemma 3 and Method of Exhaustion here for details.]

(11) From Lemma 1 again, we know that the area of this polygon is (1/2)*Q*h where Q is the perimeter of the polygon and h is the height.

(12) In the case of the circumscribed polygon (see diagram for Postulate 2), h = R.

(13) Using Postulate 2 above, we see that Q is greater than C.

(14) But then the area of the polygon is greater than K since (1/2)*Q*R is greater than (1/2)*C*R

(15) But this contradicts step #10 so we reject our assumption at step #9.

(16) Now, we apply the Law of Trichomoty (see here) and we are done.

QED

Corollary 1: For all circles, the ratio of Circumference to Diameter is constant

Proof:

(1) We know from Euclid (see here) that for any two circles C₁ and C₂ that:

Area₁/Area₂ = (Diameter₁)²/(Diameter₂)²

(2) We know from Theorem 1 above that:

Area of circle = (1/2)*radius*circumference.

(3) Combining step #1 with step #2 and using R=(1/2)D gives us:

[(1/2)*(1/2)D₁*C₁]/[(1/2)*(1/2)D₂*C₂] = [D₁]²/[D₂]²

(4) Canceling out (1/4)/(1/4) gives us:

[D₁*C₁]/[D₂*C₂] = [D₁]²/[D₂]²

(5) Multiplying both sides by (D₂/D₁) gives us:

C₁/C₂ = D₁/D₂

which means that:

C₁*D₂ = C₂*D₁

and finally that:

C₁/D₁ = C₂/D₂

QED

Definition 1: π

From the Corollary, we know that the ratio of the circumference to the diameter is constant. π is this ratio from circumference to diameter. In other words, π = C/D.

Corollary 2: The area of a circle is πr².

Proof:

(1) From the theorem, the area of a circle is (1/2)(circumference)(radius)

(2) From the definition above, π = C/D

This means that C = D*π = 2*r*π

(3) Putting this all together gives us:

area = (1/2)(circumference)(radius) = (1/2)(2*r*π)(r) = πr²

QED

References

• Archimedes, The Measure of a Circle
• William Dunham, Journey Through Genius
• David Joyce, Euclid's Elements
  

Method of Exhaustion

In today's blog, I go over the proof for Eudoxus's Method of Exhaustion. This proof is used later in Euclid's proof that the areas of circles are in proportion to the squares of their diameters.

Theorem: Method of Exhaustion

Let x,y be any two positive real numbers where x is greater than y. If we continually remove a quantity v ≥ (1/2) from x, then eventually, we will be left with a value that is smaller than y.

Proof:

(1) We will only need to prove this for v = 1/2 since:

(a) This proof is done if we can show that there exists n such that (1/2)ⁿ*x is less than y.

(b) We only need to prove it for v=1/2 since we know that if v is less than (1/2), then:

vⁿ*x is less than (1/2)*ⁿ*x.

(2) Let m be an integer such that m ≥ x/y + 1

(3) So that ym is greater than x

(a) Since ym is greater than y(x/y + 1) = x + y

(b) and x + y is greater than x since y is nonzero.

(4) There exists a value n such that 2ⁿ is greater than m.

(a) Let m = 1

(b) 2¹ = 2 is greater than 1.

(c) Assume this is true up to n.

(d) So that 2ⁿ is greater than n.

(e) Since 2ⁿ is greater than 1, we know that:
2ⁿ + 2ⁿ = 2*(2ⁿ) = 2ⁿ⁺¹ is greater than n+1.

(f) So that this is true by induction.

(5) Thus, we see that y/x is greater than 1/2ⁿ

(a) ym is greater than x (step #2)

(b) so, m is greater than x/y

(c) and 1/m is less than y/x (or in other words, y/x is greater than 1/m)

(d) Now, 2ⁿ is greater than m so 1/2ⁿ is less than 1/m which also means that it is less than y/x.

(6) Thus (1/2)ⁿx is less than y since y/x is greater than (1/2)ⁿ means that y is greater than (1/2)ⁿ*m.

QED

Now, let's look at two applications of the Method of Exhaustion.

Lemma 1: If a square is inscribed in a circle, its area is greater than half the area of the circle.





















Proof:

(1) Let EFGH be a square that is inscribed in the circle above (see here for details on this construction).

(2) Let KLMN be a square that is circumscribed around the circle above (see here for details on this construction).

(3) The area of KLMN is greater than the area of the circle.

(4) The square EFGH is equal to (1/2) the area of KLMN since:

(a) Each square KFCE, FLGC, CGMH, ECHN are (1/4) the area of KLMN.

(b) Each triangle EFC, CFG, HCG, ECH is (1/4) the area of EFGH

(c) Each triangle EFC, CFG, HCG, ECH is (1/2) the area of the corresponding square KFCE, FLGC, CGMH, ECHN.

(d) Therefore, area EFGH = (1/2) * area KLMN

(5) Finally, area EFGH = (1/2)* area KLMN which is greater than the area of (1/2) the circle (from step #3).

QED


Lemma 2: For any area A that is smaller than the area of a circle C, there exists a regular polygon P that is smaller than C but greater than A.


























Proof:

(1) Let us form a circle C on the diameter FH. [See here for details on construction]

(2) It is possible to find points E,G such EFGH forms a square that is inscribed in circle C [See here for details on this construction]

(3) The area of the square EFGH is greater than (1/2) the area of circle C [See Lemma 1 above]

(4) We can constructs points K, N, M, and L so that point K is midway between points E and F, N is midway between E and H, M is midway between H and G, and L is midway between G and F. [See here for details on this construction]

(5) Each triangle EKG, FLG, GMH, and HNE is greater in area than half the segment of the circle about it since:

(a) From each triangle, we can construct a rectangle that has an area greater than the circle segment. For example, around triangle EKG we can construct a rectangle as in the diagram above.

(b) Each half of the triangle is exactly 1/4 the size of the rectangle since each point bisects the side of the circumscribed square.

(c) So the triangle EKG is equal to (1/2) the area of the rectangle.

(d) Since the area of the rectangle is greater than the area of the segment, (c) gives us that the triangle EKG is greater than (1/2) the area of the segment.

(6) We can keep repeating this step so that each time, we remove more than (1/2) the area:

(a) We select the midpoints to each of the existing segments. [See step #4 for details]

(b) We form a triangle with these new midpoints from the points of the segment that this new point bisected.

(c) This set of triangles together forms a new regular polygon.

(d) Each triangle has more area than half the circle segment that had previously remained. [This is the same as the argument in step #5]

(11) So, then there eventually exists a polygon EKFLGMHN such that the area of this polygon is greater than the area of S. [From the Method of Exhaustion above since each time we do step #6, we are subtracting greater than 1/2 of the remaining area from the circle.]

QED

Lemma 3: If a regular polygon P₁ with area A₁ is circumscribed around a circle C, then there exists a smaller polygon P₂ with area A₂ that can also be circumscribed around circle C such that:

A₂ - C is less than (1/2)(A₁ - C)

































Proof:

(1) Let C be a circle formed from diameter BD with center E. [See here for details on construction]

(2) Circumscribe a square GHKF around circle C. [See here for details on construction]

(3) Find a point M that lies on circle C and is midway between A and B [See here for details on construction]

(4) Let ON be a line tangent to M such that N is on GA and O is on GB.

(5) ∠ GMN is a right angle [See here for details]

(6) GN is greater than MN [See the corollary here for details]

(7) MN ≅ AN since:

(a) ∠ EAN ≅ ∠ EMN since they are both right angles.

(b) AE,ME are both radii, so we have AE ≅ ME and therefore (see Theorem 1, here), ∠ EAP ≅ ∠ EMP

(c) Using subtraction of step #7b from step #7a gives us ∠ NAM ≅ ∠ NMA.

(d) And #7c gives us AN ≅ MN since congruent angles implies congruent sides (see Corollary to Theorem 1, here)

(8) So GN is greater than AN too

(9) triangle AMN and triangle GMN have the same height since they are both on the same base AG and since they share the same vertice at M.

(10) Now, we can conclude that area of GMN is greater than the area of AMN since they share the same height (step #9) but the base of GMN is greater than the base of AMN (step #8) since area = (1/2)base * height (See Lemma 2 here for details if needed)

(11) triangle GLA ≅ triangle GLB by Side-Angle-Side since:

(a) GB ≅ GA since they are the sides of the same square.

(b) ∠ BGL ≅ ∠ AGL since we can prove triangle GAE ≅ triangle GBE by Side-Side-Side.

(c) Both triangles share the same side at GL.

(12) So the area of OGN is greater than half the area of BGA.

(13) The argument in step #11 applies to each side and we can conclude that converting this polygon from a 4-sided to 8-sided regular polygon results in removing greater than (1/2) the difference between the area of the 4-sided polygon and the area of the circle.

(14) We can use the same method to divide an 8-sided polygon to a 16-sided polygon and so on.

QED


References

• Phil Schultz, Eudoxus's Proof that the area of a circle is a constant times diameter squared
  

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

• David Joyce, Euclid's Elements, Book VI, Proposition 20
  

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.

(c) AE ≅ AD.

(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:

(a) ∠BAD ≅ ∠ CAD since AD is the angle bisector.

(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

∠ ADB ≅ ∠ ADC

(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

• David Joyce, Euclid's Elements, Book I, Proposition 5
  

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:

• David Joyce, Euclid's Elements, Book III, Proposition 31