Pythagorean Theorem
by
Angie Head
This essay was inspired by a class that I am taking this quarter. The
class is the History of Mathematics. In this class, we are learning
how to include the history of mathematics in teaching a mathematics.
One
way to include the history of mathematics in your classroom is to incorporate
ancient mathematics problems in your instruction. Another way is to introduce
a new topic with some history of the topic.
Hopefully, this essay will give
you some ideas of how to include the history of the Pythagorean Theorem
in the teaching and learning of it.
We have been discussing different topics that were developed in ancient
civilizations. The Pythagorean Theorem is one of these topics. This theorem
is one of the earliest know theorems to ancient civilizations.
It was named
after Pythagoras, a Greek mathematician and philosopher. The theorem bears
his name although we have evidence that the Babylonians knew this relationship
some 1000 years earlier. Plimpton 322, a Babylonian mathematical
tablet dated back to 1900 B.C.
, contains a table of Pythagorean triples.
The Choupei, an ancient Chinese text, also gives us evidence that
the Chinese knew about the Pythagorean theorem many years before Pythagoras
or one of his colleagues in the Pythagorean society discovered and proved
it.
This is the reason why the theorem is named after Pythagoras.
Pythagoras lived in the sixth or fifth century B.C. He founded the Pythagorean
School in Crotona. This school was an academy for the study of mathematics,
philosophy, and natural science. The Pythagorean School was more than a
school; it was “a closely knit brotherhood with secret rites and observances”
(Eves 75). Because of this, the school was destroyed by democratic forces
of Italy. Although the brotherhood was scattered, it continued to exist
for two more centuries. Pythagoras and his colleagues are credited with many
contributions to mathematics. The following is an investigation of how the Pythagorean theorem has been
proved over the years.
Pythagorean Theorem
The theorem states that:
“The square on the hypotenuse of a right triangle is equal to the sum of the squares on the two legs” (Eves 8081).
This theorem is talking about the area of the squares that are built on
each side of the right triangle.
Accordingly, we obtain the following areas for the squares, where the
green and blue squares are on the legs of the right triangle and the red
square is on the hypotenuse.
 area of the green square is
area of the blue square is area of the red square is
 From our theorem, we have the following relationship:
 area of green square + area of blue square = area of red square or
As I stated earlier, this theorem was named after Pythagoras because he
was the first to prove it. He probably used a dissection type of proof similar
to the following in proving this theorem.
Pythagoras' Proof “Let a, b, c denote the legs and the hypotenuse of the given right
triangle, and consider the two squares in the accompanying figure, each
having a+b as its side.
The first square is dissected into six piecesnamely,
the two squares on the legs and four right triangles congruent to the given
triangle. The second square is dissected into five piecesnamely, the square
on the hypotenuse and four right triangles congruent to the given triangle.
By subtracting equals from equals, it now follows that the square on the
hypotenuse is equal to the sum of the squares on the legs” (Eves 81). Consider the following figure.
The area of the first square is given by (a+b)^2 or 4(1/2ab)+ a^2 + b^2.
The area of the second square is given by (a+b)^2 or 4(1/2ab) + c^2.
Since the squares have equal areas we can set them equal to another and
subtract equals. The case (a+b)^2=(a+b)^2 is not interesting. Let's do the
other case. 4(1/2ab) + a^2 + b^2 = 4(1/2ab)+ c^2
 Subtracting equals from both sides we have
concluding Pythagoras' proof. Over the years there have been many mathematicians and nonmathematicians
to give various proofs of the Pythagorean Theorem. Following are proofs
from Bhaskara and one of our former presidents, President James Garfield.
I have chosen these proofs because any of them would be appropriate to use
in any classroom. Bhaskara's First Proof
Bhaskara's proof is also a dissection proof. It is similar to the proof
provided by Pythagoras. Bhaskara was born in India.
He was one of the most
important Hindu mathematicians of the second century AD. He used the following
diagrams in proving the Pythagorean Theorem.
In the above diagrams, the blue triangles are all congruent and the yellow
squares are congruent. First we need to find the area of the big square
two different ways. First let's find the area using the area formula for
a square. Thus, A=c^2.
Now, lets find the area by finding the area of each of the components and
then sum the areas.
Area of the blue triangles = 4(1/2)ab
Area of the yellow square = (ba)^2
Area of the big square = 4(1/2)ab + (ba)^2
= 2ab + b^2 – 2ab + a^2
= b^2 + a^2 Since, the square has the same area no matter how you find it
A = c^2 = a^2 + b^2,
concluding the proof.
Bhaskara's Second Proof of the Pythagorean Theorem
In this proof, Bhaskara began with a right triangle and then he drew
an altitude on the hypotenuse. From here, he used the properties of similarity
to prove the theorem.
Now prove that triangles ABC and CBE are similar.
It follows from the AA postulate that triangle ABC is similar to triangle
CBE, since angle B is congruent to angle B and angle C is congruent to angle
E. Thus, since internal ratios are equal s/a=a/c.
Multiplying both sides by ac we get
sc=a^2. Now show that triangles ABC and ACE are similar.
As before, it follows from the AA postulate that these two triangles are
similar. Angle A is congruent to angle A and angle C is congruent to angle
E. Thus, r/b=b/c. Multiplying both sides by bc we get
rc=b^2. Now when we add the two results we get
sc + rc = a^2 + b^2.
c(s+r) = a^2 + b^2
c^2 = a^2 + b^2,
concluding the proof of the Pythagorean Theorem.
Garfield's Proof
The twentieth president of the United States gave the following proof to
the Pythagorean Theorem. He discovered this proof five years before he become
President.
He hit upon this proof in 1876 during a mathematics discussion
with some of the members of Congress. It was later published in the New
England Journal of Education.. The proof depends on calculating the
area of a right trapezoid two different ways.
The first way is by using
the area formula of a trapezoid and the second is by summing up the areas
of the three right triangles that can be constructed in the trapezoid. He
used the following trapezoid in developing his proof.
First, we need to find the area of the trapezoid by using the area formula
of the trapezoid.
A=(1/2)h(b1+b2) area of a trapezoid In the above diagram, h=a+b, b1=a, and b2=b. A=(1/2)(a+b)(a+b)
=(1/2)(a^2+2ab+b^2). Now, let's find the area of the trapezoid by summing the area of the three
right triangles.
The area of the yellow triangle is
A=1/2(ba). The area of the red triangle is
A=1/2(c^2). The area of the blue triangle is A= 1/2(ab). The sum of the area of the triangles is 1/2(ba) + 1/2(c^2) + 1/2(ab) = 1/2(ba + c^2 + ab) = 1/2(2ab + c^2).
Since, this area is equal to the area of the trapezoid we have the following
relation:
(1/2)(a^2 + 2ab + b^2) = (1/2)(2ab + c^2).
 Multiplying both sides by 2 and subtracting 2ab from both sides we get
concluding the proof.Return to Angela Head's EMT
668 Page
Prove the Pythagorean Theorem Using Triangle Similarity
When we introduced the Pythagorean theorem, we proved it in a manner very similar to the way Pythagoras originally proved it, using geometric shifting and rearrangement of 4 identical copies of a right triangle.
Having covered the concept of similar triangles and learning the relationship between their sides, we can now prove the Pythagorean theorem another way, using triangle similarity.
Problem
In a right triangle ΔABC with legs a and b, and a hypotenuse c, show that the following relationship holds:
c2 = a2+b2
Strategy
We said we will prove this using triangle similarity, so we need to create similar triangles. We have a right triangle, so an easy way to create another right triangle is by drawing a perpendicular line from the vertex to the hypotenuse:
Observe that we created two new triangles, and all three triangles (the original one, and the two new ones we created by drawing the perpendicular to the hypotenuse) are similar. Why?
All three have one right angle (In the original triangle: ∠ABC. In the two new triangles: ∠BDA and ∠BDC).
Because the two new triangles each share an angle with the original one, their third angle must be (90°the shared angle), so all three have an angle we will call α (In the original triangle: ∠BCA.
In the two new triangles: ∠BCD and ∠ABD), and an angle which is 90°α (In the original triangle : ∠BAC. In the two new triangles: ∠DBC and ∠BAD).
So all three triangles are similar, using AngleAngleAngle.
And we can now use the relationship between sides in similar triangles, to algebraically prove the Pythagorean Theorem.
Proof
 (1) m∠ABC=90° //Given, ΔABC is a right triangle
 (2) m∠BDA = m∠BDC=90° // BD was constructed as a perpendicular line to AC
 (3) ∠ABC≅∠BDA ≅∠BDC //(1), (2) transitive property of equality
 (4) ∠BCA ≅ ∠BCD // Common angle, the reflexive property of equality
 (5) m∠BAC = 90° – α // sum of angles in ΔABC is 180°, (1)
 (6) m∠DBC = m ∠BAD = 90° – α // sum of angles in ΔDBC and ΔBAD is 180°, (2)
 (7) ∠BCA ≅ ∠BCD ≅ ∠ABD // (4), (6) sum of angles in a triangle
 (8) ∠BAC≅∠DBC ≅ ∠BAD // (5), (6)
 (9) ΔABC ∼ ΔADB ∼ ΔBDC // AngleAngleAngle
 (10) a/c = (cx)/a // Similar triangles – the long leg divided by the hypotenuse
 (11) a2 = c*(cx) = c2 – cx // cross multiply (10)
 (12) c/b = b/x // Similar triangles – hypotenuse divided by the short leg
 (13) b2 = cx // cross multiply (10)
 (14) a2 + b2= c2 – cx + cx //add (11), (13)
 (15) a2 + b2= c2
Pythagorean Theorem Algebra Proof
You can learn all about the Pythagorean Theorem, but here is a quick summary:
The Pythagorean Theorem says that, in a right triangle, the square of a (a2) plus the square of b (b2) is equal to the square of c (c2):
a2 + b2 = c2
Proof of the Pythagorean Theorem using Algebra
We can show that a2 + b2 = c2 using Algebra
Take a look at this diagram … it has that “abc” triangle in it (four of them actually):
Area of Whole Square
It is a big square, with each side having a length of a+b, so the total area is:
A = (a+b)(a+b)
Area of The Pieces
Now let's add up the areas of all the smaller pieces:
 First, the smaller (tilted) square
has an area of: c2  Each of the four triangles has an area of:ab2
 So all four of them together is:4ab2 = 2ab
 Adding up the tilted square and the 4 triangles gives:A = c2 + 2ab
Both Areas Must Be Equal
 The area of the large square is equal to the area of the tilted square and the 4 triangles.
This can be written as:
 (a+b)(a+b) = c2 + 2ab
 NOW, let us rearrange this to see if we can get the pythagoras theorem:
 Start with:(a+b)(a+b) = c2 + 2ab
 Expand (a+b)(a+b):a2 + 2ab + b2 = c2 + 2ab
 Subtract “2ab” from both sides:a2 + b2 = c2
DONE!
Now we can see why the Pythagorean Theorem works … and it is actually a proof of the Pythagorean Theorem.
This proof came from China over 2000 years ago!
There are many more proofs of the Pythagorean theorem, but this one works nicely.
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Proofs of the Pythagorean Theorem  Brilliant Math & Science Wiki
Euclid's Proof
In outline, here is how the proof in Euclid's Elements proceeds. The large square is divided into a left and a right rectangle. A triangle is constructed that has half the area of the left rectangle. Then another triangle is constructed that has half the area of the square on the leftmost side.
These two triangles are shown to be congruent, proving this square has the same area as the left rectangle. This argument is followed by a similar version for the right rectangle and the remaining square.
Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the areas of the other two squares. The details follow.
Let A,B,CA, B, CA,B,C be the vertices of a right triangle with the right angle at A.A.A. Drop a perpendicular from AAA to the square's side opposite the triangle's hypotenuse (as shown below). That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs.
For the formal proof, we require four elementary lemmata:
 If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent (sideangleside).
 The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
 The area of a rectangle is equal to the product of two adjacent sides.
 The area of a square is equal to the product of two of its sides (follows from 3).
 Next, each top square is related to a triangle congruent with another triangle related in turn to one of two rectangles making up the lower square.
 Let ACBACBACB be a rightangled triangle with right angle CABCABCAB.
On each of the sides BCBCBC, ABABAB, and CACACA, squares are drawn: CBDECBDECBDE, BAGFBAGFBAGF, and ACIHACIHACIH, in that order. The construction of squares requires the immediately preceding theorems in Euclid and depends upon the parallel postulate.
From AAA, draw a line parallel to BDBDBD and CECECE. It will perpendicularly intersect BCBCBC and DEDEDE at KKK and LLL, respectively.
 Join CFCFCF and ADADAD, to form the triangles BCFBCFBCF and BDABDABDA.
Angles CABCABCAB and BAGBAGBAG are both right angles; therefore CCC, AAA, and GGG are collinear. Similarly for BBB, AAA, and HHH.
Angles CBDCBDCBDand FBAFBAFBA are both right angles; therefore angle ABDABDABD equals angle FBCFBCFBC, since both are the sum of a right angle and angle ABCABCABC.
Since ABABAB is equal to FBFBFB and BDBDBD is equal to BCBCBC, triangle ABDABDABD must be congruent to triangle FBCFBCFBC.
Since AAAKKKLLL is a straight line parallel to BDBDBD, rectangle BDLKBDLKBDLK has twice the area of triangle ABDABDABD because they share the base BDBDBD and have the same altitude BKBKBK, i.e. a line normal to their common base, connecting the parallel lines BDBDBD and ALALAL. (Lemma 2 above)
Since CCC is collinear with AAA and GGG, square BAGFBAGFBAGF must be twice in area to triangle FBCFBCFBC.
Therefore, rectangle BDLKBDLKBDLK must have the same area as square BAGF,BAGF,BAGF, which is AB2.AB^2.AB2.
Similarly, it can be shown that rectangle CKLECKLECKLE must have the same area as square ACIH,ACIH,ACIH, which is AC2.AC^2.AC2.
Adding these two results, AB2+AC2=BD×BK+KL×KC.AB^2 + AC^2 = BD imes BK + KL imes KC.AB2+AC2=BD×BK+KL×KC.
Since BD=KLBD = KLBD=KL, BD×BK+KL×KC=BD(BK+KC)=BD×BC.BD × BK + KL × KC = BD(BK + KC) = BD × BC.BD×BK+KL×KC=BD(BK+KC)=BD×BC.
Therefore, AB2+AC2=BC2AB^2 + AC^2 = BC^2AB2+AC2=BC2 since CBDECBDECBDE is a square. □_square□
Using Similar Triangles
Let ABCABCABC represent a right triangle, with the right angle located at CCC, as shown in the figure. Draw the altitude from point CCC, and call DDD its intersection with side ABABAB. Point DDD divides the length of the hypotenuse ccc into parts ddd and eee.
The new triangle ACDACDACD is similar to triangle ABCABCABC, because they both have a right angle (by definition of the altitude), and they share the angle at AAA, meaning that the third angle (((which we will call θ) heta)θ) will be the same in both triangles as well.
By a similar reasoning, the triangle CBDCBDCBD is also similar to triangle ABCABCABC. The proof of similarity of the triangles requires the triangle postulate: the sum of the angles in a triangle is two right angles, and is equivalent to the parallel postulate.
The similarity of the triangles leads to the equality of ratios of corresponding sides:
BCAB=BDBC and ACAB=ADAC.dfrac {BC}{AB} = dfrac {BD}{BC} ~~ ext{ and } ~~ dfrac {AC}{AB} = dfrac {AD}{AC}.ABBC=BCBD and ABAC=ACAD.
The fractions in the first equality are the cosines of the angle θ hetaθ, whereas those in the second equality are their sines. These ratios can be written as
BC2=AB×BD and AC2=AB×AD.BC^2 = AB imes BD ~~ ext{ and } ~~ AC^2 = AB imes AD.BC2=AB×BD and AC2=AB×AD.
Summing these two equalities results in
AC2+BC2=AB(BD+AD)=AB2.AC^2 + BC^2 = AB(BD + AD) = AB^2.AC2+BC2=AB(BD+AD)=AB2.
Therefore,
AC2+BC2=AB2. □AC^2 + BC^2 = AB^2. _squareAC2+BC2=AB2. □
Pythagorean theorem
Equation relating the side lengths of a right triangle
This article is about classical geometry. For the baseball term, see Pythagorean expectation.
Pythagorean theoremThe sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c).
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In mathematics, the Pythagorean theorem, also known as Pythagoras' theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares on the other two sides. This theorem can be written as an equation relating the lengths of the sides a, b and c, often called the “Pythagorean equation”:[1]
a
2
+
b
2
=
c
2
,
{displaystyle a^{2}+b^{2}=c^{2},}
where c represents the length of the hypotenuse and a and b the lengths of the triangle's other two sides. The theorem, whose history is the subject of much debate, is named for the ancient Greek thinker Pythagoras.
The theorem has been given numerous proofs – possibly the most for any mathematical theorem. They are very diverse, including both geometric proofs and algebraic proofs, with some dating back thousands of years.
The theorem can be generalized in various ways, including higherdimensional spaces, to spaces that are not Euclidean, to objects that are not right triangles, and indeed, to objects that are not triangles at all, but ndimensional solids.
The Pythagorean theorem has attracted interest outside mathematics as a symbol of mathematical abstruseness, mystique, or intellectual power; popular references in literature, plays, musicals, songs, stamps and cartoons abound.
Rearrangement proof
The rearrangement proof (click to view animation)
The two large squares shown in the figure each contain four identical triangles, and the only difference between the two large squares is that the triangles are arranged differently. Therefore, the white space within each of the two large squares must have equal area. Equating the area of the white space yields the Pythagorean theorem, Q.E.D.[2]
Heath gives this proof in his commentary on Proposition I.47 in Euclid's Elements, and mentions the proposals of Bretschneider and Hankel that Pythagoras may have known this proof.
Heath himself favors a different proposal for a Pythagorean proof, but acknowledges from the outset of his discussion “that the Greek literature which we possess belonging to the first five centuries after Pythagoras contains no statement specifying this or any other particular great geometric discovery to him.
“[3] Recent scholarship has cast increasing doubt on any sort of role for Pythagoras as a creator of mathematics, although debate about this continues.[4]
Other forms of the theorem
If c denotes the length of the hypotenuse and a and b denote the lengths of the other two sides, the Pythagorean theorem can be expressed as the Pythagorean equation:
a
2
+
b
2
=
c
2
.
{displaystyle a^{2}+b^{2}=c^{2}.}
If the lengths of both a and b are known, then c can be calculated as
c
=
a
2
+
b
2
.
{displaystyle c={sqrt {a^{2}+b^{2}}}.}
If the length of the hypotenuse c and of one side (a or b) are known, then the length of the other side can be calculated as
a
=
c
2
−
b
2
{displaystyle a={sqrt {c^{2}b^{2}}}}
or
b
=
c
2
−
a
2
.
{displaystyle b={sqrt {c^{2}a^{2}}}.}
The Pythagorean equation relates the sides of a right triangle in a simple way, so that if the lengths of any two sides are known the length of the third side can be found. Another corollary of the theorem is that in any right triangle, the hypotenuse is greater than any one of the other sides, but less than their sum.
A generalization of this theorem is the law of cosines, which allows the computation of the length of any side of any triangle, given the lengths of the other two sides and the angle between them. If the angle between the other sides is a right angle, the law of cosines reduces to the Pythagorean equation.
Other proofs of the theorem
This theorem may have more known proofs than any other (the law of quadratic reciprocity being another contender for that distinction); the book The Pythagorean Proposition contains 370 proofs.[5]
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