This is what Loomis claimed in his 1907 book
A New Approach to Proving the Pythagorean Theorem
C-K Shene
Current Version: February 28, 2024
Abstract
Ne’Kiya D. Jackson and Calcea Rujean Johnson presented a trigonometric proof of the
Pythagorean Theorem at the 2023 AMS Spring Southeastern Sectional Meeting claiming that
it is an impossible proof. They cited a false claim in Loomis’ 1907 book “There are no trigono-
metric proofs. ...... Trigonometry is because the Pythagorean Proposition is.” This note
presents a new method based on similarity and geometric progression with which a pure geo-
metrical proof is given. Additionally, this note also discusses some proofs in Loomis’ book and
provides more new proofs using the concept of the Lemoine Point. Finally, the Appendix has
Zimba’s original proof of the angle difference identities for sin() and cos() without using the
Pythagorean Theorem. It also includes a proof of the angle sum identities for sin() and cos()
without using the Pythagorean Theorem. Both results can be used to prove the Pythagorean
Identity. As a result, Loomis’ claim is false and the proof of Jackson-Johnson can be quickly
replaced by a purely geometrical one.
A proof of the Pythagorean Theorem using trigonometry was presented at the AMS Spring
Southeastern Sectional Meeting on March 18, 2023 by Ne’Kiya D. Jackson and Calcea Rujean
Johnson [4]. This was reported widely by the media such as The Guardian [12], Popular Mechan-
ics [8] and Scientific American [11]. Unfortunately, the authors and some reports kept suggesting
that a trigonometric proof is “impossible.” They all cited a 1907 book The Pythagorean Propo-
sition by Elisha Scott Loomis [6, second edition, pp. 244-245] (Figure 1(a)) in which Loomis
(Figure 1(b)) stated the following:
Facing forward the thoughtful reader may raise the question: Are there any proofs
based upon the science of trigonometry or analytical geometry?
There are no trigonometric proofs, because all the fundamental formulae of trigonom-
etry are themselves based upon the truth of the Pythagorean Theorem; because of this
theorem we say sin
2
A + cos
2
A = 1, etc. Trigonometry is because the Pythagorean
Theorem is [6, p.244].
This is false, because Zimba [13] showed that the validity of cos(α − β) = cos(α) cos(β) +
sin(α)sin(β) and sin(α − β) = sin(α)cos(β) − cos(α) sin(β) is independent of the Pythagorean
Identity
1
which can be proved using cos(α − β) by setting α = β. This author is unaware of other
1
The Pythagorean Identity refers to the identity sin
2
(x) + cos
2
(x) = 1, which implies the Pythagorean Theorem
immediately. Conversely, if the Pythagorean Identity holds, the Pythagorean Theorem can be obtained easily.
1
(a) The 1940 Second Edition Published by The
National Council of Teachers of Mathematics
(b) Elisha Scott Loomis (Photograph Taken 1935)
Figure 1
publications between Loomis’ book and Zimba’s paper that demonstrated similar results; however,
Zimba’s work proved that Loomis’ claim indeed is false.
We will develop a simple method based on similarity and geometric progression to prove the
Pythagorean Theorem. While this method can be applied to more general geometric shapes, we
only focus on right triangles. In what follows, Section 1 presents our method; Section 2 shows
that some classical proofs in Loomis’ book can easily be converted to use this technique, Section 3
presents Jackson and Johnson’s proof without using trigonometry; Section 4 discusses the original
trigonometrical version; Section 5 first discusses the concepts of symmedians and the Lemoine
point of a triangle, and then proceeds to offer three more proofs based on the Lemoine point, and
Section 6 has our conclusions. Section 2 is divided into three subsections: Section 2.1 discusses
proofs in which square dissection is used, Section 2.2 has simple proofs that use the given right
triangle directly, and Section 2.3 includes a proof which has a square on the hypothenuse. Finally,
the Appendix includes Zimba’s proof showing that the angle difference identities for sin() and
cos() can be derived without using the Pythagorean Identity. Following Zimba’s idea, we show
that the angle sum identities for sin() and cos() are also independent of the Pythagorean Identity.
Furthermore, from the angle sum identities the sum-to-product identities are derived from which
2
the derivatives of sin() and cos() are shown to be independent of the Pythagorean Identity. Then,
we use the double angle identities and L’H
ˆ
opital’s Rule to prove the Pythagorean Identity. Finally,
because the derivative computation of sin() and cos() are independent of the Pythagorean Identity,
a simple application of calculus yields the identity. Therefore, this chain of reasoning suggests that
“Trigonometry is because the Pythagorean Theorem is” is false.
1 The Main Idea
Given a polygonal shape A and a polygonal shape B ⊆ A, if B is similar to A with a scaling factor
ρ (i.e., any edge q of B and its corresponding edge p of A satisfying q = ρ · p, where 0 < ρ < 1),
then A(A) = A (A − B) + A (B), where A(X) denotes the area of X (Figure 2).
Note that a scaling factor ρ for length induces a scaling factor ρ
2
for area. Because B ∼ A,
A(B) = ρ
2
A(A), we have A(A) = A (A − B) + A (B) = A (A − B) + ρ
2
A(A). Therefore, we have
A(A) = A(A − B)+ A (B)
= A(A − B) + ρ
2
A(A)
= A(A − B) + ρ
2
(A(A − B) + A (B))
= A(A − B) + ρ
2
A(A − B) + ρ
2
A(B))
= A(A − B) + ρ
2
A(A − B) + ρ
4
A(A)
= A(A − B) + ρ
2
A(A − B) + ρ
4
(A(A − B) + A (B))
= A(A − B) + ρ
2
A(A − B) + ρ
4
A(A − B) + ρ
4
A(B)
= A(A − B) + ρ
2
A(A − B) + ρ
4
A(A − B) + ρ
4
(ρ
2
A(A))
= A(A − B) + ρ
2
A(A − B) + ρ
4
A(A − B) + ρ
6
A(A)
= A(A − B) + ρ
2
A(A − B) + ρ
4
A(A − B) + ρ
6
(A(A − B) + A (B))
= A(A − B) + ρ
2
A(A − B) + ρ
4
A(A − B) + ρ
6
A(A − B) + ρ
6
A(B)
.
.
.
= A(A − B)
1 + ρ
2
+ ρ
4
+ ρ
6
+ · · ·
=
A(A − B)
1 − ρ
2
(1)
Hence, if we are able to find B and ρ and compute A (A − B), it is easy to find A(A).
As for line segment length, the scaling factor is only ρ. If a point Z is selected on a line segment
XY , we have of XY = XZ + ZY (Figure 3). Let ρ = ZY /XY . Based on the idea above we have
XY = XZ + ρX Z + ρ
2
XZ + ρ
3
XZ + ρ
4
XZ + ··· =
XZ
1 − ρ
(2)
3
Figure 2: This Is How the Idea Goes for Area Computation
Figure 3: This Is How the Idea Goes for Line Segment Length Computation
2 Re-Do Some Classical Proofs
Many proofs in Loomis’ book [6] can easily be redone with the new method. The next few sub-
sections discuss how this conversion can be done easily. First, a shape A is constructed from the
given right triangle of sides a ≤ b < c with c being the hypotenuse. Second, find a sub-shape B
that is similar to A and the area of A − B can be computed easily. Third, find the scaling factor ρ.
Fourth, use our method to compute the area of figure A. Fifth, find another way to compute the area
of A without using B. Finally, equating the two results followed by some simplification yields the
desired result. However, we have to point out that for the Pythagorean Theorem, the length of the
hypotenuse c should be used in the first stage and should not be cancelled out because c is typically
not used in the second stage.
4
2.1 Proofs Involving the Use of a Square
Proof 1
In Figure 4(a) the square has side length a+b and the right triangle is repeated four times inside
the square. The scaling factor going from the outer square to the inner one is ρ = c/(a + b) and
hence we have
1
1 − ρ
2
=
(a + b)
2
(a + b)
2
− c
2
Therefore, the area of the square is
(a + b)
2
=
1
1 − ρ
2
× 4
1
2
a · b
=
(a + b)
2
(a + b)
2
− c
2
× 4
1
2
a · b
After simplifying the above, we get a
2
+ b
2
= c
2
(Loomis [6, Proof Thirty-Three, p. 48]).
Proof 2
Figure 4(b) is another commonly seen proof in which the inner square has side length b − a.
The scaling factor is ρ = (b − a)/c and 1/(1 − ρ
2
) = c
2
/(c
2
− (b − a)
2
). The area of the square is
computed as follows:
c
2
=
1
1 − ρ
2
× 4
1
2
a · b
=
c
2
c
2
− (b − a)
2
× 4
1
2
a · b
Again, simplifying the above yields a
2
+ b
2
= c
2
.
(a)
(b)
(c)
Figure 4: Proofs Involving the Use of Squares
Proof 3
In Loomis [6, Proof Two Hundred Fifteen, p. 221] a proof similar to the above one is shown.
Given a right triangle 4ABC (Figure 4(c)), construct a square of side length b on side AC and drop
5
a penpendicular from C to
←→
AB meeting it at F. Then, drop a perpendicular from D to
←→
CF meeting it
at G and perpendiculars from E to
←→
DG and
←→
AB meeting them at H and K, respectively. It is obvious
that 4AFC
∼
=
4CGD
∼
=
4DHE
∼
=
4EKA and the lengths of AK and CF are equal. Furthermore,
because 4ABC ∼ 4ACF, we have
p = AK =
a · b
c
, p + q = AF =
b
2
c
and q = KF = AF − AK =
b(b − a)
c
Hence, the sum of the areas of the four right triangles is
4
1
2
p(p + q)
=
1
2
ab
3
c
2
(3)
The scaling factor going from square ACDE to square FGHK is
ρ =
q
b
=
(b/c)(b − a)
c
=
b − a
c
and
1
1 − ρ
2
=
c
2
c
2
− (b − a)
2
The area of the outer square is
A(ACDE) =
1
1 − ρ
2
×
1
2
ab
3
c
2
=
c
2
c
2
− (b − a)
2
×
1
2
ab
3
c
2
=
2ab
3
c
2
− (b − a)
2
Because A(ACDE) = b
2
which is equal to the above result, we have
b
2
=
2ab
3
c
2
− (b − a)
2
Simplifying yields c
2
= a
2
+ b
2
.
Notes
Some proofs in Loomis [6] share the same technique, although the division of the sides of the
square may not be a : b. For example, in Loomis [6, Proof Sixty-Three, p. 137], the division of the
side c square is exactly a : b; but other rectangles and squares are needed to complete the proof.
Loomis [6, Proof Thirty-Three, p.48] is exactly the same as shown in Figure 4(b). Loomis [6,
Proof One Hundred Thirty-Three, p. 177] is similar to Figure 4(c), but the division of side c is
ac/b : c(b − a)/b. Other proofs in Loomis [6] are similar (e.g., Proofs 131–132, Proofs 134–137,
etc.) and use different ways of cutting the square of side c. These proofs can also be transformed
to use the technique presented here.
2.2 Proofs Based on Similar Right Triangles Inside or Outside the Given One
Proof 4
6
(a)
(b)
Figure 5: Very Simple Proofs Using the Given Triangle Directly
Consider 4ABC in Figure 5(a), where D is the perpendicular foot from C to side
←→
AB. Line
←→
CD
divides 4ABC into two smaller triangles both similar to 4ABC (Loomis [6, Proof One, p. 23]).
From 4CDB ∼ 4ACB, we have h = (a · b)/c and k = a
2
/c. Therefore, we have
A(4CBD) =
1
2
· h · k =
1
2
(a · b)
a
c
2
(4)
Because 4ACD ∼ 4ABC, the scaling factor ρ from 4ABC to 4ACD is ρ = h/c = b/c. Hence,
we have
A(4ABC) =
A(4CBD)
1 − ρ
2
=
1
2
a
3
b
c
2
− b
2
(5)
Because we also have A (4ABC) = (a · b)/2, the following holds:
1
2
a · b =
1
2
a
3
b
c
2
− b
2
Simplifying the above yields
1 =
a
2
c
2
− b
2
This leads to c
2
= a
2
+ b
2
, the desired result.
Proof 5
As a direct consequence of Proof 5, a very similar one was discussed in the Cut the Knot
site [1], credited to John Arioni. From D drop a perpendicular to
←→
AC meeting it at E (Figure 5(a)).
Let p = DE and q = CE. Because 4DCE ∼ 4ABC, we have
p
h
=
b
c
and
q
h
=
a
c
Because we know h = (a · b)/c, we have
p =
a · b
2
c
2
and q =
a
2
· b
c
2
7
The area of trapezoid BCED is:
A(BCED) =
1
2
(p + a) · q =
a
3
b
2c
4
(b
2
+ c
2
) (6)
The scaling factor going from 4ABC to 4ADE is ρ = p/a = (b/c)
2
and
1
1 − ρ
2
=
c
4
(c
2
− b
2
)(c
2
+ b
2
)
Then, the area of 4ABC is
A(ABC) =
1
1 − ρ
2
×
a
3
b
2c
4
(b
2
+ c
2
)
=
c
4
(c
2
− b
2
)(c
2
+ b
2
)
×
a
3
b
2c
4
(b
2
+ c
2
)
=
a
3
b
2(c
2
− b
2
)
However, because A (ABC) = (a · b)/2, we have
1
2
(a · b) =
a
3
b
2(c
2
− b
2
)
Again, we have c
2
= a
2
+ b
2
. Note that this proof is essentially applying the previous proof twice,
once reducing 4ABC to 4ACD and the other reducing 4ACD to 4ADE.
Proof 6
Proofs Three and Four in Loomis [6, p. 26] share the same idea as discussed in the first proof
in this section. We only discuss Proof Four here and Proof Three can be obtained exactly the
same way. In Figure 5(b), 4ABC is the given right triangle. Extend the hypotenuse
←→
AB to D so
that BD = BC = a, and construct a line
←→
DE perpendicular to
←→
AB meeting
←→
AC at E. It is obvious
that 4BDE
∼
=
4BCE and 4AED ∼ 4ABC. As a result, we have p = (a/b)(a + c). The area of
quadrilateral BCED is
A(BCED) = 2
1
2
a · p
=
a
2
(a + c)
b
The scaling factor ρ bringing 4AED to 4ABC is
ρ =
a
p
=
b
a + c
and
1
1 − ρ
2
=
(a + c)
2
(a + c)
2
− b
2
Consequently, we have
A(AED) =
1
1 − ρ
2
A(BCED) =
(a + c)
2
(a + c)
2
− b
2
a
2
(a + c)
b
=
a
2
(a + c)
3
b((a + c)
2
− b
2
)
However. A (AED) may also be calculated as
A(AED) =
1
2
p(a + c) =
1
2
a(a + c)
2
b
8
Both results must agree:
a
2
(a + c)
3
b((a + c)
2
− b
2
)
=
1
2
a(a + c)
2
b
A simple simplification yields c
2
= a
2
+ b
2
.
2.3 Proofs with Squares Standing on the Sides of a Right Triangle
Proof 7
There are many proofs in which a square is constructed on a side of a right triangle, and some
of these proofs can easily be adapted for our method. The following is taken from Loomis [6,
Proof Nineteen, p. 43] (Figure 6). Because of 4A
1
AA
2
∼ 4ABC and 4BB
1
B
2
∼ 4ABC, we have
p = (bc)/a and q = (ca)/b. As a result, the length of side A
1
B
1
is:
A
1
B
1
= p + c + q =
c
ab
(ab + a
2
+ b
2
)
Figure 6: A Square on Side c (i.e., AB)
The area of the trapezoid ABB
1
A
1
is
A(AA
1
B
1
B) =
1
2
(c + A
1
B
1
) · c =
c
2
2
·
(a + b)
2
ab
The scaling factor ρ is the ratio of c and A
1
B
1
ρ =
c
A
1
B
1
=
ab
ab + a
2
+ b
2
Therefore, we have
ρ
2
=
(ab)
2
(ab + a
2
+ b
2
)
2
and
1
1 − ρ
2
=
(ab + a
2
+ b
2
)
2
(a
2
+ b
2
)(a + b)
2
9
Hence, the area of 4CA
1
B
1
is calculated from the area of the trapezoid ABB
1
A
1
as follows:
A(CA
1
B
1
) =
1
1 − ρ
2
A(ABB
1
A
1
) =
(ab + a
2
+ b
2
)
2
(a
2
+ b
2
)(a + b)
2
·
c
2
2
(a + b)
2
ab
=
1
2
·
c
2
(ab + a
2
+ b
2
)
2
ab(a
2
+ b
2
)
(7)
Because of similarity, the lengths of side CA
1
and CB
1
are simply CA
1
= b/ρ and CB
1
= a/ρ.
Consequently, the area of 4CA
1
B
1
is also calculated as follows:
A(CA
1
B
1
) =
1
2
CA
1
·CB
1
=
1
2
·
a
ρ
·
b
ρ
=
1
2
·
(ab + a
2
+ b
2
)
2
ab
This result must agree with the one in Eqn. (7):
1
2
·
c
2
(ab + a
2
+ b
2
)
2
ab(a
2
+ b
2
)
=
1
2
·
(ab + a
2
+ b
2
)
2
ab
Simplifying the above yields c
2
= a
2
+ b
2
.
It does not have to use area in this particular case. Because 4ABC ∼ 4A
1
B
1
C, the altitude
from from C to
←→
AB and the altitude from C to
←−→
A
1
B
1
are h = (a · b)/c and h + c = (a · b + c
2
)/c,
respectively, and hence the scaling factor ρ can also be computed as follows:
ρ =
h
h + c
=
ab
ab + c
2
Because ρ = ρ, c
2
= a
2
+ b
2
follows immediately.
3 Jackson and Johnson’s Proof without Trigonometry
Proof 8
This section will re-do the proof of Ne’Kiya D. Jackson and Calcea Rujean Johnson. The
construction is similar, but the proof is completely geometrical without the use of trigonometry.
Given a right triangle 4ABC with
6
A = α,
6
B = β > α,
6
C = 90
◦
, a = BC, b = CA and c =
AB (Figure 8). The case of α = β = 45
◦
will be addressed separately. Given a line segment
Y
0
Z
0
of length x, construct a triangle 4XY
0
Z
0
so that
6
Y
0
= α and
6
Z
0
= α + 90
◦
. Note that this
construction does not work if α = β = 45
◦
because X is at infinity and the area of 4XY
0
Z
0
is not
finite. From Z
0
construct a line perpendicular to
←−→
Y
0
Z
0
meeting
←→
XY
0
at Y
1
and then construct a line
perpendicular to
←−→
Y
1
Z
0
at Y
1
meeting
←→
XZ
0
at Z
1
. Let p = Y
0
Y
1
, q = Z
0
Z
1
, r = Y
1
Z
1
and h = Z
0
Y
1
.
Because 4ABC ∼ 4Y
0
Y
1
Z
0
, we have h = x(a/b) and p = x(c/b). Because 4ABC ∼ 4Z
0
Z
1
Y
1
,
we have q = h(c/b) = x(ac/b
2
) and r = x(a/b)
2
. As a result, the area of trapezoid Y
0
Z
0
Z
1
Y
1
is
A(Y
0
Z
0
Z
1
Y
1
) =
1
2
(x + r) · h =
1
2
x + x
a
b
2
·
xa
b
=
x
2
2
a(a
2
+ b
2
)
b
3
(8)
10
Figure 7
Figure 8: Jackson and Johnson’s Proof: Part 1
Because 4XY
0
Z
0
∼ 4XY
1
Z
1
and r/x = (a/b)
2
, the scaling factor from 4XY
0
Z
0
to 4XY
1
Z
1
is
ρ = (a/b)
2
. Hence, we have
ρ
2
=
a
b
4
and
1
1 − ρ
2
=
b
4
b
4
− a
4
=
b
4
(b
2
− a
2
)(b
2
+ a
2
)
The area of 4XY
0
Z
0
is:
A(4XY
0
Z
0
) =
1
1 − ρ
2
·A (Y
0
Z
0
Z
1
Y
1
) =
b
4
(b
2
− a
2
)(b
2
+ a
2
)
x
2
2
a(a
2
+ b
2
)
b
3
=
x
2
2
ab
b
2
− a
2
(9)
Then, we determine the lengths of XY
0
and XZ
0
. Because we know Y
0
Y
1
= p = x(c/b) and
ρ = (a/b)
2
, our method (Eqn. (2)) yields
XY
0
=
p
1 − ρ
= x
bc
b
2
− a
2
and XZ
0
=
q
1 − ρ
= x
ac
b
2
− a
2
(10)
Construct a line perpendicular to
←→
XY
0
at Y
0
meeting
←→
XZ
0
at X
0
. It is not difficult to see that
4X
0
Y
0
Z
0
is an isoceles with
6
Y
0
=
6
Z
0
= β and
6
X
0
= 2α (Figure 9).
The length k of the altitude on side Y
0
Z
0
is k = (x/2)(b/a), and we have
A(4X
0
Y
0
Z
0
) =
1
2
(x · k) =
x
2
2
2
b
a
(11)
The length t of side
X
0
Y
0
is t = (x/2)(c/a). Therefore, the area of triangle 4XY
0
X
0
is
A(4XY
0
X
0
) =
1
2
t · XY
0
=
1
2
·
x
2
·
c
a
x
b · c
b
2
− a
2
=
x
2
2
2
·
b
a
·
c
2
b
2
− a
2
(12)
11
Figure 9: Jackson and Johnson’s Proof: Part 2
The area of 4XY
0
X
0
may also be calculated as the sum of the areas of 4X
0
Y
0
Z
0
and 4XY
0
Z
0
:
A(4XY
0
X
0
) = A(4XY
0
Z
0
) + A (4X
0
Y
0
Z
0
) =
x
2
2
2
b
a
+
x
2
2
·
ab
b
2
− a
2
=
x
2
2
·
b
2a
·
a
2
+ b
2
b
2
− a
2
(13)
Because the areas computed by Eqn. (12) and Eqn. (13) are the same, we have
x
2
2
2
·
b
a
·
c
2
b
2
− a
2
=
x
2
2
·
b
2a
·
a
2
+ b
2
b
2
− a
2
After a simple simplification, we have the desired result is c
2
= a
2
+ b
2
.
If α = β = 45
◦
, the altitude on the hypotenuse is c/2. The area of the right triangle can be
computed in two ways: a
2
/2 and ((c/2)·c)/2. Therefore, a
2
/2 = ((c/2)·c)/2 implies a
2
+a
2
= c
2
.
4 Jackson and Johnson’s Original Proof
Proof 9
The original proof of Jackson and Johnson used trigonometry based on side lengths XY
0
and
XX
0
and the law of sines that is independent of the Pythagorean Theorem and the Pythagorean
Identity [7, 9]. We know the following from the previous section:
XY
0
= x ·
bc
b
2
− a
2
and t =
x
2
c
a
12
To compute sin(2α), we need XX
0
= t + XZ
0
. Our technique gives XZ
0
as follows:
XZ
0
=
1
1 − ρ
· q = x ·
ac
b
2
− a
2
Therefore, we have
XX
0
= t + XZ
0
=
x
2
c
a
+ x ·
ac
b
2
− a
2
= (x · c)
a
2
+ b
2
2a(b
2
− a
2
)
From the right triangle 4XY
0
X
0
, with the help of Ean (10) we have sin(2α) as follows:
sin(2α) =
XY
0
XX
0
=
x ·
bc
b
2
−a
2
(x · c)
a
2
+b
2
2a(b
2
−a
2
)
=
2ab
a
2
+ b
2
(14)
From the given triangle we have sin(β) = b/c. From 4X
0
Y
0
Z
0
, the law of sines gives
sin(2α)
x
=
sin(β)
t
=
(b/c)
(x/2) · (c/a)
=
2ab
x · c
2
Hence, we have the second way of computing sin(2α):
sin(2α) =
2ab
c
2
(15)
The results from Eqn (14) and Eqn (15) must be equal. Then, it is obvious that the desired result
a
2
+ b
2
= c
2
holds. Because we know sin(2α) = 2ab/(a
2
+ b
2
), the above discussion yields the
Pythagorean Theorem and the double angle formula for sin(x) at the same time. Obviously, the
only difference between Jackson and Johnson’s original proof and the proof in the previous section
is the use of length and trigonometry vs. the use of area.
5 Possibly New Proofs
This section presents our (possibly) new proofs based on the concept and properties of the Lemoine
point. Section 5.1 introduces the concepts of symmedian and the Lemoine point and some proper-
ties. Then, Section 5.2 presents three more proofs.
5.1 Symmedians and the Lemoine Point of a Triangle
A vertex of a triangle has an angle bisector that bisects the angle of that vertex and a median that is
the line joining that vertex and the midpoint of that vertex’s opposite side. In Figure 10, the black
solid line is the angle bisector of angle C, the red dashed line is the median, and the blue dashed
line is the line symmetric to the median with respect to the angle bisector. This line is referred to
as the symmedian of the median at that vertex. Note that the angle between the angle bisector and
13
Figure 10: The Symmedian at a Vertex
the median is equal to the angle between the angle bisector and the symmedian. Or, the bisector
bisects the angle between the median and the symmedian.
Each triangle has three vertices and hence three symmedians. It is known that these three
symmedians are concurrent. The point where the three symmedian lines meet is referred to as
the Lemoine point, the Grebe point or the Symmedian point (Figure 11). In this note we shall use
“Lemoine point”exclusively. This point plays a significant role in modern triangle geometry.
Figure 11: The Lemoine/Grebe/Symmedian Point
Suppose a triangle 4ABC has all three squares on its sides (Figure 12). Extending the outer
side of each square creates a similar right triangle 4A
0
B
0
C
0
. It is clear that 4ABC ∼ 4A
0
B
0
C
0
.
Because the corresponding sides are parallel, their intersection points are collinear (i.e., meeting
at the line at infinity) and by Desargues’ Theorem the lines joining the corresponding vertices
are concurrent at a point K. This point K is exactly the Lemoine point of 4ABC and 4A
0
B
0
C
0
(Gallatly [2, Chap X, p.86]).
In fact, K is the homothetic center of 4ABC and 4A
0
B
0
C
0
. It is also known as the center of
similarity or center of similitude of 4ABC and 4A
0
B
0
C
0
. Let the distances from K to sides a,
b and c be p
a
, p
b
and p
c
, respectively. An important property of K is p
a
: p
b
: p
c
= a : b : c or
p
a
/a = p
b
/b = p
c
/c (Honsberger [3, p. 59]). This property can be used as a characterization of
14
Figure 12: The Lemoine Point Construction
the Lemoine/Grebe/Symmedian point.
For a right triangle, the Lemoine point is the midpoint of the altitude on the hypotenuse. Sup-
pose 4ABC is a right triangle with
6
C = 90
◦
. We need to show that (1) the altitude on the hy-
potenuse is an symmedian and (2) the midpoint of this symmedian is the Lemoine point.
Figure 13: The Lemoine/Grebe/Symmedian Point of a Right Triangle
The first thing we need to show is that the altitude
←→
CD is actually a symmedian. This is not
difficult to do. Suppose
6
A and
6
B of 4ABC be α and β (Figure 13). The line joining the midpoint
O of side
←→
AB and C is a median. Because 4ABC is a right triangle, O is the center of its circumcircle
whose radius is c/2. Hence, 4OAC is an isosceles with
6
OAC =
6
OCA = α. Let the altitude on the
hypotenuse be
←→
CD. Because
6
BCD = α, the altitude is symmetric to the median
←→
CO with respect
to the angle bisector of
6
C. As a result, the altitude on the hypotenuse is an symmedian.
Now we need to show the Lemoine point is the midpoint of CD. Let K be a point on the altitude
and the distances from K to side a, b and c be p
a
, p
b
and p
c
. For convenience, let p
c
= w = r · h,
where h is the length of the altitude and 0 < r < 1. In this way, p
c
is measured based on the ratio
15
related to h. Because 4ABC ∼ 4CKE, we have
p
a
a
=
CK
c
=
h(1 − r)
c
Because 4ABC ∼ 4KCF, we have
p
b
b
=
h(1 − r)
c
If K is the Lemoine point, we must have
p
a
a
=
p
b
b
=
p
c
c
=
r · h
c
This implies
h(1 − r)
c
=
h · r
c
Therefore, if K is the Lemoine point, r = 1/2 and the Lemoine point is the midpoint of the altitude
on the hypotenuse.
5.2 New Proofs Based on the Lemoine Point
Proof 10
We showed that K is the midpoint of the altitude CD. Hence, p
c
= h/2. Because h = (a · b)/c,
we have p
c
= h/2 = ((a · b)/c)/2 = (a · b)/(2c). Because of p
a
/a = p
b
/b = p
c
/c = (a · b)/(2c
2
),
we have p
a
, p
b
and p
c
as follows:
p
a
=
a
2
b
2c
2
, p
b
=
ab
2
2c
2
and p
c
=
ab
2c
(16)
Using the areas of the three trapezoids A (AA
0
B
0
B), A (BB
0
C
0
C) and A (CC
0
A
0
A) the area of
4A
0
B
0
C
0
is calculated easily with our method. Note that p
c
and the p
c
+ c are the lengths of the
altitude of 4ABC and 4A
0
B
0
C
0
on the hypotenuse. Because 4ABC ∼ 4A
0
B
0
C
0
, ρ = c/c
0
=
p
c
/(p
c
+ c) and hence the scaling factor ρ going from 4A
0
B
0
C
0
to 4ABC is
ρ =
p
c
p
c
+ c
=
ab
2c
ab
2c
+ c
=
a · b
a · b + 2c
2
and
1 − ρ
ρ
=
2c
2
ab
(17)
Because a
0
= a/ρ, b
0
= b/ρ and c
0
= c/ρ, the areas of trapezoids AA
0
C
0
C, BC
0
B
0
B and AA
0
B
0
B
are as follows:
A(AA
0
C
0
C) =
1
2
(a + a
0
) · a =
1
2
a +
a
ρ
· a =
a
2
2
1 +
1
ρ
A(BC
0
B
0
B) =
1
2
(b + b
0
) · b =
1
2
b +
b
ρ
· b =
b
2
2
1 +
1
ρ
A(AA
0
B
0
B) =
1
2
(c + c
0
) · c =
1
2
c +
c
ρ
· c =
c
2
2
1 +
1
ρ
16
The area sum of all three trapezoids is
A(outer ring of 4ABC) =
1 +
1
ρ
a
2
+ b
2
+ c
2
2
Therefore, the area of 4A
0
B
0
C
0
according to our method is
A(A
0
B
0
C
0
) =
1
1 − ρ
2
a
2
+ b
2
+ c
2
2
1 +
1
ρ
=
1
2
a
2
+ b
2
+ c
2
ρ(1 − ρ)
(18)
The area of 4A
0
B
0
C
0
may also be computed as follows:
A(4A
0
B
0
C
0
) =
1
2
A
0
C
0
· B
0
C
0
=
1
2
a
ρ
·
b
ρ
=
1
2
·
a · b
ρ
2
This result must agree with the one shown in Eqn. (18) and we have the following:
1
2
a
2
+ b
2
+ c
2
ρ(1 − ρ)
= A(A
0
B
0
C
0
) =
1
2
·
a · b
ρ
2
Simplifying yields
a
2
+ b
2
+ c
2
1 − ρ
=
a · b
ρ
or a
2
+ b
2
+ c
2
=
1 − ρ
ρ
· (a · b)
and after plugging the value of ρ Eqn. (17) followed by a very simple simplification we have
c
2
= a
2
+ b
2
.
Proof 11
It is worthwhile to note that with the property of the Lemoine point for right triangle in hand,
a simpler proof is possible. Recall the results in Eqn (16), the area of 4ABC is the sum of three
smaller triangles 4KAB, 4KBC and 4KCA (Figure 14):
A(ABC) =
1
2
(p
a
· a + p
b
· b + p
c
· c) =
a · b
2
a
2
2c
2
+
b
2
2c
2
+
1
2
=
a · b
2
·
a
2
+ b
2
+ c
2
2c
2
Because the above is equal to (ab)/2, after a simple simplification we have c
2
= a
2
+ b
2
. This is a
direct proof of the Pythagorean Theorem.
It is worthwhile to note that if K is the midpoint of the altitude on the hypotenuse, then p
a
, p
b
and p
c
can be computed using similar triangles as in Eqn (16) and hence the same idea yields the
Pythagorean Theorem. In this way, the concept of the Lemoine can be completely avoided.
Proof 12
Our next question is: can K be selected as an arbitrarily point on the altitude? Of course, K
cannot be C and D. In this way, the distance p
c
can be measured with respect to the length of the
17
Figure 14: A Proof that Only Uses the Lemoine Point
altitude. Again, let h be the length of the altitude CD and p
c
= r · h (Figure 13). We have already
obtained the following at the beginning of this section:
p
a
a
=
p
b
b
=
(1 − r)h
c
and
p
c
c
=
r · h
c
and hence
p
a
=
a
c
h(1 − r), p
b
=
b
c
h(1 − r) and p
c
= h · r
The area of 4ABC is the sum of the areas of three triangles:
A(4ABC) = A (4KBC)+ A(4KCA) + A (4KAB)
=
1
2
a
a · h
c
(1 − r)
+
1
2
b
b · h
c
(1 − r)
+
1
2
c · h · r
=
1
2
a
2
h
c
−
a
2
h
c
r +
b
2
h
c
−
a
2
h
c
r + c · h · r
=
1
2
h
c
a
2
+ b
2
− h · r
a
2
c
+
b
2
c
− c
=
1
2
h
c
a
2
+ b
2
− r
a
2
+ b
2
− c
2
Because the area of 4ABC is also computed as (h · c)/2, we have
1
2
h
c
a
2
+ b
2
− r
a
2
+ b
2
− c
2
=
1
2
(c · h)
A simple simplification yields:
(a
2
+ b
2
) − r(a
2
+ b
2
− c
2
) = c
2
18
Hence, we have
(1 − r)
a
2
+ b
2
− c
2
= 0
Because 0 < r < 1, a
2
+ b
2
− c
2
= 0 and hence the Pythagorean Theorem holds.
Note that this proof still works when we set r = 0. In this way, the values for p
a
and p
b
are still
correct and p
c
= 0.
Proof 13
The previous proof actually provides another proof without the use of Lemoine point. In Fig-
ure 15 we have K = D and p
c
= 0. Because 4CDE ∼ 4ABC ∼ 4CDF, we have
p
a
h
=
a
c
and
p
b
h
=
b
c
We know that h = (a · b)/c. Plugging h into p
a
and p
b
yields:
p
a
= h ·
a
c
=
a · b
c
a
c
=
a
2
b
c
2
p
b
= h ·
b
c
=
a · b
c
b
c
=
a · b
2
c
2
The area of 4ABC is the sum of the areas of the rectangle CEDF and the two right triangles 4DBE
and 4ADF. The area of rectangle CEDF is
A(CEDF) = p
a
· p
b
=
a
2
b
c
2
·
a · b
2
c
2
=
a
3
b
3
c
4
(19)
Figure 15: A Special Case for a Right Triangle
Let x and y be the length of segments EB and FA, respectively. Because 4BDE ∼ 4BAC, we
have x/p
a
= a/b. Because 4ADF ∼ 4ABC, we have y/p
b
= b/a. Hence, x and y in terms of a, b
19
and c are
x = p
a
·
a
b
=
a
2
b
c
2
· h =
a
2
b
c
2
·
a · b
c
=
a
3
c
2
y = p
b
·
b
a
=
ab
2
c
2
· h =
ab
2
c
2
·
a · b
c
=
b
3
c
2
The areas on 4DBE and 4ADF are computed as follows:
A(4DBE) =
1
2
x · p
a
=
1
2
·
a
3
c
2
·
a
2
b
c
2
=
1
2
a
5
b
c
4
A(4DAF) =
1
2
y · p
b
=
1
2
·
b
3
c
2
·
ab
2
c
2
=
1
2
ab
5
c
4
The area of 4ABC is calculated as follows:
A(4ABC) = A (CEDF) + A(4DBE) + A(4ADF)
=
a
3
b
3
c
4
+
1
2
a
5
b
c
4
+
1
2
ab
5
c
4
=
a · b
c
4
·
a
4
+ b
4
+ 2a
2
b
2
2
=
a · b
c
4
·
(a
2
+ b
2
)
2
2
This area computation must agree with the known one (a · b)/2:
a · b
c
4
·
(a
2
+ b
2
)
2
2
=
1
2
(a · b)
Then we have (a
2
+ b
2
)
2
= c
4
. However, this is equivalent to (a
2
+ b
2
)
2
− (c
2
)
2
= 0 and hence we
have
(a
2
+ b
2
− c
2
)(a
2
+ b
2
+ c
2
) = 0
Because a
2
+ b
2
+ c
2
cannot be 0, we must have a
2
+ b
2
− c
2
= 0 and the Pythagorean Theorem
follows.
6 Conclusions
We developed an easy and effective way for proving the Pythagorean Theorem. This method is
based on a simple principle of similarity. Given a shape A and a similar shape B ⊆ A, if the
scaling factor from A to B is ρ (0 < ρ < 1), then the area of A is computed as A (A) = A (A −
B)/(1 − ρ
2
). Note that even though this method is only applied to right triangles, it can be used
with general shapes. This method is applied to several classical proofs in Lommis [6] and to
new proofs. In particular, the use of trigonometry in Jackson and Johnson’s proof [4] is eliminated
20
becoming a geometrical one. With the help of the Lemoine Point, we have a short and simple direct
proof and another one based on our method. The Appendix includes Zimba’s proof of the angle
difference identities of sin() and cos() being independent of the Pythagorean Theorem. We also
show that a similar technique can be used to prove the angle sum identities being independent of the
Pythagorean Theorem. Then, the computation of the derivatives of sin() and cos() is derived from
the angle sum identities, and, finally, with the help of L’H
ˆ
opital’s Rule the double angle identities
are used to proof sin
2
(x) + cos
2
(x) = 1. Consequently, this manuscript successfully demonstrated
that many fundamental formulae of trigonometry are independent of the Pythagorean Theorem and
the sin
2
(x) + cos
2
(x) = 1 identity.
A Proofs of Some Important Trigonometric Identities
We present Zimba’s proof in [13] that the angle difference identities for sin() and cos() are indepen-
dent of the Pythagorean Theorem and the Pythagorean Identity. With a similar technique, we prove
that the angle sum identities for sin() and cos() are also independent of the Pythagorean Theorem.
Furthermore, the derivative computation of sin() and cos() share the same property and the double
angle identities can be used to prove the Pythagorean Identity. This firmly shows that Loomis’s
claim is false and that even though analytic geometry uses the Cartesian Coordinate System many
fundamental results are independent of the Pythagorean Theorem and the Pythagorean Identity.
A.1 Zimba’s Proof of the Angle Difference Identities
Without loss of generality, we assume 0 < β ≤ α < 90
◦
in this section because the main focus is
a right triangle. Consider Figure 16. Line
←→
OQ makes an angle of α − β with the x-axis, where
OQ = 1. Let line
←→
OP make an angle of β with
←→
OQ, where P is the perpendicular foot from Q to
←→
OP.
Thus,
←→
OP makes an angle of α with the x-axis. From P and Q drop perpendiculars to the x-axis
meeting it at S and T . Therefore, we have QT = sin(α − β) and OT = cos(α − β). From 4OPQ
we have PQ = sin(β) and OP = cos(β).
In 4OPS, because sin(α) = PS/PO = PS/cos(β) we have PS = sin(α) cos(β). Similarly, we
have OS = cos(α)cos(β). From Q drop a perpendicular to
←→
PS meeting it at R. Note that
6
P of
4PQR is α. In 4PQR, because sin(α) = QR/QP = QR/sin(β) we have QR = sin(α)sin(β).
Similarly, we have PR = cos(α) sin(β). Consequently, the desired results are as follows:
sin(α − β) = QT = PS − PR = sin(α)cos(β) − cos(α)sin(β)
cos(α − β) = OS + ST = OS + RQ = cos(α)cos(β) + sin(α) sin(β)
If α = β, we have the following:
1 = cos(0) = cos(α − α) = cos
2
(α) + sin
2
(α)
21
Figure 16: Proof of the Angle Difference Identities
A.2 The Angle Sum Identities
We shall prove the angle sum identities for sin() and cos() based on Zimba’s approach. From O
construct a line
←→
OP that makes an angle of α + β with the x-axis and OP = 1 (Figure 17). From
O construct a line
←→
OQ that makes an angle α with the x-axis such that Q is the perpendicular foot
from P to
←→
OQ. In this way, the angle between
←→
OP and
←→
OQ is β. Let the perpendicular feet from P
and Q to the x-axis be S and T . From Q construct a perpendicular to
←→
PS meeting it at R. Hence, we
have sin(α + β) = PS, cos(α+ β) = OS, sin(β) = PQ and cos(β) = OQ.
From 4OQT , because sin(α) = QT /QO = QT / cos(β) we have QT = sin(α)cos(β). Simi-
larly, we have OT = cos(α)cos(β). From 4PQR, because sin(α) = QR/QP = QR/ sin(β) we have
QR = sin(α)sin(β). Similarly, we have PR = cos(α)sin(β). Therefore, we have:
sin(α + β) = PR + RS = sin(α)cos(β) + cos(α) sin(β)
cos(α + β) = OT − ST = cos(α)cos(β) − sin(α)sin(β)
Consequently, the angle sum identities for sin() and cos() are independent of the Pythagorean
Theorem and the Pythagorean Identity.
A.3
d sin(x)
dx
and
d sin(x)
dx
Are Independent of the Pythagorean Theorem
The angle sum and angle difference identities give the following:
sin(α + β) = sin(α)cos(β) + cos(α) sin(β)
sin(α − β) = sin(α)cos(β) − cos(α) sin(β)
22
Figure 17: Proof of the Angle Sum Identities
Subtracting the second from the first yields
sin(α + β) − sin(α − β) = 2 cos(α)sin(β)
Let p = α + β and q = α − β. Then, α = (p + q)/2 and β = (p − q)/2. Plugging p and q into the
above identity gives one of the sum-to-product identities:
sin(p) − sin(q) = 2 cos
p + q
2
sin
p − q
2
Then, the derivative of sin() is computed as follows:
d sin(x)
dx
= lim
h→0
sin(x + h) − sin(x)
h
= lim
h→0
2cos
2x+h
2
sin
h
2
h
=
lim
h→0
cos
2x + h
2
·
lim
h→0
sin(h/2)
h/2
= cos(x)
As h → 0, the first term approaches cos(x) while the second approaches 1. Note that lim
h→0
sin(h)/h =
1 does not depend on the Pythagorean Theorem. Because cos(x) = sin(π/2 −x), by the Chain Rule
we have
d cos(x)
dx
=
d sin(π/2−x)
dx
= cos(π/2 − x)
d(π/2−x)
dx
= − cos(π/2 − x) = −sin(x) and the calcula-
tion of
d sin(x)
dx
and
d sin(x)
dx
is independent of the Pythagorean Theorem and the Pythagorean Identity.
23
A.4 The Double Angle Identities
We have shown that sin(α±β), cos(α±β) and the derivatives of sin() and cos() are all independent
of the Pythagorean Theorem and the Pythagorean Identity. We now prove sin
2
(x) + cos
2
(x) = 1
with the double angle identities:
sin(2α) = 2 sin(α)cos(α)
cos(2α) = cos
2
(α) − sin
2
(α)
Because of the following:
sin(x) = 2 sin
x
2
cos
x
2
cos(x) = cos
2
x
2
− sin
2
x
2
we have
sin
2
(x) + cos
2
(x) =
sin
2
x
2
+ cos
2
x
2
2
With the same technique, we have:
sin
2
(x) + cos
2
(x) =
sin
2
x
2
+ cos
2
x
2
2
=
sin
2
x
4
+ cos
2
x
4
2
2
=
sin
2
x
2
2
+ cos
2
x
2
2
2
2
.
.
.
=
sin
2
x
2
n
+ cos
2
x
2
n
2
n
We need to prove the following:
lim
n→∞
h
sin
2
x
2
n
+ cos
2
x
2
n
i
2
n
= 1
The left-hand side of the above can be rewritten as
h
sin
2
x
2
n
+ cos
2
x
2
n
i
2
n
= exp
2
n
ln
sin
2
x
2
n
+ cos
2
x
2
n
= exp
ln
sin
2
x
2
n
+ cos
2
x
2
n
1
2
n
!
For convenience, let h = 1/2
n
. Therefore, as n → ∞, h → 0 and the above becomes
h
sin
2
x
2
n
+ cos
2
x
2
n
i
2
n
= exp
ln(sin
2
(xh) + cos
2
(xh))
h
24
As h → 0, the numerator approaches ln(sin
2
(0) + cos
2
(0)) = ln(1) = 0 and the denominator ap-
proaches 0. As a result, we have an indefinite form of 0/0 and L’H
ˆ
opital’s Rule is needed to
compute the limit. The derivative of ln(sin
2
(xh) + cos
2
(xh)) with respect to h is
d(ln(sin
2
(xh) + cos
2
(xh)))
dh
=
1
sin
2
(xh) + cos
2
(xh)
d(sin
2
(xh) + cos
2
(xh))
dh
=
1
sin
2
(xh) + cos
2
(xh)
(2sin(xh)cos(xh)x + 2cos(xh)(−sin(xh))x)
= 0
The derivative of the denominator is 1. As a result, we have
lim
n→∞
h
sin
2
x
2
n
+ cos
2
x
2
n
i
2
n
= exp(0) = 1
Consequently, sin
2
(x) + cos
2
(x) = 1 holds.
A.5 A Simple Calculus Based Proof
Finally, we offer a very simple proof based on calculus. Let function f (x) be defined as follows:
f (x) = cos
2
(x) + sin
2
(x)
Differentiate this function yields:
d f (x)
dx
=
d(cos
2
(x) + sin
2
(x))
dx
= 2cos(x)(− sin(x)) + 2 sin(x) cos(x) = 0
Therefore, f (x) is a constant function for some c:
f (x) = sin
2
(x) + cos
2
(x) = c
Because sin(0) = 0 and cos(0) = 1, we have
f (x) = sin
2
(x) + cos
2
(x) = 1
This proves the Pythagorean Identity. Note that the computation of the derivatives of sin(x) and
cos(x) is independent of the Pythagorean Theorem and the Pythagorean Identity. Consequently,
the above proof is valid.
B Updating History
1. First Draft: August 21, 2023
25
2. Typos Corrected and Abstract and Appendix Added: September 19, 2023
3. Typos/Diagrams Corrected + Images & New Material Added: November 3, 2023
4. Partially Rewritten: January 15, 2024.
5. A Minor Typo in Section A.5 Corrected: February 28, 2024.
References
[1] Alexander Bogomolny, Cut the Knot, available at http://www.cut-the-knot.org/
pythagoras/Proof100.shtml (retrieved August 10, 2023).
[2] William Gallatly, The Modern Geometry of the Triangle, 2nd edition, Francis Hodgson, Lon-
don, 1910.
[3] Ross Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, The
Mathematical Association of American, 1995.
[4] Ne’Kiya D. Jackson and Calcea Rujean Johnson, An Impossible Proof of Pythagoras, AMS
Spring Southeastern Sectional Meeting, March 18, 2023 (https://meetings.ams.org/
math/spring2023se/meetingapp.cgi/Paper/23621).
[5] Zsolt Lengv
´
arszky, Proving the Pythagorean Theorem via Infinite Dissections, The American
Mathematical Monthly, Vol. 120 (2013), No. 8 (October), pp. 751–753.
[6] Elisha Scott Loomis, The Pythagorean Proposition, 2nd edition, The National Council of
Teachers of Mathematics, 1940. A scanned PDF file can be found at https://files.eric.
ed.gov/fulltext/ED037335.pdf.
[7] MathTrain, How High Schools Proved Pythagoras Using Just Trig! (and Some Other Stuffs),
YouTube video, 2023 (https://www.youtube.com/watch?v=nQD6lDwFmCc).
[8] Darren Orf, Teens Have Proven the Pythagorean Theorem With Trigonom-
etry. That Should Be Impossible, Popular Mechanics, March 31,
2023 (https://www.popularmechanics.com/science/math/a43469593/
high-schoolers-prove-pythagorean-theorem-using-trigonometry/).
[9] Polymathematic, Pythagoras Would Be Proud: High School Students’ New Proof of the
Pythagorean Theorem [TRIGONOMETRY], YouTube video 2023 (https://www.youtube.
com/watch?v=p6j2nZKwf20).
[10] Ching-Kuang Shene, [CK’s Geometry Talks] The Pythagorean Theorem: I, II and III. Avail-
able at EP4: A 100+ Years Old Incorrect Claim (https://youtu.be/sdli0rR9ot0), EP5: A
New Approach (https://youtu.be/EXjwoVTPWMI) and EP6: Trigonometric Proof of Jack-
son and Johnson (https://youtu.be/3rRST-NHmGA).
26
[11] Leila Sloman, 2 High School Students Prove Pythagorean Theorem. Here’s What That Means,
Scientific American, April 10, 2023 (https://www.scientificamerican.com/article/
2-high-school-students-prove-pythagorean-theorem-heres-what-that-means/).
[12] Ramon Antonio Vargas, US teens say they have new proof for 2,000-year-old mathemati-
cal theorem, The Gaurdian, March 24, 2023 (https://www.theguardian.com/us-news/
2023/mar/24/new-orleans-pythagoras-theorem-trigonometry-prove).
[13] Jason Zimba, On the Possibility of Trigonometric Proofs of the Pythagorean Theorem, Forum
Geometricorum: A Journal on Classical Euclidean Geometry, Vol. 9 (2009), pp. 275–278.
27
Calcea Johnson and Ne'Kiya Jackson are two high school seniors from New Orleans. The students presented their groundbreaking findings at the American [Mathematical Society's Spring Southeastern Sectional Meeting on March 18 2023](https://meetings.ams.org/math/spring2023se/meetingapp.cgi/Paper/23621).
This is what Loomis claimed in his 1907 book
There's a really good video on the proof by [MindYourDecisions](https://www.youtube.com/watch?v=juFdo2bijic).
Worth watching this [60 Minutes episode](https://www.youtube.com/watch?v=VHeWndnHuQs) on the proof and its authors.
[Jason Zimba proved 14 years ago that the identity](https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf) $cos^2 x+ sin^2 x = 1$ can be derived independently of the
Pythagorean theorem, despite common beliefs to the contrary.
Nice! But which of the two calculus/limits-based approaches is “easier” to understand? :
limits of infinite sequences, etc. ?
limits involved in calculating derivatives?
In A.3 heading: one of the two “sin” should be a “cos”!
A.5 is probably the only thing here worth writing down. And this proof has been known since the beginnings of calculus, I guess.
But it is so trivial that no one would publish a paper on it, or on its independence of the Pythagorean Theirem.
Michael Malione, inspired by the work of Jackson and Johnson, came up with a proof that does not require the law of sines, just the basic definitions of the sine and cosine ratios appearing in a right triangle.
### Simplified Proof
Given a right triangle, △ABC with legs \(AC\) and \(BC\), and hypotenuse \(AB\). Prove that \( \sin^2(A) + \cos^2(A) = 1 \), where \( \sin(A) = \frac{BC}{AB} \) and \( \cos(A) = \frac{AC}{AB} \).
### Main Triangle
Drop an altitude \( CD \) from point \( C \) to point \( D \) on segment \( AB \), and then another altitude \( DE \) (for △BCD) from point \( D \) to point \( E \) on segment \( BC \). Continue constructing altitudes back and forth between the two lines in this manner, constructing similar triangles, ad infinitum.
### Constructed Altitudes
Note the triangle similarities:
\[ △ABC \sim △ACD \sim △CDE \sim △DEF \sim \ldots \]
Now, using our trigonometry definitions, with these similar triangles, we have:
\[ \sin(A) = \frac{BC}{AB} = \frac{CD}{AC} = \frac{DE}{CD} = \frac{EF}{DE} = \ldots \]
and
\[ \cos(A) = \frac{AC}{AB} = \frac{AD}{AC} = \frac{DF}{DE} = \ldots \]
Reconstructing our segment \( AB \) as an infinite sum of lengths \( AD, DF, FH, \) etc., we have, after rearranging the \( \cos(A) \) expression from above, that \( AD = (AC) \cos(A) \) and \( DF = (DE) \cos(A) \), and so on for each additional segment along \( AB \).
We also have that \( DE = (CD) \sin(A) \) and \( CD = (AC) \sin(A) \). Putting this all together reveals that \( AB \) is the sum of a geometric sequence, with a ratio of \( \sin^2(A) \) between successive terms.
\[ DF = (DE) \cos(A) = (CD) \sin(A) \cos(A) = (AC) \sin^2(A) \cos(A) \]
So with \( AD = (AC) \cos(A) \), and \( DF = (AC) \cos(A) \sin^2(A) \), etc., we have an infinite geometric series for \( AB \):
\[ AB = (AC) \cos(A) \left(1 + \sin^2(A) + \sin^4(A) + \sin^6(A) + \ldots \right) \]
Since \( |\sin^2(A)| < 1 \) for all right triangles, the series converges to \( \frac{1}{1-r} \) where \( r = \sin^2(A) \), and the above equation becomes:
\[ AB = (AC) \cos(A) \frac{1}{1 - \sin^2(A)} \]
Then
\[ 1 = \frac{AC}{AB} \cos(A) \frac{1}{1 - \sin^2(A)} = \frac{\cos^2(A)}{1 - \sin^2(A)} \]
And finally:
\[ \cos^2(A) = 1 - \sin^2(A) \]
\[ \sin^2(A) + \cos^2(A) = 1 \]