In a right-angled triangle the square
of the hypotenuse is equal to the sum of the squares of the other two sides.
a2 + b2 = c2
Various cultures at the time knew of this
relationship, however Pythagoras was the first to leave a written proof of this
theorem. We will look at his proof and others as well later.
So, given any two we can solve for the third. To
do so, rearrange the equation so the unknown is on one side and everything else
is on the other side, then take the square root.
If you think of each side of the triangle as the
side of its own square, then this theorem is
stating that the number of sub-squares in smaller
squares adds up to the number of sub-squares in
the largest square.
16 sub-squares + 9 sub-squares = 25
sub-squares
Note, this is an illustration/demonstration
of Pythagorean's theorem, not a proof of his theorem.
Now let's investigate ways to prove
Pythagorean's theorem.
First a short proof.
The following is a figure of two squares and 4
resulting triangles. Once we prove that all four triangles are congruent, then
the proof follows immediately. (The top corner of the blue square has the Greek
letter delta (δ), and my apologies for the busy diagram.)
The proof (in paragraph form):
First we need to establish that the 4
triangles shown are congruent.
Once this is accomplished, we directly compare
the areas to get the result.
Notice each triangle is a right triangle,
since each has one angle of the outer square. Since the sum of the angles of any
triangle equals 180°, then
β + δ = 90°
(triangle 2). Looking at the angles α,
β, and the corner angle of the square with side c (which is 90° since it is part
of that square) and noting that they add up to 180° (the straight angle), we
have α + β + 90° = 180° which
means
α + β
= 90°.
Since β + δ = 90°
we have β + δ = α + β. So, δ = α. Now α + γ = 90° (triangle 1) .
Since α + β = 90°
= α + γ , β must equal γ. So we have triangle 1 congruent to triangle 2, which
means corresponding sides of these triangles must also be congruent (This
argument can be used around the square to show all triangles are congruent, and
would make a good exercise.)
We've shown that the four triangles are congruent.
Now, the area of
the larger square is (a + b)2. If we subtract the area of the inner
blue square (c2) the remaining area must be the area of all four
congruent triangles which is 4•( ½ a b) = 2ab.
Putting this all
together we have:
(a + b)2 - c2
= 2ab
a2 + 2ab + b2 - c2 = 2ab
a2 + b2
= 2ab +c2 – 2ab
a2 + b2
= c2
And, we've proven the Pythagorean Theorem.
Observations:
Construction is commonly required to carry
out geometry proofs. The construction must be valid however.
First, how do we create these two squares?
Obviously we must have the inner square smaller than the outer square. Now, what
is the range of sizes the inner square must be to allow all four of its corners
to touch the outer square and where do we place it such that all four corners
will simultaneously touch the outer square with no crossings? This construction
is important for the proof. Is any of this obvious? Hopefully not. Except for
axioms, nothing else can be assumed; proof is required.
As a start, imagine both squares sharing the
same center. The inner square will rotate about this center with a radius equal
to ½ the distance of either one of its diagonals. The question is at what point
will one corner of the inner square touch the side of the outer square? How big
does this radius have to be (smallest and largest) to cause this? And finally,
do all other corners touch the outer square in the same way?
Use the following figures in the discussion
that follows.
Figure 1 shows the starting arrangement of
the two squares.
Assign d to the length of one side of
the outer square. From the center we have the shortest distance to the side of
the outer square as ½d.
Both squares share the same center and we will
rotate the inner square clockwise until its corners touch the outer square.
Figure 3 shows when they touch; figure 4 shows the smallest length of the inner
diagonal to be d that still allow the inner circle's corners to touch the
outer circle.
Now the diagonals of both the inner and outer
squares bisect their respective corner angles. This means that the corners
of the inner square all start from the same relative angle between their
respective outer square corners. As we rotate the inner square all of its
corners pass through the same angle until they touch the outer square. We
can drop a perpendicular from the center to the sides of the outside square.
Doing so we have two sides and an included angle congruent in each such
triangle, which means the lengths of each corresponding line segment created by
the touching of the corners of the inner square are congruent, so this
construction will work. Circle theorems will make this argument (not
a proof as is) more rigorous.
Advanced: Proof of
the Pythagorean Theorem
(purpose: improves geometry visualization and
deductive reasoning skills)
The theorem:
In a right-angled triangle the square of the
hypotenuse is equal to the sum of the squares of the other two sides.
This proof is the classical proof of the
Pythagorean Theorem, using a theorem of Euclid; proving Euclid's theorem leads
us directly to the proof of Pythagorean Theorem. (visualization skills are
necessary here)
Euclid's theorem: In a right
triangle, the square of one of the two shorter sides has the same area as the
rectangle whose sides are the projection of that side onto the hypotenuse and
the hypotenuse itself.
Below is a right triangle drawn on it's
hypotenuse.
Euclid's theorem states that the regions of
matching color have the same areas. That is square 1 has the same area as
rectangle 2 and square 3 has the same area as rectangle 4.
The outline of the proof of Euclid's theorem is
as follows:
draw a line from the bottom left corner of
rectangle 1 to vertex B.
use this new triangle to relate square 1
and rectangle 2 (the insight required is rotating the triangle.)
Observe:
The blue triangle (ABD)
in the figure below has the base AD and the height from A, AC (AC is
perpendicular to to AD, both sides are adjacent sides of square 1) . Since
the area of a triangle is ½ ● base
● height, in this case we have ½ ●
AD ● AC = ½
● b ● b = ½ b2
(AD=AC=b)
Now rotate the blue
triangle 90 clockwise so that its base lies along AC.
I've changed the color
of the blue triangle to red so can focus on its new position. These two
triangles are congruent (not because we visually rotated them!) but since
they have identical corresponding sides. The figure is getting a bit noisy,
so let's pull out the items of interest, we'll come back to this figure.
Look at the figure following this one.
We now have a similar
situation a in step #1, the triangle ACF shares its base and height with
rectangle AFGH. So the area of this triangle must be equal to the area of
the rectangle AFGH. Since the triangle's area is the same in both cases, the
areas of square 1 and rectangle 2 must be equal. In particular we have b2
= cq.
A similar argument shows
the equality of the areas of square 3 and rectangle 4. a2 = cp.
(a good exercise, try it!)
Using these results we
have:
a2 + b2
= cp + cq
= c ( p + q )
and since c = p + q we have
= c ( c ) = c2
And we've proven the Pythagorean theorem.
