### Graphical proof of the Pythagorean theorem

To prove the Pythagorean theorem, construct two identical squares from the sum of two arbitrary lengths, a and b:

The four shaded right triangles in the left square are congruent with the four shaded right triangles in the right square.  The shaded areas can, therefore, be removed from both squares while preserving equality.  (When equals are subtracted from equals, the remainders are equal.)

QED:

$\LARGE c^{2}=a^{2}+b^{2}$
One of the most useful equations in trigonometry comes directly from the Pythagorean theorem:

$\sin^{2} \Theta +\cos ^{2}\Theta =1$

Let theta be the angle between the horizontal axis and a point on a unit circle.
Since the vertical component is the sine of theta, and the horizontal component is the cosine of theta, it follows that the sum of their squares is equal to one squared, or one.

A new page has been added to this site.  It presents seemingly difficult problems with simple solutions. Quick Math

### Three dimensional Pythagorean theorem

Consider a diagonal, d, of a rectangular box with mutually orthogonal edges, a, b, and c.
The square of diagonal, e, of the face bounded by edges a and b is:

$e^{2}=a^{2}+b^{2}$

Since edge c is perpendicular to the plane of a, b, and e:

$d^{2}=e^{2}+c^{2}$

$\therefore d^{2}=a^{2}+b^{2}+c^{2}$

Any vector is the diagonal of a rectangular box with edges equal to its three components in any arbitrary orthogonal coordinate system.  Thus, the square of the length of any vector is the sum of the squares of its three Cartesian components.

The corollary is also true.  Vectors have equal length if the sum of the squares of their components are equal.  This is the basis for the equation of a sphere in Cartesian coordinates.

$x^{2}+y^{2}+z^{2}=r^{2}$

The surface of a sphere is the locus of points which satisfy the above equation for a constant radius, r.

### Pythagorean theorem in n-dimensions

A defining characteristic of a Euclidean space is that the square of the length of a vector is the sum of the squares of any set of orthogonal components:

$\left \| x \right \|^{2}=\sum_{i=1}^{n}(x_{i})^{2}$

### Parallelogram "law"

The parallelogram law states that the sum of the squares of the diagonals equal the sum of the squares of the sides.  This can be demonstrated from the above Pythagorean principle.

The vector diagram above shows that the long diagonal is the vector sum of vectors A and B.  The short diagonal is the vector sum of A and negative B.  A component of a vector sum is simply the sum of each vector's component along that same coordinate direction.

Consider the components of these diagonals along the horizontal, x, axis.

$\left | A+B \right |_{x}=\left | A_{x} \right |+\left | B_{x} \right |$

$\left | A-B \right |_{x}=\left | A_{x} \right |-\left | B_{x} \right |$

Squaring each diagonal's x component:

$\left | A+B \right |_{x}^{2}=\left | A_{x} \right |^{2}+2\left | A_{x} \right |\left | B_{x} \right |+\left | B_{x} \right |^{2}$

$\left | A-B \right |_{x}^{2}=\left | A_{x} \right |^{2}-2\left | A_{x} \right |\left | B_{x} \right |+\left | B_{x} \right |^{2}$

Adding the squares of both diagonals' x component, the center terms on the rhs cancel, leaving twice the square of the x components of both A and B.

The squares of each diagonal's vertical component have the exact same format.  (With the x subscript replaced by y.)

The sum of the squares of both components of both diagonals is:

$\left | A+B \right |^{2}+\left | A-B \right |^{2}=2(\left | A_{x} \right |^{2}+\left | A_{y} \right |^{2})+2(\left | B_{x} \right |^{2}+\left | B_{y} \right |^{2})$

$\therefore \left | A+B \right |^{2}+\left | A-B \right |^{2}=2\left | A \right |^{2}+2\left | B \right |^{2}$

(While the above demonstration used coordinate directions x and y in the plane defined by A and B, the same argument would be valid for any arbitrary Cartesian coordinate system.)

The Pythagorean theorem can also be demonstrated from a special case of the parallelogram law, specifically, when the parallelogram is a rectangle.

$c^{2}+d^{2}=2a^{2}+2b^{2}$

For a rectangle, both diagonals are equal, and the angles between a and b are right angles.

$c=d$

So that:

$2c^{2}=2a^{2}+2b^{2}$

$\therefore c^{2}=a^{2}+b^{2}$

The rectangle also demonstrates another well known and sometimes useful feature of right angles.  Any chord of a circle which subtends a right angle is also a diameter of the circle.

Like all parallelograms, the diagonals bisect each other, and in the rectangle the diagonals are also equal.  Therefore, the intersection of the diagonals is equidistant from the four corners, defining the center of the circumscribed circle.  Stated another way, the point which bisects the hypotenuse is equidistant from all corners of a right triangle.

The rectangle also proves that the area of a right triangle is equal to one half the product of the two sides.  (A diagonal of a rectangle divides the rectangle into two equal right triangles.)  This well known property can be used to derive the area of a circle.

### A two dimensional theorem with a three dimensional proof

Given two triangles with their vertices on three arbitrary lines radiating from a point, the intersections of their respective (extended) sides form a straight line.

While this is extremely difficult to prove using plane geometry, it turns out to be quite simple with solid geometry.

Consider the above diagram to be a projection of a three-sided pyramid with vertex, O, and edges R1, R2, and R3.  Triangles a and b then represent planes cutting through the pyramid.

Triangle sides a1 and b1 are both in the plane defined by R1 and R2, and since they are not parallel, must intersect.

Since a1 is in the plane of triangle a, and b1 is in the plane of triangle b, point 1 must lie on the intersection of the planes of the two triangles.

The same argument can be made for points 2 and 3.  The intersection of two planes is always a straight line and any projection of a straight line is also a straight line. QED

Special cases:
If one of the sides is parallel to the respective side of the other triangle, then the line between the other two intersections is also parallel to the parallel sides.

If two sides are parallel to their respective sides on the other triangle, then all respective sides are parallel.

The take away here is that a change in perspective can sometimes reveal an easy solution to a difficult problem.

This web site is all about exploring alternative perspectives.

### Using the Pythagorean theorem to compute PI

The Pythagorean theorem can be used to derive a simple iterative computation of the value of PI.  PI can be defined as the arc length of a semi-circle with unit radius.  From a diameter of the unit circle, two equal chords can be constructed as the sides of an isosceles triangle with the diameter as the base.  Each of these chords define the base of yet another isosceles triangle which doubles the number of chords subtending the semi-circle.

In the above diagram, the Pythagorean theorem defines the general relationship between the new chord length, the height of the isosceles triangle, h, and the old half-chord length.

$S_{new}^{2}=\left ( \frac{S_{old}}{2} \right )^{2}+ h^{2}$

The height of each isosceles triangle is simply:

$h=1-g$

Where g is given by:

$g=\sqrt{1-\left (\frac{S_{old}}{2} \right )^{2}}$

After back substituting for g and h, and performing some algebra, we can write:

Since at any step of the iteration, the square of each half-chord can be defined as a function of the square of the half-chord of the previous step, this iterative algorithm is ideally suited to spreadsheet computation.