Legendre's Proof

1)      Legendre’s Proof of the Pythagorean Theorem

References: 

Loomis, Elisha S. (1968). The Pythagorean proposition. Washington, DC: National Council of Teachers of Mathematics.

Brodie, Scott. (1996). The pythagorean theorem is equivalent to the parallel postulate. Retrieved from http://www.cut-theknot.org/triangle/pythpar/PTimpliesPP.shtml

To prove the Pythagorean Theorem using Legendre’s method, we use the properties of a right triangle and its altitude.

1.      Let ABC be any right triangle, with right angle C.

2.      Draw the altitude CF from the right angle ACB to the hypotenuse AB.

3.      Label  the sides of the triangle a, b, c, and the vertices opposite each side A, B, C.

4.      Let BF = x, FA = y, and CF = h.

5.      Then triangle BFC triangle CFA, and triangle BCA are right triangles, and we may apply the Pythagorean assumption three times:

        I.            a² + b² = c² (outer triangle ABC)

      II.            x² + h² = a² (left inner triangle BFC)

    III.            h² + y² = b² (right inner triangle BCA)

6.      We also know that c = (x+y), and we can square both sides to find that c² = (x + y)² = x² + y² + 2xy

7.      Using substitution for a² + b² = c² (a² = x² + h², b² = h² + y², and c² = x² + y² + 2xy) therefore,

8.      x² + h² + h² + y² = x² + y² + 2xy

9.      Subtracting x² + y² from each side, 2h² = 2xy

10.  Divide both sides by 2, and h² = xy

11.  Divide both sides by h, and h = xy/h

12.  Divide both sides by x, and h/x = y/h

13.  We can set this common ratio h/x = y/h = k. Then h = kx, y = kh

14.  Using substation, since b² = h² + y²_, then b² = (kx)² + (kh)²= k²x² + k²h²

15.  Pulling out a greatest common factor of k², b² = k²(x² + h²)

16.  Since x² + h² = a², b² = k²a²

17.  Dividing both sides by a² and taking the square root of both sides, we find that b/a = k = h/x = y/h.

18.   Using a similar process, we can show that b/c = h/a = y/b,

Thus the corresponding sides of the two small triangles are proportional, as are the corresponding sides of the original triangle and either of the small triangles.
To complete the proof, we must show that corresponding angles are proportional. We must show a case where this is true, and cannot simply assume based on our previous discovery of similarity in corresponding sides. We can use an isosceles right triangles to complete the proof.
Assume triangle ABC is an isosceles triangle, therefore we know that a = b.
Using the properties of similar triangles, we can set up a proportion a/b = x/h
Since a = b, and since a/b = x/h, we have x = h; similarly, h = y.

Thus triangles BFC and AFC are likewise isosceles, and their base angles FBC, BCF, FCA, and FAC are all equal. Since angles BFC and CFA are right, we conclude that triangles BFC and CFA are equiangular with each other, and with triangle ACB.
But angle BCF and angle FCA sum to one right angle. Therefore, the equal angles FBC and FAC sum to one right angle, and the angles of the original triangle sum to two right angles.

Bhaskara's First Proof

1)      Bhaskara’s First Proof of the Pythagorean Theorem

References: 

Loomis, Elisha S. (1968). The Pythagorean proposition. Washington, DC: National Council of Teachers of Mathematics.

Bogomolny, Alexander. (1996). The pythagorean theorem [Cut the knot]. Retrieved from http://www.cut-the-knot.org/pythagoras/

To complete Bhaskara’s Proof, you must draw the following diagram. The outside shape is a square with side of length c. Inside the large square are 4 right triangle with base length b and altitude length a. In the center is a small square with length a – b.

To prove the Pythagorean Theorem using Bhaskara’s method, we must find the area of the large square in two ways.

First, area of large square equals c².

Second, we find the area of the four congruent triangles with legs with length a and b, and hypotenuse with length c, and the area of the small square with length a – b.

The area of the 4 congruent triangles:

4 (½ ab) = 2ab

The area of the small interior square:

(a – b)²  = a² – 2ab + b²

When we sum the area of the four triangles and the small square, we find the total area of the large square equal to:

a² – 2ab + b² + 2ab = a² + b²

Since the area of the large square is equal to the sum of the areas of its parts:

c² = a² + b²

President Garfield

1)         President Garfield’s Proof of the Pythagorean Theorem

References: 

Loomis, Elisha S. (1968). The Pythagorean proposition. Washington, DC: National Council of Teachers of Mathematics.

Multiplication By Infiniti. (2011). US president garfield’s proof of the Pythagorean theorem. Retrieved from http://tetrahedral.blogspot.com/2011/04/us-president-garfields-proof-of.html

In the figure shown below, we have taken an arbitrary right triangle with sides of length a, b, and hypotenuse of length c. We draw a second copy of this same triangle ABC as pictured below.

 

First, we need to find the area of the trapezoid by using the area formula of the trapezoid.

Formula for the area of a Trapeziod  A = ½ h(b₁ + b₂)

Given the above diagram, h= a + b, b₁=a, and b₂=b.

Using substitution and the area formula for a trapezoid:

A=½(a+b)(a+b)
A =½(a²+2ab+b²).

Now, let's find the area of the trapezoid by summing the area of the three right triangles.

The area of the yellow triangle:
A_y = ½ ab

The area of the white triangle:
A_w = ½c²

The area of the blue triangle:
A_b = ½ ab

We find the sum of the three triangles:
½ ab + ½c²+ ½ ab = ½ (ab + c² + ab) = ½ (2ab + c²).

Since the area of the trapezoid equals the sum of the areas of the three triangles, we can set up the following equality:

½(a² + 2ab + b²) = ½ (2ab + c²).

Multiplying both sides by 2:

a² + 2ab + b² = 2ab + c²

And, subtracting 2ab from both sides:

a² + b² = c²

Thus, we are able to prove the Pythagorean Theorem.

How do we define the pythagorean theorem without diagrams or pictures?

The Pythagorean Theorem describes the relationship in Euclidean Geometry between the three sides of any right triangle.

Definitions:

Right triangle – a triangle with a right angle

Hypotenuse – longest side of a right triangle, opposite the right angle

Legs of a right triangle – the two sides opposite the two acute angles

In any right triangle, the area of the square whose side is the length of the hypotenuse of the right triangle is equal to the sum of the two areas of the squares whose sides are the lengths of the two legs.

In the below diagram, the legs of the triangle are AB and CB, and the hypotenuse is AC. If you add the area of the squares adjacent to AB and CB together, the total will equal the area of the square adjacent to AC.