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.