[Math Lair] The Pythagorean Proposition: Algebraic Proofs, Proof #1

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[Figure 1]

Fig. 1

In rt. tri. ABH, draw HC perp. to AB. The tri's ABH, ACH and HCB are similar. For convenience, denote BH, HA, AB, HC, CB and AC by a, b, h, x, y and h-y resp'y. Since, from three similar and related tri­angles, there are possible nine sim­ple proportions, these proportions and their resulting equations are:
(1) a : x = b : h - y ∴ ah - ay = bx.
(2) a : y = b : x ∴ ax = by.
(3) x : y = h - y : x ∴ x² = hy - y².
(4) a : x = h : b ∴ ab = hx.
(5) a : y = h : a ∴ a² = hy.
(6) x : y = b : a ∴ ax = by.
(7) b : h - y = h : b ∴ b² = h² - hy
(8) b : x = h : a ∴ b² = h² - hy
(9) h - y : x = b : a ∴ ah - ay = bx. See Versluys, p. 86, fig. 97, Wm. W. Rupert.

Since equations (1) and (9) are identical, also (2) and (6), and (4) and (8), there remain but six different equations, and the problem becomes, how may these six equations be combined so as to give the desired relation h² = a² + b², which geometrical­ly interpreted is AB² = BH² + HA².

In this proof One, and in every case here­after, as in proof Sixteen, p. 41, the symbol AB², or a like symbol, signifies AB².

Every rational solution of h² = a² + b² af­fords a Pythagorean triangle. See "Mathematical Mon­ograph, No. 16, Diophantine Analysis," (1915), by R. D. Carmichael.

1st.--Legendre's Solution

a. From no single equation of the above nine can the desired relation be determined, and there is but one combination of two equations which will give it; viz., (5) a² = hy; (7) b² = h² - hy; adding these gives h² = a² + b²

This is the shortest proof possible of the Pythagorean Proposition.

b. Since equations (5) and (7) are implied in the principle that homologous sides of similar tri­angles are proportional it follows that the truth of this important proposition is but a corollary to the more general truth——the law of similarity.

c. See Davis Legendre, 1858, p. 112,

   Journal of Education, 1888, V. XXV, p. 404, fig. V.

   Heath's Math. Monograph, 1900, No. 1, p. 19, proof III, or any late text on geometry.

d. W. W. Rouse Ball, of Trinity college, Cam­bridge, England seems to think Pythagoras knew of this proof.

2nd.--Other Solutions

a. By the law of combinations there are pos­sible 20 sets of three equations out of the six dif­ferent equations. Rejecting all sets containing (5) and (7), and all sets containing dependent equations, there are remaining 13 sets from which the elimina­tion of x and y may be accomplished in 44 different ways, each giving a distinct proof for the relation h² = a² + b².

b. See the American Math. Monthly, 1896, V. III, p. 66 or Edward's Geometry, p. 157, fig. 15.

Some Noted Proofs | Algebraic Proofs, Proof #36
For more information on this work, see The Pythagorean Proposition by Elisha Scott Loomis.