# The Pythagorean Proposition: Geometric Proofs, Proof #15

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### Fifteen

Fig. 116

Take HR = HE and FS = FR = EQ = DP.

Draw RU par. to AH, ST par. to FH, QP par. to BH, and UP par. to AB. Extend GA to M, making AM = AG, and DB to L and draw CN par. to AH and KO par. to BH.

Place rect. GT in position of EP. Obvious that:  Sq. AK = parts (1
+ 2 + 3) + (4 + 5 of rect. HP).  ∴ Sq. upon AB = sq. upon BH + sq. upon AH.
∴ h² = a² + b².

a. Math. Mo., 1858-9, Vol. I, p. 231, where this dissection is credited to David W. Hoyt, Prof. Math and Mechanics, Polytechnic College, Phila., Pa.; also to Pliny Earle Chase, Phila., Pa.

b. The Math. Mo. was edited by J. D. Runkle, A.M., Cambridge Eng. He says this demonstration  is essentially the same as the Indian demonstration found in "Bija Gauita" and referred to as the figure of "The Brides Chair."

c. Also see said Math. Mo., p. 361, for an­other proof; and Dr. Hutton (tracts, London, 1812, in his History of Algebra).

Algebraic Proofs, Proof #89 | Geometric Proofs, Proof #31
For more information on this work, see The Pythagorean Proposition by Elisha Scott Loomis.