Fig. 133

Extend GA making AD = AO. Extend DB to N, draw CL and KM. Extend BF to S making FS = HB, complete sq. SU, draw HP par. to AB, PR par. to AH and draw SQ.

Then, obvious, sq.
AK = 4 tri. BAN + sq. NL

= rect. AR + rect. TR + sq.
GQ = rect. AR + rect. QF

+ sq. GQ + (sq. TF = sq. ND)
= sq. HG + sq. HD. ∴ sq.
upon AB = sq. upon BH + sq.
upon AH. ∴ h² = a² + b².
Q.E.D.

a. This proof is credited to Miss E. A.
Coolidge, a blind girl. See Journal of Education,

V. XXVIII, 1888, p. 17, 26th proof.

b. The reader will note that this proof employs exactly the same dissection and arrangement as found in the solution by the Hindu mathematician, Bhaskara. See fig. 324, proof Two Hundred Twenty- Five.

Geometric Proofs, proof #31

For more information on this work, see The Pythagorean Proposition by Elisha Scott Loomis.

For more information on this work, see The Pythagorean Proposition by Elisha Scott Loomis.