by Euc. I. 14, (d) BA is in the same straight line with A H. Now, by definition of a square and Axiom 11, angle DBC is equal to angle FB A. To each of these add angle ABC. Then, by Axiom 2a, (e) the whole angle DBA is equal to the whole angle FBC. And, by definition of a square, AB, BD are equal to FB, BC, each to each. by Euc. I. 4, (f) the triangle A B D is equal to the triangle FBC. Again, the parallelogram B L, and the triangle ABD, are on the same base BD, and between the same parallels BD and A L. by Euc. I. 41, (g) BL is double of A B D. And, the square GB and the triangle CBF are on the same base FB, and between the same parallels, FB, A C. by Euc. I. 41, (h) GB is double of CBF. But, by (g) and (h), and Axiom 6a, (k) the parallelogram BL is equal to the square G B. In the same way it can be shown that (7) the parallelogram CL is equal to the square H C. ... The whole square BDEC described on BC is equal to the two squares ABFG described on A B, and A CKH described on A C. Q. E. D. EXERCISE.-Prove, as above (1), that the parallelogram CL is equal to the square HC, by joining AE and BK (Fig. 3). THEOREM (Euclid I. 48). Repeat. The enunciations of Euc. I. 8 and I. 47; Axioms 1 and 2a. General Enunciation. If the square described on one of the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by those two sides is a right angle. Particular Enunciation. Given. The triangle ABC; such that the square described on BC is equal to the squares described on AB, A C. (a) draw A D at right angles to AC (by Euc. I. II). (b) Make A D equal to BA (by Euc. I. 3). Join DC. Proof. A To each of these add the square on A C. by Axiom 2a, (d) the squares on DA, A C are equal to the squares on BA, A C. Again, DAC is a right angle (a). ... by Euc. I. 47, (e) the square on DC is equal to the squares on DA, A C. (f) And, the square on BC is equal to the squares on BA, AC. Hence, by (d), (e), and (ƒ), and Axiom 1, The square on DC is equal to the square on B C. (g). The side DC is equal to the side BC. Again, D A C, B A C are two triangles, having .. by Euc. I. 8, angle DAC is equal to angle BA C. But angle DAC is a right angle (a), .. angle B A C is a right angle. EXERCISES. Q. E. D. I. Prove that BA, AD in the last theorem are in the same straight line. II. Prove that the square on the side subtending an acute angle of a triangle, is less than the squares on the sides containing the acute angle. (a) When the triangle is right-angled. (b) When the triangle is not right-angled. III. Prove that the square on the side subtending an obtuse angle of a triangle is greater than the squares on the sides containing the obtuse angles. THE END. ELFE BROTHERS, CHARTErhouse buildings, aldersgate, e.c |