And, for the same reason, BAH is one and the same straight line. Next, because the angles DBC and FBA are each a right angle (cons.), therefore the angle DBC = the angle FBA; add to each the angle ABC, and therefore the angle DBA = the angle FBC (ax. 2). the And because in the triangles ABD and FBC we have the sides AB and BD and their angle ABD in the former sides FB and BC and their angle FBC in the latter, each to each, therefore the triangle ABD = the triangle FBC (I. 4). Again, because the parallelogram BL and the triangle ABD are upon the same base BD, and between the same parallels AL and BD, therefore the parallelogram BL is double the triangle ABD (I. 41). And, for the same reason, the square GB is double the triangle FBC. But, as we have proved, the triangle ABD the triangle FBC; therefore the parallelogram BL = the square GB (ax. 6). = In the same manner, by joining AE and BK, it is proved that the parallelogram CL the square HC, and therefore the whole square BDEC = the sum of the squares GB and HC (ax. 2). But BDEC is the square on BC, subtending the right angle BAC, and GB and HC are the squares on BA and AC, the sides containing the right angle BAC; Therefore, it is proved, as required, that The square described on the side BC subtending the right angle the sum of the squares described on the sides BA and AC, containing that angle. Wherefore, In any right-angled triangle, &c. Exercises. Q. E. D. 1. Prove, as stated in the above proposition, that the parallelogram CL the square HC. = 2. Prove that if in the figure of Prop. 47 the points E and K are joined by the line EK, and the points D and F by the line DF, the triangles KCE and FBD are equal to each other. If the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it, the angle contained by these two sides is a right angle. Let ABC be a triangle, with the square described upon BC= the sum of the squares described upon BA and AC. Then it is to be proved that The angle BAC is a right angle. B CONSTRUCTION.-From A draw AD at right angles to AC (I. 11), make AD equal to AB (I. 3), and join DC. PROOF.-Because AD = AB (cons.), therefore the square on AD = the square on AB; and if the square on AC be added to each of these equals, then the squares on AD and AC = the squares on AB and AC (ax. 2). And because the angle DAC is a right angle (cons.), therefore the square on DC = the squares on AD and AC (I. 47). Also the square on BC=the squares on AB and AC (hyp.) ; therefore the square on DC = the square on BC (ax. Ì), and therefore DCBC. Next, because in the triangles BAC and DAC we have the sides BA, AC, and CB in the former = the sides DA, AC, and CD in the latter, each to each (cons. and proof above), therefore the angle BAC = the angle DAC (I. 8), and because the angle DAC is a right angle (cons.), Therefore, it is proved, as required, that, The angle BAC is a right angle. Wherefore, If the square, &c. Q. E. D. ADDENDUM. BOOK I. CLASSIFICATION OF PROPOSITIONS IN BOOK L I. Straight Lines. Props. 2, 3, 10, 11, 12, 14. II. Angles in connection with Straight Lines. III. Parallel Lines. Props. 27, 28, 30, 31, 33. IV. Angles in connection with Parallel Lines. Prop. 29. √. Construction of Triangles. Props. 1, 7, 22. VI. Angles in connection with Triangles. Props. 5, 16, 17, 18, 21, 32, 48. VII. Sides in connection with Triangles. VIII. Comparison of Triangles. IX. Triangles and Parallel Lines. X. Triangles and Parallelograms. Prop. 41 XI. Properties and Comparison of Parallelograms. Props. 34, 35, 36, 42, 43, 44, 45. XII. Construction of the Square. |