Book I. to the triangle DBC, having the angle GHM equal to the angle E and because the angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM; add c 29. 1. : to each of these the angle KHG; therefore the angles FKH, D F GL E are equal to two B CK HM right angles; and because at the point H in the straight line GH, the two straight lines KH, HM, upon the opposite sides of GH make the adjacent angles equal to two right angles, KH is in the same d 14. 1. straight lined with HM: and because the straight line HG e 30. 1. с meets the parallels KM, FG, the alternate angles MHG, HGF, are equal: Add to each of these the angle HGL: Therefore the angles MHG, HGL, are equal to the angles HGF, HGL: But the angles MHG, HGL, are equal to two right angles; wherefore also the angles HGF, HGL are equal to two right angles, and FG is therefore in the same straight line with GL: And because KF is parallel to HG, and HG to ML; KF is parallel to ML: and KM, FL are parallels; wherefore KFLM is a parallelogram; and because the triangle ABD is equal to the parallelogram FH, and the triangle DBC to the parallelogram GM; the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done. COR. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal b 44. 1. figure, viz. by applying to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle. b Book I. PROP. XLVI. PROB. To describe a square upon a given straight line. Let AB be the given straight line; it is required to describe a square upon AB. d E From the point A draw a AC at right angles to AB; and a 11. 1. make AD equal to AB, and through the point D draw DE b 3. 1. parallel to AB, and through B draw BE parallel to AD; there- c 31. 1. fore ADEB is a parallelogram; whence AB is equal to DE, d 34. 1. and AD to BE: But BA is equal to AD; therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral, likewise all its angles are right angles; because the straight line AD meeting the parallels AB, DE, the angles BAD, ADE are equal e to two right angles; but BAD is a right angle; therefore also ADE is a right angle; but the opposite angles A of parallelograms are equal; there D B fore each of the opposite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB. Which was to be done. COR. Hence every parallelogram that has one right angle has all its angles right angles. PROP. XLVII. THEOR. In any right angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle. Let ABC be a right angled triangle having the right angle BAC; the square described upon the side BC is equal to the squares described upon BA, AC. e 29. 1. On BC describe the square BDEC, and on BA, AC the a 46. 1. Book I. b 31. 1. b squares GB, HC, and through A draw AL parallel to BD, or CE, and join AD, FC; then, because each of the angles BAC, BAG is a right an G angle, add to each the an- f 4. 1. g 41. 1. f e 2. Ax. angle DBA is equal to the whole FBC; and because the two sides AB, BD are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equal to the base FC, and the triangle ABD to the triangle FBC: Now the parallelogram BL is double g the triangle ABD, because they are upon the same base BD, and between the same parallels BD, AL; and the square GB is double the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. h 6. Ax. But the doubles of equals are equal to one another: Therefore the parallelogram BL is equal to the square GB: And, in the same manner, by joining AE, BK, it is demonstrated, that the parallelogram CL is equal to the square HC: Therefore the whole square BDEC is equal to the two squares GB, HC; and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC: Wherefore the square upon the side BC is equal to the squares upon the sides BA, AC. Therefore, " in any right angled triangle,” &c. Q. E. D. PROP. XLVIII. THEOR. 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. If the square described upon BC, one of the sides of the Book I. triangle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle. D From the point A draw AD at right angles to AC, and a 11. 1. make AD equal to BA, and join DC: Then because DA is equal to AB, the square of DA is equal to the square of AB. To each of these add the square of AC; therefore the squares of DA, AC, are equal to the squares of BA, AC: But the square of DC is equal to the squares of DA, AC, because DAC is a right angle; and the square of BC, by hypothesis, is equal to the squares of BA, AC; therefore the square of DC is equal to the square B b A b 47. 1. of BC; and therefore also the side DC is equal to the side BC. And because the side DA is equal to AB, and AC common to the two triangles, DAC, BAC, the two DA, AC are equal to the two BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle e 8. 1. BAC: But DAC is a right angle; therefore also BAC is a right angle. Therefore," if the square," &c. Q. E. D. THE ELEMENTS OF EUCLID. BOOK II. DEFINITIONS. I. Book II. EVERY right angled parallelogram, (or rectangle,) is said to be contained by any two of the straight lines which contain one of the right angles. II. In every parallelogram, any of the parallelograms about a diameter, together with the Thus the two complements, is call- A E gnomon, which is more HF G D briefly expressed by the • letters AGK, or EHC, B 'which are at the opposite K C 'angles of the parallelograms which make the gnomon.' PROP. I. THEOR. If there be two straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.. |