Book I. having one of its angles CEF equal to the given angle D. Which was to be done. PROP. XLIII. THEOR. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another. Α H D K E F Let ABCD be a parallelogram, of which the diameter is AC, let EH, FG be the parallelograms about AC, that is, through which AC passes, and let BK, KD be the other parallelograms which make up the whole figure ABCD, which are therefore called the complements: the complement BK is equal to the complement KD. Because ABCD is a pa B G C a 34. 1. rallelogram, and AC its diameter, the triangle ABC is equal a to the triangle ADC: and because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK: By the same reason, the triangle KGC is equal to the triangle KFC: Then, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to KFC; the triangle AEK, together with the triangle KGC is equal to the triangle AHK together with the triangle KFC: But the whole triangle ABC is equal to the whole ADC; therefore the remaining complement BK is equal to the remaining complement KD. Wherefore, the complements," &c. Q. E. D. PROP. XLIV. PROB. To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle. It is required to apply to the straight line AB, a parallelogram equal to the triangle C, and having an angle equal to D. straight line H A with AB, and produce FG to H; and through A draw AH b 31. 1. parallel to BG or EF, and join HB. Then, because the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE are together equal to two right angles; wherefore the c 29. 1. angles BHF, HFE are less than two right angles: But straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet if produced: Therefore HB, FE, will meet if pro-d 12. Ax. duced; let them meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB to the points L, M: Then HLKF is a parallelogram, of which the diameter is HK, and AG, ME are the parallelograms about HK; and LB, BF are the complements; therefore LB is equal to BF: But BF is equal to the triangle C; where- e 43. 1. fore LB is equal to the triangle C; and because the angle GBE is equal to the angle ABM, and likewise to the angle f 15. 1. D; the angle ABM is equal to the angle D: Therefore the parallelogram LB, which is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. Which was to be done. c PROP. XLV. PROB. To describe a parallelogram equal to a given rec- See N. tilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. a Join DB, and describe the parallelogram FH equal to the a 42. 1. triangle ADB, and having the angle HKF equal to the angle E; b and to the straight line GH apply the parallelogram GM equal b 44. 1. 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 to each of these the angle KHG; therefore the angles FKH, KHG are equal c 29. 1. to the angles are equal to two A D F GL E are equal to two 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. c 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. b 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. On BC describe the square BDEC, and on BA, AC the e 29. 1. a 46. 1. Book I. b 31. 1. 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 C c 30. Def. gle, the two straight lines d 14. 1. is in the same straight line d f 4. 1. G 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 s 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. |