Again, because EKHA is a parallelogram, and AK its diameter, therefore the triangle AEK is equal to the triangle AHK; (1. 34.) and for the same reason, the triangle KGC is equal to the triangle KFC. Wherefore the two triangles AEK, KGC are equal to the two triangles AHK, KFC, (ax. 2.) but the whole triangle ABC is equal to the whole triangle ADC; therefore the remaining complement BK is equal to the remaining complement KD. (ax. 3.) Wherefore the complements, &c. Q.E.D. PROPOSITION XLIV. PROBLEM. 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. Make the parallelogram BEFG equal to the triangle C, (1. 42.) and having the angle EBG equal to the angle D, so that BE be in the same straight line with AB; produce FG to H, through A draw AH parallel to BG or EF, (1. 31.) and join HB. Then because the straight line HF falls upon the parallels AH, EF, therefore the angles AHF, HFE are together equal to two right angles; (1. 29.) wherefore the angles BHF, HFE are less than two right angles: but straight lines which with another straight line make the two interior angles upon the same side less than two right angles, do meet if produced far enough: (ax. 12.) therefore HB, FE shall meet, if produced; let them be produced and meet in K, through K draw KL parallel to EA or FH, and produce HA, GB to meet KL in the points L, M. Then HLKF is a parallelogram, of which the diameter is HK; and AG, ME, are the parallelograms about HK; also LB, BF are the complements; therefore the complement LB is equal to the complement BF; (1. 43.) but the complement BF is equal to the triangle C; (constr.) wherefore LB is equal to the triangle C. And because the angle GBE is equal to the angle ABM, (1. 15.) and likewise to the angle D; (constr.) therefore the angle ABM is equal to the angle D. (ax. 1.) Therefore to the straight line AB, the parallelogram LB is applied, equal to the triangle C, and having the angle ABM equal to the angle D. Q.E.F. PROPOSITION XLV. PROBLEM. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rec◄ tilineal angle. It is required to describe a parallelogram that shall be equal to the figure ABCD, and having an angle equal to E. Describe the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to the angle E; (1. 42.) to the straight line GH, apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. (1. 44.) Then the figure FKML shall be the parallelogram required. Because the angle E is equal to each of the angles FKH, GHM, therefore the angle FKH is equal to the angle GHM; add to each of these equals the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal to two right angles; (1. 29.) therefore also KHG, GĤM 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 it, make the adjacent angles equal to two right angles, therefore HK is in the same straight line with HM. (1.14.) And because the line HG meets the parallel KM, FG, therefore the angle MHG is equal to the alternate angle HGF; (1.29.) add to each of these equals 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; (1.29.) therefore also the angles HGF, HGL are equal to two right angles, and therefore FG is in the same straight line with GL. (1. 14.) And because KF is parallel to HG, and HG to ML, therefore KF is parallel to ML; (1. 30.) and KM has been proved parallel to FL, wherefore the figure FKML is a parallelogram; and since the triangle ABD is equal to the parallelogram HF, and the triangle BDC to the parallelogram GM; therefore 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. Q. E. F. 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 figure; viz. by applying to the given straight line a parallelogram equal to the first triangle ABD, (1.44.) and having an angle equal to the given angle. PROPOSITION XLVI. PROBLEM. 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. From the point A draw AC at right angles to AB; (1. 11.) through the point D draw DE parallel to AB, and through B, draw BE parallel to AD; (1.31). whence AB is equal to DE, and AD to BE; (1.34.) therefore the four lines BA, AD, DE, EB are equal to one another, therefore the angles BAD, ADE are equal to two right angles; (1.29.) but BAD is a right angle; (constr.) therefore also ADE is a right angle. But the opposite angles of parallelograms are equal; (1.34.). therefore each of the opposite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular, and it has been proved to be equilateral; therefore the figure ADED is a square, (def. 30.) and it is described upon the given straight line AB. Q. E. F. COR. Hence, every parallelogram that has one right angle, has all its angles right angles. PROPOSITION XLVII. THEOREM. 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. Then the square described upon the side BC, shall be equal to the squares described upon BA, AC On BC describe the square BDEC, (1.46.) Then because the angle BAC is a right angle, (hyp.) the two straight lines AC, AG upon the opposite sides of AB, make add to each of these equals the angle ABC, therefore the whole angle DBA is equal to the whole angle FBC. (ax.2.) And because the two sides AB, BD, are equal to the two sides FB, BC, each to each, and the included angle ABD is equal to the included angle FBC, therefore the base AD is equal to the base FC, (1. 4.) and the triangle ABD to the triangle FBC. Now the parallelogram BL is double of the triangle ABD, (1.41.) because they are upon the same base BD, and between the same parallels BD, AL; also the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal to one another; (ax. 6.) therefore the parallelogram BL is equal to the square GB. Similarly, by joining AE, BK, it can be proved, that the parallelogram CL is equal to the square HC. Therefore the whole square BDEC is equal to the two squares GB, HC; (ax. 2.) and the square BEDC is described upon the straight line BC, and the squares GB, HC, upon AB, ‘AC: therefore the square upon the side BC is equal to the squares upon the sides AB, AC. Therefore, in any right-angled triangle, &c. Q. E. D. PROPOSITION XLVIII. THEOREM. 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 the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other two sides AB, AC. Then the angle BAC shall be a right angle. D B From the point A draw AD at right angles to AC, make AD equal to AB, and join DC. (1. 11.) therefore the square of AD is equal to the square of AB; therefore the squares of AD, AC are equal to the squares of AB, AC: but the squares of AD, AC are equal to the square of DC, (1.47.) because the angle 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 of BC; and therefore the side DC is equal to the side BC. And because the side AD is equal to the side AB, and AC is common to the two triangles DAC, BAĆ; the two sides DA, AC, are equal to the two BA, AC, each to each; and the base DC has been proved to be equal to the base BC; therefore the angle DAC is equal to the angle BAC; (1. 8.) but DAC is a right angle; therefore also BAC is a right angle. Therefore, if the square described upon, &c. Q. E. D. |