B E XXXIX.-Equal triangles on the same, base and on Let there be equal triangles ABC, DEF. for, if it is not, through A draw A D E but the triangle ABC is equal to DEF: therefore also the triangle DEF is equal to GEF, the greater to the less, which is impossible; therefore AG is not parallel to BF: and no other line is parallel to it but AD; AD is therefore parallel to BF. XLI.-If a parallelogram and triangle be upon the same base, and between the same parallels, the parallelogram shall be double of the triangle. Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE. Join AC; then the A triangle ABC is equal to the triangle EBC. But the parallelogram ABCD is double of the triangle ABC, because the diameter AC divides it into two equal parts; wherefore ABCD is also double of EBC. B DE XLII. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given rectilineal angle. Bisect BC in E, join AE, and at the point E in the straight line EC make the angle CEF equal to D; and through A draw AG parallel to EC, and through C draw CG parallel to EF: therefore FECG is a parellelogram: and because BE is equal to EC the triangle ABE is likewise equal to the triangle AEC, since they are upon equal bases, BE, EC, and between the same parallels BC, AG: therefore the triangle ABC is double of AEC. And the parallelogram FECG B A F G N E is likewise double of AEC; therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the given angle D. XLIII.—The complements about the diameter of any parallelogram, are equal to one another. Let ABCD be a parallelogram, of which the diameter is AC, and EH, FG, the parallelograms about AC, and BK, KD, the other parallelograms which make up the whole figure ABCD, called the A H complements: BK is equal to KD. D E K F B G Because ABCD is a parallelogram, the triangle ABC is equal to ADC: and because EKHA is a parallelogram, the triangle AEK is equal to AHK; so the triangle KGC is equal to KFC: then, because the triangle AEK is equal to AHK, and the triangle KGC to KFC; AEK together with KGC is equal to AHK together with KFC: but ABC is equal to ADC; therefore BK is equal to KD. XLIV.-To a given straight line to apply a parallelogram, equal to a given triangle, with an angle equal to a given rectilineal angle. K F E D G C B M H A Let AB be the line, C the triangle, and D the angle. Make the parallelogram BEFG equal to C, having EBG equal to 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, and join HB. Because HF falls on the parallels AH, EF, AHF, HFE, are together equal to two right angles; so BHF, HFE, are less than two right angles: but HB, FE, meet if produced; let them meet at K; through K draw KL, parallel to EA or FH, and produce HA, GB, to L, M: then HLKF is a parallelogram; the diameter is HK; and AG, ME, are the parallelograms about HK; and LB, BF, the complements; therefore LB is equal to BF; but BF is equal to C; wherefore LB is equal to C; and because the angle GBE is equal to ABM, and to D, ABM is equal to D: therefore LB is applied to AB, is equal to C, and has ABM equal to D. XLV. 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 figure, and E the angle. Join DB, and describe the parallelogram FH equal to the triangle ADB, having the angle HKF equal to E; and to GH apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to E; and because E is equal to each of FKH, GHM, FKH is equal to GHM: add to each KHG; therefore the angles FKH, KHG, are equal to KHG, GHM; but FKH, KHG, are equal to two right angles; therefore also KHG, GHM, are equal to two right angles; and because at H in GH, KH, HM, on the opposite sides of it make the adjacent angles equal to two right angles, KH is in the same straight line with HM, and because HG meets the parallels KM, FG, the angles MHG, HGF, are equal: add to each HGL: then MHG, HGL, are equal to HGF, HGL; but MHG, HGL, are equal to two right angles; wherefore HGF, HGL, F G 1. B ск н м are equal to two right angles, and FG is in the same straight line with GL; 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 HF, and DBC to GM, the whole ABCD is equal to the whole KFLM; therefore KFLM has been made equal to ABCD, having FKM equal to the given angle E. XLVI.-To describe a square upon a given straight line. Let AB be the given straight line. From A draw AC at right angles to AB; and make AD equal to AB, and through D draw DE parallel to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram; and AB is equal to DE, and AD to BE, but BA is equal to AD; therefore BA, AD, DE, EB, are all equal, and ADEB is equilateral, and all its angles are right angles; because AD meeting the parallels AB, DE, the angles BAD, ADE, are equal to R two right angles; but BAD is a right anglo; therefore also ADE is a right angle; but each of the opposite angles ABE, BED, is a right angle; wherefore ADEB is rectangular, and it is equilateral; it is therefore a square, and is on AB. XLVII. In any right-angled triangle, the square on the side subtending the right angle is equal to the squares on the sides which contain the right angle. Let ABC be a right-angled triangle having the right angle BAC. On BC describe the square BDEC, and on BA, AC, the squares GB, HC; through A. draw AL parallel to BD, or CE, and join AD, FC; because each of BAC, G H K BAG, is a right angle AC, AG, make the adjacent angles equal two right angles; CA is in the same straight line with AG; so AB and AH are in the same F straight line; and because the angle DBC is equal to FBA, the whole DBA is equal to FBC; and because AB, BD, are equal to FB, BC, and the angle DBA to FBC; AD is equal to FC, and the triangle ABD B to FBC; now BL is double of ABD, and GB is double of FBC, on the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal: therefore BL is equal to GB; and so by joining AE, BK, it is shown that CL is equal to HC; therefore the whole square BDEC is equal to the two squares GB, HC; and BDEC is on BC, and the squares GB, HC, upon BA, AC; wherefore the square où BC is equal to the squares on BA, AC. XLVIII.—If the squares on one side of a triangle equal the squares on the other two sides, these two sides contain a right angle. B A D If the square BC, one of the sides of ABC, be equal to the squares on BA, AC, BAC, is a right angle. From A draw AD at right angles to AC, and, make AD equal to BA, and join DC; because DA is equal to AB, the square of DA is equal to the square of AB: to each 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; and the square of BC is equal to the squares of BA, AC; therefore the square of DC is equal to the square of BC;and DC is equal to BC; because DA is equal to AB, and AC common, DA, AC, are equal to BA, AC; and DC is equal to BC; therefore the angle DAC is equal to BAC, but DAC is a right angle ; therefore also BAC is a right angle. |