Sch.-1. This is the last of the three criteria for inferring the equality of triangles; the first criterion, P. 4, is—that the two sides and the included angle of each triangle be equal; the second, P. 8, that the three sides of one triangle equal the three sides of the other; and the third, P. 26, that in each triangle two angles be equal and one side, either adjacent to the equal angles, or opposite to one of them. 2. Of the six parts of a triangle any three, except the three angles, being given equal to one another, the equality of the other parts may be readily proved. APP.-1. The discovery of this Proposition is ascribed to THALES, who made use of it for measuring inaccessible distances, as from A to B. At A set out a line AC perpendicular to AB; and at C observe the angle ACB, F A B and by P. 23 make the ang. ACF equal to ang. ACB; the lines BA and CF produced meet in the point F; and AF will be found equal to AB. For in triangles ACB, ACF, by construction, we have the angles at A equal, C the angles ACB and ACFequal, and the side AC common therefore, by P. 26, AF will equalAB; if now we measure AF, we have a measurement equal to that of AB. E Or, produce AC to D, and make CD D equal to CA; then at D draw DE perp. to AD, and produce BC, till it cuts DE in E; in this case DE may be proved equal to the distance from A to B. 2. By the theory of Representative Values the distance of two stations A and B may be found. Suppose AC to measure 70 feet or yards, ang. A 909, ang. ACB 66°; at A make a right angle, and on AC set of 70 equal parts ; at C draw an angle of 66°; the line CB will cut off B, the distance from A: let the distance from A to B be applied to the same scale, and AB will be found equal to about 150 feet or yards, according to the unit of length fixed on for AC. 3. When neither A nor B is accessible, a process rather different has to be employed; select two stations C and D, from both of which A and B are visible; and measure the line CD, suppose 150 ; take the angles BCD 60?, and BDA 2279; ADC 450; and ACB 45? : it is required to ascertain the distance AB. From a scale make CD 150 ; draw the angles BCD 60%, and BDA 22}"; the lines CB and DB meet in B: again draw the angles ACB 45?, and ADC 45° the lines DA and CA meet in A: take the distance AB in the compasses, and apply it to the same scale; AB will be found to be about 158. с PROP. 27.-THEOR. If a st. line falling upon two other st. lines makes the alternate angles equal to one another; these two lines shall be parallel. Con.-Pst. 2.- DEM.-P. 16, Def. 35. E, 1 Hyp. 1. , Let the st. line EF fall on the two st. lines A E B L. F | the st. line CD. EFD ; C. 1 Sap. If AB and CD are not ll, 2 Def. 35. on being produced they will meet towards A, C, or B, D; 3.Pst.2 & Sup let them be produced towards B and D, and, if possible, meet in the · G, and form the A EFG. D. 1 byC.3&P16 ::: GEF is a A, .. the ext. 2 AEF > the int. and opp. _ EFG ; when produced towards A,C: Q.E.D. Let EH fall on AB, CD, and make the alt. Zs B 2 Conc. then the lines AB and CD F are parallel. C. 1 by P. 12. At G draw GA I AB; 2 P.3, Pst. 1 take GD= AF,& join F,D. C D. I by C. 2. In As AGF, DFG, G D • GD=AF, GF common, H. 2 Нур. and / AFG = FGD; 3 P. 4. AG = DF, L GDF= L GAF: 4 C. 1. But / GAF is a rt. angle, 5 Def. 10. GDF is a rt. ang., and DF perp. to CD. 6 Нур. Moreover the par. to CD is to be drawn from F; 7 D. 3. and since the perpendiculars AG and DF are equal, 8 Ax. 12, n. the parallel to CD must pass through A : 9 Conc. AB is parallel to CD. Son.—Of the angles which two st. lines, one at each extremity of a third line, make with it, the alternate angles are those which are on opposite sides and at opposite extremities of the third line. PROP. 28. THEOR. If a st. line falling upon two other st. lines makes the exterior angle equal to the interior and opposite angle upon the same side of the line; or makes the int. angles upon the same side together equal to two rt. angles; the two st. lines shall be parallel. DEM.-P. 15, Ax. 1, P. 27, P. 13, Ax. 3. B 2. make the ext. 2 EGB the int. and opp. L GHD; H. D + GHD = two right F angles ; 21 Conc. then the st line AB is || the st. line CD. CD: D. 1 by Hyp. 2. For, :: _ EGB = L GHD, 2 P. 15. and L EGB = L AGH; 3 Ax. 1. .. the alt. L AGH = the alt. LGHD 4 P. 27. and ... also st. line AB is || CD. 5 Hyp. 3. Again, :: Ls BGH + GĦD = two rt. angles, 6 P. 13. and Z8 AGH + BGH = two rt. angles ; 7 Ax. 1. i. ZS AGH + BGH = Ls BGH + GÜD: 8 Sub. Take away the common _ BGH, 9 Ax. 3. the rem. L AGH = the rem. [ GHD: 10. Cons. But Zs AGH and GHD are alternate angles ; 11 P. 27. the st. line AB is || the st. line CD. 12 Rec. Therefore, if a st. line falling upon two other, &c. Q.E.D. SCH.-1, When one st. Jine cats, or falls upon, two other st. lines,-of the four angles formed on each side of the incident line, taken in succession, each pair consists of an ext. and an int. adjacent angle; but taken alternately, each pair consists of an ext. and an int. opposite angle. 2. The principle really assumed in the twelfth axiom is,-that two st. lines which intersect each other cannot both be parallel to the same st. line. PROP. 29.-THEOR.—(Converse of the 27th, and the 28th.) If a line fall upon two parallel st. lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite angle upon the same side ; and likewise the two interior angles upon the same side together equal to two rt, angles. DEM.-Ax. 4, P.13, Ax. 12, Def 35, P. 15, Ax. 1. Let the st. line EF fall E 2 Conc. 1. then the alt. [ AGH A B = the alt. L GHD; G 2. the ext. 2 EGB = the int. and opp. LGHD; с D side = two rt. angles. \F D. 1 by Sup. 1. If _ AGHis# LGHD, one is > the other : 2. Let L AGH be > L GHD; 3 Add. and to each angle let the _ BGH be added ; 4 Ax. 4, then _s AGH + BGH are > Z8 BGH + GHD 5 P. 13. But ZS AGH + BGH are = two rt. angles ; 6 Ax. 1. Zs BGH + GHD are < two rt. angles ; 7) Ax. 12. Now on this condition the st. lines AB and CD, being produced, will meet 8 Hyp. Def.35 but :: AB is parallel to CD, they never meet ; 9 Conc. 1. .. LAGH is not #, i.e., is = L GHD : 10 P.15&Ax.1 But _ AGH LEGB, . . L EGB = L GHD; 11 Add. to each angle add _ BGH ; 12 Ax. 2. then _s EGB + BGH = Zs GHD + BGH: 13 P. 13. But Z8 EGB + BGH = two rt. angles. 14 Ax. 1. .. Ls GHD + BGH= two rt. angles. 15 Rec. Wherefore, if a line fall on two parallel st. lines, &c. Q.E.D. 2 SCH-1. It has been objected to the twelfth axiom that it is not self-evident, and that it forms the converse to P. 27 : now, both the assumed axiom and its converse should be so obvious as not to require demonstration. 2. "The Twelfth Aciom may be expressed in any of the following ways: -Two diverging rt. lines cannot be both parallel to the same rt. line ; or, If a rt. line intersect one of two parallel rt. lines, it must also intersect the other : : or, Only one rt. line can be drawn through a given point parallel to a given rt. line."-LARDNER, P. 316. APP.-ERATOSTHENES, who died B. C. 196, at the age of 80, applied the last three Propositions to the measuring of the circumference of the earth, and followed the method still employed. (See Dictionary of Greek and Roman Biography, vol. ii., p. 44-47.) The principle with which he set out was, that two rays of light proceeding from the centre of the sun to two points on the earth, are physically parallel. He learned that “at Syene, in Upper Egypt, deep wells were enlightened to the bottom on the day of the summer soistice, and that vertical objects cast no shadows. He concluded, therefore, that Syene was on the tropic, and its latitude equal to the obliquity of the ecliptic, which he had determined to be 23° 51' 20": he presumed that it was in the same longitude as Alexandria, in which he was out about 30, which is not enough to produce what would at that time have been a sensible error. By observations made at Alexandria, he determined the zenith of that place to be distant by the fiftieth part of the circumference from the solstice; which was equivalent to saying that the arc of the meridian between the two places is 7° 12'.” The distance from Alexandria to Syene, Eratosthenes gives as 5000 stadia. From these data it is a very simple operation to deduce the earth's circumference; for if an arc of 7° 12' measures 5000 stadia, the question to be solved, is, -How many stadia are in 360°? For the demonstration of the process, we suppose the circumference GSH to be representative of the earth’s circumference ; Alexandria to be at the point A, and Syene at the point S; at A the style AC is erected perpendicular to the horizon : DS and EC are rays of light from the sun when in the E Solstice, the ray DS perpendicular to Syene, C and continued towards F, the earth's centre. Now at A, take the angle ACG, made by AC the perpendicular to the horizon, and by the ray EG; then, because the rays DF and EG are parallel, and the line CF meets them, the alternate angles ACG and SFA are equal. Thus, if we measure the one, we ascertain the other. The arc which measures angle SFA is SA; and if the actual distance F in miles from S to A be measured, we have the elements for finding the earth's circumference; for as the degrees in the Arc AS are to the distance measured from A to S, so Are 360° to the measure in units of length H of the whole circumference. G PROP. 30.-THEOR.(similar to Ax. 1.) Straight lines which are parallel to the same st. line are parallel to eucha other. DEM.-P. 29, Ax. 1, P. 27. Let the st. lines AB & G 2 Conc, then AB shall be also í A -B · to CD. H C. hy Sup. Let the st. line GHK cut E F the lines AB, EF, & KI C the Is AB, EF, 2 P. 29. ... the LAGH = alt. Z GHF; 3 C. & Hyp And ::: GHK also cuts the lis EF, CD, 41 P. 29, ... the ext. GHF = int. opp. GKD: 5 D. 2. But L AGH or AGK = L GHF; 6) Ax. 1. .. alt L AGK = alt. _ GKD ; 7 P. 27. and . , AB is || to CD. 8) Rec. Wherefore, st. lines which are parallel, &c. Q.E.D. SoH.-The corollary to this proposition, that two lines parallel to the same line cannot pass through the same point, is equivalent to the twelfth axiom. PROP. 31.-PROB. Sol.-Pst. 1, P. 23, Pst. 2.-DEM.--P. 27. F 2 Quæs. to draw through • A a st. line || to BC. C. i| by Sup. In BC take any · D, and B C D & Pst. 1. join A, D ; 2 P. 23. At · A in AD make the _ DAE= | ADC; 3 Pst. 2. and produce EA to F; 4 Sol. then EF, through A, is || to BC. D. 1 by C. 1, 2. ::: st. line AD meets two st. lines EF and BC, and makes alt / EAD = alt. L ADC, 2 P. 27. .:. EF is || BC. 3) Rec. Wherefore, through a given point, &c. Q.E.F. Sch. The Twelfth Axiom may very readily be demonstrated by the principles nou established. E. 11 Hyp. 1 Let AC falling on AB, CD, make the int. _8 ACD, E ; В D H 3) Conc. 1. then AB and CD being C 2 G produced shall intersect, F 2. and AE and CD shall be parallele. >> |