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, 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, the angles ACB and ACFequal, and the side AC common therefore, by P. 26, AF will equalAB; if now we measure AF, we have & T measurement equal to that of AB.

OR, produce AC to D, and make CD 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 tro 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.

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3. When neither Anor 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 2219; ADC 450; and ACB 452 : it is required to ascertain the distance AB.

From a scale make CD 150 ; draw the angles BCD 609, and BDA 221" ; the lines CB and DB meet in B: again draw the angles ACB 45?, and ADC 459 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.

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

AB and CD,
2 „ 2. and make the alt. l.

AEF = the alt. /

3 Conc. then the st. line AB is

ll the st. line CD.

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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 3°, 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 Solstice, the ray DS perpendicular to Syene, and continued towards É, the earth's centre. Now at A, take the angle ACG, made by

NA 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 anglos 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 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, 80 Are 360° to the measure in units of length of the whole circumference.

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