take equal things, the remainder will be equal, and the reverse in respect to unequal things. 4. The whole is greater than any of its parts. 5. Two right lines do not contain a space. 6. All the angles within a circle cannot amount to more nor less than 360 degrees, nor in a semi-circle to more nor less than 180 degrees. 7. The value, or measure, of an angle is not affected or changed by the lines whereby it is formed being either lengthened or shortened. 8. Two lines standing at an angle of 90 degrees from each other will not be affected by any change of position of the entire figure in which they meet, but will still be mutually perpendicular.

After thus much preparation, we may conclude the student to be ready to proceed in the solution of problems, which we shall study to exhibit in the most simple, as well as in a progressive manner.

PROBLEM I.

To describe an equilateral triangle upon a given line. Let A B (fig. 1.) be the given line, with an opening of your compasses equal to its length; from each end, A and B, draw the arcs CD and EF, to whose point of intersection at C draw the lines AC and B C.

PROBLEM II.

To divide an angle equally. Fig. 2. Let BAC be the given angle, measure off equal distances from A to B, and from A to C; then with the opening B C draw alternately from B and from C the arcs which intersect at D: a line drawn from A to D will bisect the angle BA C.

PROBLEM III.

To bisect a given line. Fig. 3. Let A B be the given line; from each end (or nearer, if space be wanting), with an opening of your compasses rather more than half the length of A B, describe the arcs which intersect above at C, and below at D: draw

the line CD, passing through the points of intersection, and the line A B will be divided into two equal parts. Observe, this is an easy mode of ere cting a perpendicular upon any given line.

PROBLEM IV.

To raise a perpendicular on a given point in a line. Fig. 4. With a moderate opening of your compasses, and placing one of its legs a little above or below the given

line, describe a circle passing through the given point A on the line B C; then draw a line from the place where the circle cuts at D, so as to pass through E, the centre to F on the opposite side of the circle: the line FA will be the perpendicular required.

PROBLEM V.

From a given point to let fall a perpendicular on a given line. Fig. 5. From the given point A draw the segment B C, passing under the line DE; bisect BC in F, and draw the perpendicular A F.

THEOREM VI.

The opposite angles made by intersecting lines are equal; (fig. 6.) as is shown in this figure: 0, 0, are equal; p, p, are equal; 8, 8, are equal.

PROBLEM VII.

To describe a triangle with three given lines. Fig. 7. Let A B, BC, and CD, be the three given lines; assume either of them, say A B, for a base, then with an opening equal to B C, draw the segment from the point B of the base, and with the opening CD make a segment from C: the intersection of the two segments will determine the lengths of the two lines B C and CD, and of the angle ABC.

PROBLEM VIII.

To imitate a given angle at a given point. Fig. 8. Let A B C be the given angle, and O the point on the line OD whereon it is to be imitated. Draw the line AC, and from O measure towards D with an opening equal to A B: then from O make a segment with an opening equal to B C, and from K make a segment with an opening equal to A C: their intersection at E will give the point through which a line from O will make an angle with O D equal to the angle

ABC.

THEOREM IX.

All right lines severally parallel to any given line are mutually parallel, as shown in fig. 9, where AB, CD, EF, and GH, being all parallel to I K, are all parallels to each other severally.

N. B. They all make equal angles with the oblique line O P.

PROBLEM X.

To draw a parallel through a given point. Fig. 10. From the end, on any part of the given line A B, draw an oblique line to the

given point C. Measure the angle made by ABC, and return another of equal measurement upon the line BC, so as to make the angle BCD equal to ABC: the line CD will be parallel to the line A B. Or, as in fig. 11, you may from any points, say CD, in the line A B draw two semicircles of equal dimensions; the tangent EF will be parallel to A B. Or you may, according to Problem 5, draw a perpendicular from the given point to the given line, and draw another line through the given point at right angles with the perpendicular proceeding from it to the line whose parallel was to be made, and which will be thus found. See fig. 12.

THEOREM XI.

Parallelograms of equal base and altitude are reciprocally equal. Fig. 13. The paral lelogram No. 1 is rectangular: No. 2 is inclined, so as to hang over a space equal to the length of its own base; but the line A B, which is perpendicular thereto, divides it into two equal parts: let the left half, ABE, be cut off, and it will, by being drawn up to the right, be found to fit into the dotted space ACD. This theorem might be exemplified in various modes; but we presume the above will suffice to prove its validity.

THEOREM XII.

Triangles of equal base and altitude are reciprocally equal. Fig. 14. As every parallelogram is divisible into two equal and similar triangles, it follows that the same rule answers for both those figures under the position assumed in this proposition: we have shown this by fig. 15.

PROBLEM XIII.

To make a parallelogram equal to a given triangle, with a given inclination or angle. Fig. 16. Let BAC be the given triangle, and ED F the given angle. On the line D F measure a base equal to BC, the base of the triangle. Take BG equal to half the altitude of the triangle for the altitude of the parallelogram, and set it off on the line ED. Draw F H parallel to ED, and HE parallel to DF, which will complete the parallelogram E F D H, equal to the triangle ВАС.

PROBLEM XIV.

To apply a parallelogram to a given right line, equal to a given triangle, in a given right lined figure. Fig. 17. Let A B be the given line to which the parallelogram is to be an

nexed. Let C be the triangle to be com muted, and D the given angle. Make BEFG equal to C, on the angle EBG: continue A B to E: carry on F E to K, and make its parallel HAL, bounded by FH, parallel to E A: draw the diagonal H K and GM both through the point B; then KL; and the parallelogram B M AL will be equal to the triangle C, and be situated as desired.

PROBLEM XV.

To make a parallelogram, on u given inclination, equal to a right-lined figure. Fig. 18. Let A BCD be the right-lined figure, and FKH the given angle or inclination; draw the line D B, and take its length for the altitude, FK, of the intended parallelogram, applying it to the intended base line K M: now take half the greatest diameter of the triangle DC B, and set it off from K to M, and set off half the greatest diameter of the triangle DA B, and set it off from H to M: make GH and LM parallel to FK, and FG parallel to K H. The parallelogram FKGH will be equal in area to the figure A B C D, and stand at the given inclination or angle.

PROBLEM XVI.

To describe a square on a given line. Fig. 19. Raise a perpendicular at each end of the line A B equal to its length; draw the line CD, and the square is completed.

THEOREM XVII.

The square of the hypothenuse is equal to both the squares made on the other sides of a right-angled triangle. Fig. 20. This comprehends a number of the foregoing propositions, at the same time giving a very beautiful illustration of many. Let ABC be the given right-angled triangle; on each side thereof make a square. For the sake of arithmetical proof, we have assumed three measurements for them: viz. the bypothenuse at 5, one other side at 4, and the last at 3. Now the square of 5 is 25. The square of 4 is 16, and the square of 3 is 9: it is evident the sum of the two last sides make up the sum of the hypothenuse's square; for 9 added to 16 make 25. But the mathematical solution is equally simple and certain. The squares are lettered as follow: BDCE, FG BA, and AHGK. Draw the following lines: FC, BK, AD, AL, and AE. We have already shown, that parallelograms and triangles of equal base and altitude are respectively equal.

The two sides F B, BC, are equal to the two sides A B, B D, and the angle DAB is equal FBC: the triangle ABD must therefore be equal to the angle F BC. But the parallelogram B L is double the triangle ABD. The square G B is also double the triangle FBC: consequently the parallelogram BL is equal to the square G B. The square HC in like manner is proved to be equal to the parallelogram CL, which completes the solution. Euclid, 47th of 1st Book.

PROBLEM XVIII.

To divide a line so that the rectangle contained under the whole line, and one segment, be equal to the square of the other segment. Fig. 21. On the given line AB describe the square A B C D; bisect AC in E, and with the distance EB extend AC to F, measuring from E. Make on the excess FA the square FH, and continue GH to K. The square FH will be equal to the parallelogram HD.

PROBLEM XIX.

To make a square equal to a given rightlined figure. Fig. 22. Let A be the given right-lined figure: commute it to a parallelogram, B D, as already shown (prob. 15.): add the lesser side ED to BE, so as to proceed to F: bisect BF in G, and from that point describe the semicircle BHF. Continue D E to H, which will give HE for the side of a square equal in area to the parallelogram BD, and to the original given figure A.

in C; the perpendicular DC will divide the figure into two equal and similar parts.

PROBLEM XXIII.

In a given circle to describe a triangle equiangular to a given triangle. Fig. 26. Let A B C be the circle, and D E F the triangle given. Draw the line GH, touching the circle in A: make the angle HAC equal to DEF, and G A B equal to DFE: draw BC, and the triangle BAC will be similar to the triangle DEF.

PROBLEM XXIV.

About a given circle to describe a triangle similar to a given triangle. Fig. 27. Let ABC be the given circle, and DEF the given triangle: continue the line EF both ways to G and H, and having found the centre, K, of the circle, draw a radius, K B, at pleasure; then from K make the angle BKA equal to D E C, and BKC equal to DFH; the tangents LN perpendicular to KC, MN perpendicular to K B, and ML perpendicular to KA, will form the required triangle.

PROBLEM XXV.

To describe a circle about a given triangle. Fig. 28. In the given triangle ABC, bisect any two of the angles; the intersection of their dividing lines, BD and CD, will give the centre D, whence a circle may be, described about the triangle, with the radius DC.

PROBLEM XXVI.

PROBLEM XXIX.

To describe a circle within a given square. Fig. 31. Divide the square into four equal parts, by the lines A C, BD, whose intersection at E, shows the centre of a circle to be drawn with any one of the converging lines, say EA, as a radius.

PROBLEM XXX.

To describe a square on a given circle.` Fig. 31. Divide the circle into four equal parts, (or quadrants) by the lines A C, BD; draw the tangents GH, FK, parallel to A C, and G F, HK, parallel to BD; which will give the required square.

PROBLEM XXXI.

To make an isosceles triangle, having each of the angles at the base double that at the summit. Fig. 32. Cut any given line, as A B, into extreme and mean proportions, (as in Problem 18); then, from A, as the centre, draw a circle B D E, with the opening A B, and apply the line BD within its circumference, equal to AC, the greater portion of A B; join CD, ABD will be the isosceles triangle sought.

PROBLEM XXXII.

To describe a regular pentagon. Fig. 33. Make the isosceles triangle ACD within the circle ABCDE; the base CD will give the fifth part of the circumference.

PROBLEM XXXIII.

To describe a regular pentagon about a circle. Fig. 33. This is done by drawing parallels to the lines AB, BC, CD, DE, EA; making them all tangents to the circle; on the same principle, a square, a hexagon, &c. may be drawn around a circle, from a similar figure inscribed within it.

PROBLEM XXXIV.

To describe a circle around a pentagon. Fig. 33. Bisect any two angles of a pentagon, and take their point of intersection, G, as a centre, using either of the converging lines, DG, or E G, for a radius. Where a circle is to be described within a pentagon, you must bisect any two of the faces, and raise perpendiculars at those points, which will meet in the centre either of the converging lines serving for a radius.

PROBLEM XXXV.

To inscribe a regular hexagon within a circle. Fig. 34. The radius of a circle being equal to one-sixth of its circumference, es

To change a pentagon into a triangle. Fig. 36. Continue the base line A B to C, and from the centre D let a perpendicular fall on A B, bisecting it in E. Measure from B a space equal to four times E B. Through the centre D draw D F, parallel and equal to EC; draw FC: the parallelogram contained under ECDF will equal the area of the pentagon. Or the pentagon may be changed to a triangle by adding to A B four times its own length, and drawing a line from the centre, to the produced termination of A B; the angle at the centre would then be obtuse.

PROBLEM XXXIX.

To draw a spiral line from a given point. Fig. S7. Draw the line A B through the given point C, and from C draw the semicircle D E, then shift to D for a centre, and make the semi-circle A E in the opposite side of the line: shift again from D to C for a centre, and draw the semi-circle FG; and then continue to change the centres alternately, for any number of folds you may require; the centre C serving for all above, the centre D for all below, the line A B.

With respect to the application of geometry to its pristine intent, namely, the mea

to SURVEYING; under which head it will be found practically exemplified. We trust sufficient has been here said to show the utility and purposes of this important science, and to prove serviceable to such persons as may not have occasion for deep research, or for extensive detail.

GEORGIC, a poetical composition upon the subject of husbandry, containing rules therein, put into a pleasing dress, and set off with all the beauties and embellishments of poetry.

GEORGINA, in botany, a genus of the Syngenesia Superflua class and order. Receptacle chaffy; no down; calyx double; the outer many-leaved; inner oneleaved, eight-parted. There are three species.

GERANIUM, in botany, crane's bill, a genus of the Monadelphia Decandria class and order. Natural order of Gruinales. Gerania, Jussieu. Essential character: calyx five-leaved; corolla five-petalled, regular; nectary five honied glands, fastened to the base of the longer filaments; fruit five-grained, beaked; beaks simple, naked, neither spiral nor bearded. There are thirty-two species.

GERARDIA, in botany, so called in honour of John Gerarde, our old English botanist, a genus of the Didynamia Angiospermia class and order. Natural order of Personatæ. Scrophulariæ, Jussieu. Es sential character: calyx five-cleft; corolla two-lipped, lower lip three-parted, the lobes emarginate, the middle segments twoparted; capsule two-celled, gaping. There are ten species.

GERMINATION. When a seed is placed in a situation favourable to vegetation, it very soon changes its appearance; the radicle is converted into a root, and sinks into the earth; the plumula rises above the earth, and becomes the trunk or stem. When these changes take place, the seed is said to germinate; the process itself has been called germination, which does not depend upon the seed alone; something external must affect it. Seeds do not germinate equally and indifferently in all places and seasons, they require moisture and a certain degree of heat, and every species of plant seems to have a degree of heat peculiar to itself, at which its seeds begin to germinate; air also is necessary to the germination of seeds; it is for want of air that seeds which are buried at a very great depth in the earth, either thrive but indif.

quently preserve, however, their germinating virtues for many years within the bowels of the earth; and it is not unusual, upon a piece of ground being newly dug to a considerable depth, to observe it soon after covered with several plants which had not been seen there in the memory of man. Were this precaution frequently repeated, it would perhaps be the means of recovering certain species of plants which are regarded as lost; or which, perhaps, never coming to the knowledge of botanists, might hence appear the result of a new creation. Light is supposed to be injurious to the process which affords a reason for covering seeds with the soil in which they are to grow, and for carrying on the business of malting in darkened apartments; malting being nothing more than germination, conducted with a particular view.

GEROPOGON, in botany, a genus of the Syngenesia Polygamia Equalis class and order. Natural order of Compositæ Semiflosculosæ, or compound flowers, with semi-florets or ligulate florets only. Cicho. raceæ, Jussieu. Essential character: calyx simple; receptacle with bristle-shaped chaffs; seeds of the disk, with a feathered down of the ray, with five awns. There are three species.

GESNERIA, in botany, so named in honour of Conrad Gesner, of Zurich, the famous botanist and natural historian, a, genus of the Didynamia Angiospermia class and order. Natural order of Personatæ. Campanulaceæ, Jussieu. Essential character: calyx five-cleft, sitting on the germ; corolla incurved and recurved; capsule inferior, two-celled. There are twelve species.

GETHYLLIS, in botany, a genus of the Hexandria Monogynia class and order. Natural order of Spathacea. Narcissi, Jussieu. Essential character: calyx none; corolla six-parted; berry club-shaped, radicle, one-celled. There are four species.

GEUM, in botany, English avens, or herb bennet, a genus of the Icosandria Po lygynia class and order. Natural order of Senticosa. Rosacea, Jussieu. Essential character: calyx ten-cleft; petals five; seeds with a kneed awn. There are nine species, natives of Europe and North America.

GHINIA, in botany, so named in memory of Lucas Ghini, a famous physician and botanist of Bologna, a genus of the Diandria Monogynia class and order

Na