take 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 giren 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 C D and EF, to whose point of intersection at C draw the lines A PROBLEM II. - To divide an angle equally. Fig. 2. Let B A C 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 BAC. PROBLEM III. Tb bisect a given line. Fig. 3. Let A B be the given I me ; 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 C D, 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 er« 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 compatses, 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 K, the centre to F on the opposite side of the circle: the line F A will be the perpendicular required. PROBLEM v. From a given point to let fail a perpendicular on a given line. Fig. 5. From the given point A draw the segment B C, passing under the line D E; bisect B C 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: o, o, are equal; p, p, are equalj s, s, are equal. PROBLEM VII. To describe a triangle with three giren lines. Fig. 7. Let A B, B C, and C D, 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 C D, 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 O D whereon it is to be imitated. Draw the line A C, and from O measure towards D with an opening equal to A B : then from 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 B C, so as to make the angle BCD equal to ABC: the line C D will be parallel to the line A B. Or, as in fig. 11, you may from any points, say C D, 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 XT. Parallelograms of equal base and altitude are reciprocally equal. Fig. 13. The parallelogram 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 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 B A C he the given triangle, 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 B E F G equal to C, on the angle E B G: continue A B to E: carry on F E to K, and make its parallel HAL, bounded by F H, parallel to E A: draw the diagonal H K and G M both through the point B; then K L; and the parallelogram B M A L will be equal to the triangle C, and be situated as desired. PROBLEM XV. To make a parallelogram, on a given inclination, equal to a right-lined figure. Fig. 18. Let A B C D be the right-lined figure, and F K H 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 D C B, and set it off from K to M, and set off half the greatest diameter of the triangle DAB, and set it off from H to M: make G H and L M parallel to F K, and F G 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 sq uare on a given line. Fig. 19. Raise a perpendicular at each end of the line A B equal to its length; draw the line C D, and the square is completed. THEOREM XVII. The square of the hypothermic 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 hypothenuse 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'i square; for 9 added to 16 make are respectively equal.
The two sides F B, B C, are equal to the two sides A B. B D, and the angle DAB is equal F B C: the triangle A B D must therefore be equal to the angle F B C But the parallelogram B L' is double the triangle A B D. The square G B is also double the triangle F p, C: consequently the parallelogram B L is equal to the square G B. The square H C in like manner is proved to be equal to the parallelogram C L, which completes the solution. Euclid, 47th of 1st Book.PROBLEM XVIII. To divide a and continue G H to K. The square F H will be equal to the parallelogram H D.PROBLEM XIX. To make a square equal to a giren 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 PROBLEM XX. To find the centre of a given circle. Fig. 23. Draw at pleasure the chord A B, bisect it in D by means of a diameter, which being bisected will give F for the centre of the circle. PROBLEM XXI. To complete a circle upon a I draw the line A C, and bisect it in D ; draw also the perpendicular B E through D, draw B A, and on it make the angle B A E, equal to D B A; this will give the point of intersection E for the centre, whence the rircle may be completed. It matters not whether the segment be more or less than a semicircle.PROBLEM XXII. To cut a given circumference into two equal 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. PROBLEM XXIV. About a giren circle to describe a triangle similar to a 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, B D and C D, will give the centre D, whence a circle may be described about the triangle, with the radius DC PROBLEM XXVI. To inscribe a circle in a given triangle. Fig. 29. In the triangle ABC, divide the angles ABC, and B C A, equally by the lines B D, CD. Their junction at D, will give a point whence the circle E C F may be described, with the radius D F perpendicular to B C. PROBLEM XXVII. To inscribe a square in a given circle. Fig. 30. Draw the diameter AC, and, perpendicular thereto, the diameter B D: the lines A B, B C, C D, and D A, will form a correct square. PROBLEM XXVIII. To describe a circle around a square. Fig. 30. In the square ABCD, draw the diagonals A C, B D, their intersection at E will give the centre of a circle, whose radius may be any one of the four converging lines; say EA, that will enclose the square. PROBLEM XXIX. To describe a circle within a given square. Fig. 31. Divide the square into four equal 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, 15 D; draw the tangents GH, FK, parallel to A C, and GF.HK, parallel loBl); 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. St. 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 B D within its circumference, equal to A C, the greater portion of AB ; 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, 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 PROBLEM XXXV. To inscribe a regular hexagon within a circle. Fig. 34. The radius of a circle being equal to one-sixth tablishes a very easy mode of setting off the six sides as follows: draw the diameter AB, set one leg of your compasses at A, and draw the segment D F, and from B draw the segment CE; thus dividing the circle into six equal portions; draw lines joining them, and the figure will be complete. PROBLEM XXXVI. To form a quindecagon, or figure of 15 equal sides, within a circle. An equilateral triangle being inscribed within a circle, by assuming the distance between three points of a hexagon, say from A to C in the last figure for a side, let one point of such triangle be applied to each angle of a pentagon in succession ; its two other points will divide the opposite sides in three equal parts, as the figure changes place within the pentagon. PROBLEM WWII. To change a circle to a triangle. Fig. 35. Draw the tangent A B equal to 3$ diameters A D of the circle, and from the centre C draw C B, and C A: the triangle CAB will be equal in contents to the circle A D. PROBLEM XXXVIII. 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 E C; draw F C: 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 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 GERANIUM, in botany, GERARDIA, in botany, so called in honour of John Gerarde, our old English botanist, a genus of the Didynamia Angio.'permia class and order. Natural order of Personatse. Scrnphularia;, Jussieu. Essential 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 isplaced 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 ferently, or do not rise at all. GEROPOGON, in botany, a genus of the Syngenesia Polygamia jEqualis class and order. Natural order of Composite Semiflosculoss, or compound flowers, with semi-florets or ligtilate florets only. Cichoraces, Jussieu. Essential character: calyx simple; receptacle with bristle shaped 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 Personate. Campanulacea*, Jussieu. Essential character: calyx five-cleft, sitting on the germ; corolla incurved and recurved; capsule inferior, two-celled. There are twelve species. GETHYLLIS, in GEUM, in botany, English ovens, or herb bennet, a genus of the Icosandria Polygyria 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 |