Geomantia, a kind of divination, performed by means of a number of little points, or dots, made on paper at random; and considering the various lines and figures, which those points present; and thence forming a pretended judgment of futurity, and deciding any question proposed.

The word is formed of the Greek γη, terra, "earth;" and μαντεία, "divination;" it being the ancient custom to cast little pebbles on the ground, and thence to form their conjectures; instead of the points afterwards made use of.

Polydore Virgil defines geomancy a kind of divination performed by means of clefts or chinks made in the ground; and takes the Persian Magi to have been the inventors thereof.

GEOMETRY Geometry

Originally signified no more than the art of measuring the earth, or any distances or dimensions within it; but at present it denotes the science of magnitude in general; comprehending the doctrine and relations of whatever is susceptible of augmentation or diminution, considered in that light.

Hence to geometry may be referred the consideration not only of lines, surfaces, and solids; but also of time, velocity, number, weight, &c.

This science had its rise among the Egyptians, who were in a manner compelled to invent it, to remedy confusion which generally happened in their lands, from the inundations of the river Nile, which carried away all boundaries, and effaced all the limits of their possessions. Thus this invention, which at first consisted only in measuring the lands, that every person might have what belonged to him, was called geometry, or the art of measuring land; and it is probable that the draughts and schemes, which they were annually compelled to make, helped them to discover many excellent properties of these figures; which speculations continued to be gradually improved, and are so to this day.

Part I. General Principles of Geometry.

Art. 1. A Point is that which is not made up of parts, or which is of itself indivisible.

2. A line is a length without breadth, as B—

3. The extremities of a line are points; as the extremities of the line AB, are the points A and B, fig. 1.

4. If the line AB be the nearest distance between its extremes A and B, then it is called a straight line, as AB; but if it be not the nearest distance, then it is called a curve line, as ab, fig. 1.

5. A surface is that which is considered as having only length and breadth, but no thickness, as fig. 2.

6. The terms or boundaries of a surface are lines.

7. A plain surface is that which lies equally between its extremes.

8. The inclination between two lines meeting one another (provided they do not make one continued line), or the opening between them, is called an angle; thus the inclination of the line AB to the line CB(fig.3.) meeting one another at B, or the opening between the two lines AB and CB, is called an angle.

9. When the lines forming the angle are right lines, then it is called a right-lined angle, as fig. 4.; if one of them be right and the other curved, it is called a mixed angle, as fig. 5; if both of them be curved, it is called a curve-lined angle, as fig. 6.

10. If a right line AB fall upon another DC, (fig. 7.) so as to incline neither to one side nor to the other, but make the angles ABD, ABC, on each side equal to one another; then the line AB is said to be perpendicular to the line DC, and the two angles are called right-angles.

11. An obtuse angle is that which is greater than a right one, as fig. 8.; and an acute angle, that which is less than a right one, as fig. 9.

12. If a right line DC be fastened at one of its ends C, and the other end D be carried quite round, then the space comprehended is called a circle; the curveline described by the point D, is called the periphery or circumference of the circle; the fixed point C is called the centre of it; fig. 10.

13. The describing line CD is called the radius, viz. any line drawn from the centre to the circumference; whence all radii of the same or equal circles are equal.

14. Any line drawn through the centre, and terminated both ways by the circumference, is called a diameter, as BD is a diameter of the circle BADE. And the diameter divides the circle and circumference into two equal parts, and is double the radius.

15. The circumference of every circle is supposed to be divided into 360 equal parts, called degrees; and each degree is divided into 60 equal parts, called minutes; and each minute into 60 equal parts, called seconds; and these into thirds, fourths, &c. these parts being greater or less according as the radius is.

16. Any part of the circumference is called an arc, or arc; and is called an arc of as many degrees as it contains parts of the 360, into which the circumference was divided: thus if AD be the \( \frac{1}{8} \) of the circumference, then the arc AD is an arc of 45 degrees.

17. A line drawn from one end of an arc to the other, is called a chord, and is the measure of the arc; thus the right line AB is the chord of the arc ADB, fig. 11.

18. Any 18. Any part of a circle cut off by a circle, is called a segment; thus the space comprehended between the chord AB and circumference ADB (which is cut off by the chord AB) is called a segment. Whence it is plain,

1/2. That all chords divide the circle into two segments.

2/3. The less the chord is, the more unequal are the segments, and e contra.

3/4. When the chord is greatest, viz. when it is a diameter, then the segments are equal, viz. each a semicircle.

19. Any part of a circle (less than a semicircle) contained between two radii and an arc, is called a sector; thus the space contained between the two radii AC, BC, and the arc AB, is called the sector, fig. 12.

20. The right sine of any arc, is a line drawn perpendicular from one end of the arc, to a diameter drawn through the other end of the same arc; thus (fig. 13.) AD is the right sine of the arc AB, it being a line drawn from A, the one end of the arc AB, perpendicular to CB, a diameter passing through B, the other end of the arc AB.

Now the sines standing on the same diameter, still increase till they come to the centre, and then becoming the radius, it is plain that the radius EC is the greatest possible sine, and for that reason it is called the whole sine.

Since the whole sine EC must be perpendicular to the diameter FB (by def. 2.), therefore producing the diameter EG, the two diameters FB, EG, must cross one another at right angles, and so the circumference of the circle must be divided by them into four parts, EB, BG, GF, and FE, and these four parts are equal to one another (by def. 10.) and so EB a quadrant, or fourth part of the circumference; therefore the radius EC is always the sine of the quadrant, or fourth part of the circle EB.

Sines are said to be of so many degrees, as the arc contains parts of the 360°, into which the circumference is supposed to be divided; so the radius being the sine of a quadrant, or fourth part of the circumference, which contains 90 degrees (the fourth part of 360°), therefore the radius must be the sine of 90 degrees.

21. The part of the radius comprehended between the extremity of the right sine and the lower end of the arc, viz. DB, is called the versed sine of the arc AB.

22. If to any point in the circumference, viz. B, there be drawn a diameter FCB, and from the point B, perpendicular to that diameter, there be drawn the line BH; that line is called a tangent to the circle in the point B; which tangent can touch the circle only in one point B, else if it touched it in more, it would go within it, and so not be a tangent but a chord, (by art 17.)

23. The tangent of any arc AB, is a right line drawn perpendicular to a diameter through the one end of the arc B, and terminated by a line CAH, drawn from the centre through the other end A; thus BH is the tangent of the arc AB.

24. And the line which terminates the tangent, viz. CH, is called the secant of the arc AB.

N° 137.

25. What an arc wants of a quadrant is called the complement of that arc; thus AE, being what the arc AB wants of the quadrant EB, is called the complement of the arc AB.

26. And what an arc wants of a semicircle is called the supplement of that arc; thus since AF is what the arc AB wants of the semicircle BAF, it is the supplement of the arc AB.

27. The sine, tangent, &c. of the complement of any arc, is called the co-sine, co-tangent, &c. of that arc; thus the sine, tangent, &c. of the arc AE, is called the co-sine, co-tangent, &c. of the arc AB.

28. The sine of the supplement of an arc is the same with the sine of the arc itself; for drawing them according to the definitions, there results the self-same line.

29. A right lined angle is measured by an arc of a circle described upon the angular point as a centre, comprehended between the two legs that form the angle; thus (fig. 14.) the angle ABD is measured by the arc AD of the circle CADE that is described upon the point B as a centre; and the angle is said to be of as many degrees as the arc is; so if the arc AD be 45 degrees, then the angle ABD is said to be an angle of 45 degrees.

Hence the angles are greater or less, according as the arc described about the angular point, and terminated by the two legs, contain a greater or a less number of degrees.

30. When one line falls perpendicularly on another, as AB on CD, fig. 15. then the angles are right (by the 10th def.); and describing a circle on the centre B, since the angles ABC ABD are equal, their measures must be so too. i.e., the arcs AC AD must be equal; but the whole CAD is a semicircle, since CD, a line passing through the centre B, is a diameter; therefore each of the parts AC AD is a quadrant, i.e. 90 degrees; so the measure of a right angle is always 90 degrees.

31. If one line AB fall any way upon another, CD, then the sum of the two angles ABC ABD is always equal to the sum of two right angles; fig. 16. For on the point B, describing the circle CAD, it is plain, that CAD is a semicircle (by the 14th); but CAD is equal to CA and AD the measure of the two angles; therefore the sum of the two angles is equal to a semicircle, that is, to two right angles. (by the last.)

Cor. 1. From whence it is plain, that all the angles which can be made from a point in any line, towards one side of the line, are equal to two right angles.

2. And that all the angles which can be made about a point, are equal to four right ones.

32. If one line AC crosses another BD in the point E, then the opposite angles are equal, viz. BEA to CED, and BEC equal to AED; fig. 17. For upon the point E, as a centre, describing the circle ABCD, it is plain ABC is a semicircle, as also BCD (by the 14th); therefore the arc ABC is equal to the arc BCE; and from both taking the common arc BC, there will remain AB equal to CD, i.e. the angle BEA equal to the angle CED (by art. 29.) After the same manner we may prove, that the angle BEC is equal to the angle AED.

33. Lines which are equally distant from one another, are called parallel lines; as AB, CD, fig. 18.

34. If 34. If a line GH crosses two parallels AB, CD, (fig. 19.) then the external opposite angles are equal, viz. GEB equal to CFH, and AEG equal to HFD. For since AB and CD are parallel to one another, they may be considered as one broad line, and GH crossing it; then the vertical or opposite angles GEB CFH are equal (by art. 32.), as also AEG and HFD by the same.

35. If a line GH crosses two parallels AB, CD, then the alternate angles, viz. AEF and EFD, or CFE and FEB, are equal; that is, the angle AEF is equal to the angle EFD, and the angle CFE is equal to the angle FEB, for GEB is equal to AEF (by art. 32.), and CFH is equal to EFD (by the same); but GEB is equal to CFH (by the last); therefore AEF is equal to EFD. The same way we may prove FEB equal to EFC.

36. If a line GH crosses two parallel lines AB, CD, then the external angle GEB is equal to the internal opposite one EFD, or GEA equal to CFE. For the angle AEF is equal to the angle EFD (by the last); but AEF is equal to GEB (by art. 32.), therefore GEB is equal to EFD. The same way we may prove AEG equal to CFE.

37. If a line GH crosses two parallel lines AB, CD, then the sum of the two internal angles, viz. BEF and DFE, or AEF and CFE, are equal to two right angles; for since the angle GEB is equal to the angle EFD (by art. 36.), to both add the angle FEB, then GEB and BEF are equal to BEF and DFE; but GEB and BEF are equal to two right angles (by art. 31.), therefore BEF and DFE are also equal to two right angles. The same way we may prove that AEF and CFE are equal to two right angles.

38. A figure is any part of space bounded by lines or a line. If the bounding lines be straight, it is called a rectilineal figure, as fig. 20. If they be curved, it is called a curvilinear figure, as fig. 21. and fig. 22.; if they be partly curve lines and partly straight, it is called a mixt figure, as fig. 23.

39. The most simple rectilinear figure is that which is bounded by three right lines, and is called a triangle, as fig. 24.

40. Triangles are divided into different kinds, both with respect to their sides and angles: with respect to their sides, they are commonly divided into three kinds, viz.

41. A triangle having all its three sides equal to one another, is called an equilateral triangle, as fig. 25.

42. A triangle having two of its sides equal to one another, and the third side not equal to either of them, is called an Isosceles triangle, as fig. 25.

43. A triangle having none of its sides equal to one another, is called a scalene triangle, as fig. 27.

44. Triangles, with respect to their angles, are divided into three different kinds, viz.

45. A triangle having one of its angles right, is called a right-angled triangle, as fig. 28.

46. A triangle having one of its angles obtuse, or greater than a right angle, is called an obtuse-angled triangle, as fig. 29.

47. Lastly, a triangle having all its angles acute, is called an acute angled-triangle, as fig. 30.

48. In all right-angled triangles, the sides compr-

Vol. VII. Part II. and CD (fig. 39.) be crossed by a third line EF, and the alternate angles AEF and EFD be equal, the lines AB and CD will be parallel; for if they are not parallel, they must meet one another on one side of the line EF (suppose at G), and so form the triangle EFG, one of whose sides GE being produced at A, the exterior angle AEF must (by this article) be equal to the sum of the two angles EFG and EGF; but, by supposition, it is equal to the angles EFG alone; therefore the angle AEF must be equal to the sum of the two angles EFG and EGF, and at the same time equal to the angle EFG alone, which is absurd; so the lines AB and CD cannot meet, and therefore must be parallel.

60. In any triangle ABC, all the three angles taken together are equal to two right angles. To prove this, you must produce BC, one of its legs, to any distance, suppose to D; then by the last proposition, the external angle, ACD, is equal to the sum of the two internal opposite ones CAB and ABC; to both add the angle ACB, then the sum of the angles ACD and ACB will be equal to the sum of the angles CAB and CBA and ACB. But the sum of the angles ACD and ACB is equal to two right ones (by art. 32.), therefore the sum of the three angles CAB and CBA and ACB is equal to two right angles; that is, the sum of the three angles of any triangle ACB is equal to two right angles.

Cor. 1. Hence in any triangle given, if one of its angles be known, the sum of the other two is also known: for since (by the last) the sum of all the three is equal to two right angles, or a semicircle, it is plain, that taking any one of them from a semicircle or 180 degrees, the remainder will be the sum of the other two. Thus (in the former triangle ABC) if the angle ABC be 40 degrees, by taking 40 from 180 we have 140 degrees; which is the sum of the two angles BAC, ACB: the converse of this is also plain, viz. the sum of any two angles of a triangle being given, the other angle is also known by taking that sum from 180 degrees.

2. In any right-angled triangle, the two acute angles must just make up a right one between them; consequently, any one of the oblique angles being given, we may find the other by subtracting the given one from 90 degrees, which is the sum of both.

61. If in any two triangles, ABC (fig. 40.) DEF (fig. 41.) two legs of the one, viz. AB and AC, be equal to two legs of the other, viz. to DE and DF, each to each respectively, i.e. AB to DE and AC to DF; and if the angles included between the equal legs be equal, viz. the angle BAC equal to the angle EDF; then the remaining leg of the one shall be equal to the remaining leg of the other, viz. BC to EF; and the angles opposite to equal legs shall be equal, viz. ABC equal to DEF (being opposite to the equal legs AC and DF), also ACB equal to DFE (which are opposite to the equal legs AB and DE). For if the triangle ABC be supposed to be lifted up and put upon the triangle DEF, and the point A on the point D; it is plain, since BA and DE are of equal length, the point E will fall upon the point B; and since the angles BAC EDF are equal, the line AC will fall upon the line DF; and they being of equal length, the point C will fall upon the point F; and so the line BC will exactly agree with the line EF, and the triangle ABC will in all respects be exactly equal to the triangle DEF; and the angle ABC will be equal to the angle DEF, also the angle ACB will be equal to the angle DFE.

Cor. 1. After the same manner it may be proved, that if in any two triangles ABC, DEF (see the preceding figure), two angles ABC and ACB of the one, be equal to two angles DEF and DFE of the other, each to each respectively, viz. the angle ABC to the angle DEF, and the angle ACB equal to the angle DFE, and the sides included between these angles be also equal, viz. BC equal to EF, then the remaining angles, and the sides opposite to the equal angles, will also be equal each to each respectively; viz. the angle BAC equal to the angle EDF, the side AB equal to DE, and AC equal to DF: for if the triangle ABC be supposed to be lifted up and laid upon the triangle DEF, the point B being put upon the point E, and the line BC upon the line EF, since BC and EF are of equal lengths, the point C will fall upon the point F, and since the angle ACB is equal to the angle DFE, the line CA will fall upon the line FD, and by the same way of reasoning the line BA will fall upon the line ED; and therefore the point of intersection of the two lines BA and CA, viz. A, will fall upon the point of intersection of the two lines ED and FD, viz. D, and consequently BA will be equal to ED, and AC equal to DF, and the angle BAC equal to the angle EDF.

Cor. 2. It follows likewise from this article, that if any triangle ABC (fig. 42.) has two of its sides AB and AC equal to one another, the angles opposite to these sides will also be equal, viz. the angles ABC equal to the angle ACB. For suppose the line AD bisecting the angle BAC, or dividing it into two equal angles BAD and CAD, and meeting BC in D, then the line AD will divide the whole triangle BAC into two triangles ABD and DAC; in which BA and AD two sides of the one, are equal to CA and AD two sides of the other, each to each respectively, and the included angles BAD and DAC are by supposition equal; therefore (by this article) the angle ABC must be equal to the angle ACB.

62. Any angle, as BAD (fig. 43.) at the circumference of a circle BADE, is but half the angle BCD at the centre standing on the same arc BED. To demonstrate this, draw through A and the centre C the right line ACE, then the angle ECD is equal to both the angles DAC and ADC (by art. 59.); but since AC and CD are equal (being two radii of the same circle), the angles subtended by them must be equal also (by art. 62. cor. 2.), i.e. the angle CAD equal to the angle CDA; therefore the sum of them is double any one of them, i.e. DAC and ADC is double of CAD, and therefore ECD is also double of DAC; the same way it may be proved, that ECB is double of CAB; and therefore the angle BCD is double of the angle BAD, or BAD the half of BCD, which was to be proved.

Cor. 1. Hence an angle at the circumference is measured by half the arc it subtends; for the angle at the centre (standing on the same arc) is measured by the whole arc (by art. 29.); but since the angle at the centre is double that at the circumference, it is plain the General angle at the circumference must be measured by only half the arc it stands upon.

Cor. 2. Hence all angles, ACB, ADB, AFB, &c. (fig. 44.) at the circumference of a circle, standing on the same chord AB, are equal to one another; for by the last corollary they are all measured by the same arc, viz. half the AB which each of them subtends.

Cor. 3. Hence an angle in a segment greater than a semicircle is less than a right angle: thus, if ADB be a segment greater than a semicircle (see the last figure), then the arc AB, on which it stands, must be less than a semicircle, and the half of it less than a quadrant or a right angle; but the angle ADB in the segment is measured by the half of AB, therefore it is less than a right angle.

Cor. 4. An angle in a semicircle is a right angle. For since ABD (fig. 46.) is a semicircle, the arc AED must also be a semicircle: but the angle ABD is measured by half the arc AED, that is, by half a semicircle or quadrant; therefore the angle ABD is a right one.

Cor. 5. Hence an angle in a segment less than a semicircle, as ABD (fig. 45.), is greater than a right angle: for since the arc ABD is less than a semicircle, the arc AED must be greater than a semicircle, and so it is half greater than a quadrant, i.e. than the measure of a right angle; therefore the angle ABD, which is measured by half the arc AED, is greater than a right angle.

63. If from the centre C of the circle ABE (fig. 47.) there be let fall the perpendicular CD on the chord AB, then that perpendicular will bisect the chord AB in the point D. To demonstrate this, draw from the centre to the extremities of the chord the two lines CA, CB; then, since the lines CA and CB are equal, the angles CAB, CBA, which they subtend, must be equal also (by art. 62. cor. 2.), but the perpendicular CD divides the triangle ACB into two right-angled triangles ACD and CDB, in which the sum of the angles ACD and CAD in the one is equal to the sum of the angles DCB and CDB in the other, each being equal to a right angle (by cor. 2. of art. 61.) but CAD is equal to CBD, therefore ACD is equal to BCD. So in the two triangles ACD and BCD, the two legs AC and CD in the one, are equal to the two legs BC and CD in the other, each to each respectively, and the included angles ACD and BCD are equal; therefore the remaining legs AD and BD are equal (by art. 61.), and consequently AB bisected in D.

64. If from the centre C of a circle ABE, there be drawn a perpendicular CD on the chord AB, and produced till it meet the circle in F, then the line CF bisects the arch AB in the point F; for (see the foregoing figure) joining the points A and F, F and B by the straight lines AF, FB, then in the triangles ADF, BDF, AD is equal to DB (by art. 63.), and DF common to both; therefore AD and DF, two legs of the triangle ADF, are equal to BD and DF, two legs of the triangle BDF, and the included angles ADF BDF are equal, being both right; therefore (by art. 61.) the remaining legs AF and FB are equal; but in the same circle equal lines are chords of equal arches, therefore the arches AF and FB are equal. So the whole arch AFB is bisected in the point F by the line CF.

Cor. 1. From art. 63. it follows, that any line bisecting a chord at right angles is a diameter; for since (by art. 63.) a line drawn from the centre perpendicular to a chord, bisects that chord at right angles; therefore, conversely, a line bisecting a chord at right angles, must pass through the centre, and consequently be a diameter.

Cor. 2. From the two last articles it follows, that the fine of any arc is the half of the chord of twice the arc; for (see the foregoing scheme) AD is the fine of the arc AF, by the definition of a fine, and AF is half the arc AFB, and AD half the chord AB (by art. 63.); therefore the corollary is plain.

65. In any triangle, the half of each side is the fine of the opposite angle; for if a circle be supposed to be drawn through the three angular points A, B, and D of the triangle ABD, fig. 48. then the angle DAB is measured by half the arch BKD (by cor. 1. of art. 62.), but the half of BD, viz. BE, is the fine of half the arch BKD, viz. the fine of BK (by cor. 2. of the last), which is the measure of the angle BAD; therefore the half of BD is the fine of the angle BAD: the same way, it may be proved, that the half of AD is the fine of the angle ABD, and the half of AB is the fine of the angle ADB.

66. The fine, tangent, &c. of any arch is called also the fine, tangent, &c. of the angle whose measure the arc is: thus, because the arc GD (fig. 49.) is the measure of the angle GCD; and since GH is the fine, DE the tangent, HD the verified fine, CE the secant, also GK the co-fine, BF the co-tangent, and CF the co-secant, &c. of the arch GD; then GH is called the fine, DE the tangent, &c. of the angle GCD, whose measure is the arch GD.

67. If two equal and parallel lines, AB and CD (fig. 50.) be joined by two others, AC and BD; then these shall also be equal and parallel. To demonstrate this, join the two opposite angles A and D with the line AD; then it is plain this line AD divides the quadrilateral ACDB into two triangles, viz. ABD, ACD, in which AB a leg of the one, is equal to DC a leg of the other, by supposition, and AD is common to both triangles; and since AB is parallel to CD, the angle BAD will be equal to the angle ADC (by art. 36.); therefore in the two triangles BA and AD, and the angle BAD, is equal to CD and DA, and the angle ADC; that is, two legs and the included angle in the one is equal to two legs and the included angle in the other; therefore (by art. 61.) BD is equal to AC, and since the angle DAC is equal to the angle ADB, therefore the lines BD AC are parallel (by cor. art. 59.)

Cor. 1. Hence it is plain, that the quadrilateral ABDC is a parallelogram, since the opposite sides are parallel.

Cor. 2. In any parallelogram the line joining the opposite angles (called the diagonal) as AD, divides the figure into two equal parts, since it has been proved that the triangles ABD ACD are equal to one another.

Cor. 3. It follows also, that a triangle ACD on the same base CD, and between the same parallels with a parallelogram ABDC, is the half of that parallelogram.

Cor. 4. Hence it is plain, that the opposite sides of a parallelogram are equal; for it has been proved, that \(ABDC\) being a parallelogram, \(AB\) will be equal to \(CD\), and \(AC\) equal to \(BD\).

68. All parallelograms on the same or equal bases, and between the same parallels, are equal to one another; that is, if \(BD\) and \(GH\) (fig. 51.) be equal, and the lines \(BH\) and \(AF\) be parallel, then the parallelograms \(ABDC\), \(BDFE\), and \(EFHG\), are equal to one another. For \(AC\) is equal to \(EF\), each being equal to \(BD\) (by cor. 4. of 67.) To both add \(CE\), then \(AE\) will be equal to \(CF\). So in the two triangles \(ABE CDF\), \(AB\) a leg of the one, is equal to \(CD\) a leg in the other; and \(AE\) is equal to \(CF\), and the angle \(BAE\) is equal to the angle \(DCF\) (by art. 36.) therefore the two triangles \(ABE CDF\) are equal (by art. 61.) and taking the triangle \(CKE\) from both, the figure \(ABKC\) will be equal to the figure \(KDFE\); to both which add the little triangle \(KBD\), then the parallelogram \(ABDC\) will be equal to the parallelogram \(BDFE\). The same way it may be proved, that the parallelogram \(EFHG\) is equal to the parallelogram \(EFDB\); so the three parallelograms \(ABDC\), \(BDFE\), and \(EFHG\), will be equal to one another.

Cor. Hence it is plain, that triangles on the same base, and between the same parallels, are equal; since they are the half of the parallelograms on the same base and between the same parallels (by cor. 3. of last art.)

69. In any right-angled triangle, \(ABC\) (fig. 52.), the square of the hypotenuse \(BC\), viz. \(BCMH\), is equal to the sum of the squares made on the two sides \(AB\) and \(AC\), viz. to \(ABDE\) and \(ACGF\). To demonstrate this, through the point \(A\) draw \(AKL\) perpendicular to the hypotenuse \(BC\), join \(AH\), \(AM\), \(DC\), and \(BG\); then it is plain that \(DB\) is equal to \(BA\) (by art. 53.), also \(BH\) is equal to \(BC\) (by the same); so in the two triangles \(DBC ABH\), the two legs \(DB\) and \(BC\) in the one are equal to the two legs \(AB\) and \(BH\) in the other; and the included angles \(DBC\) and \(ABH\) are also equal; (for \(DBA\) is equal to \(CBH\), being both right;) to each add \(ABC\), then it is plain that \(DBC\) is equal to \(ABH\); therefore the triangles \(DBC ABH\) are equal (by art. 61.), but the triangle \(DBC\) is half of the square \(ADBE\) (by cor. 3. of 67.), and the triangle \(ABH\) is half the parallelogram \(BKLH\) (by the same), therefore half the square \(ABDE\) is equal to half the parallelogram \(BKLH\). Consequently the square \(ABDE\) is equal to the parallelogram \(BKLH\). The same way it may be proved, that the square \(ACGF\) is equal to the parallelogram \(KCML\). So the sum of the squares \(ABDE\) and \(ACGF\) is equal to the sum of the parallelograms \(BKLH\) and \(KCML\), but the sum of these parallelograms is equal to the square \(BCMH\); therefore the sum of the squares on \(AB\) and \(AC\) is equal to the square on \(BC\).

Cor. 1. Hence in a right-angled triangle, the hypotenuse and one of the legs being given, we may easily find the other, by taking the square of the given leg from the square of the hypotenuse, and the square root of the remainder will be the leg required.

Cor. 2. Hence the legs in a right-angled triangle being given, we may find the hypotenuse, by taking the sum of the squares of the given legs, and extracting the square root of that sum.

7c. If upon the line \(AB\) (fig. 53.) there be drawn a semicircle \(ADB\), whose centre is \(C\), and on the point \(C\) there be raised a perpendicular to the line \(AB\), viz. \(CD\); then it is plain the arc \(DB\) is a quadrant, or contains 90 degrees; suppose the arc \(DB\) to be divided into 9 equal arcs, each of which will contain 10 degrees, then on the point \(B\) raising \(BE\) perpendicular to the line \(AB\), it will be a tangent to the circle in quadrant, viz. \(B_{10}\), \(B_{20}\), \(B_{30}\), \(B_{40}\), &c. you draw the fine, tangent, &c. (as in the scheme) we shall have the fine, tangent, &c. to every 10 degrees in the quadrant; and the same way we may have the fine, tangent, &c. to every single degree in the quadrant, by dividing it into 90 equal parts beginning from \(B\), and drawing the fine, tangent, &c. to all the arcs beginning at the same point \(B\). By this method they draw the lines of fines, tangents, &c. of a certain circle on the scale; for after drawing them on the circle, they take the length of them, and set them off in the lines drawn for that purpose. The same way, by supposing the radius of any number of equal parts, (suppose 1000, or 10,000, &c.) it is plain the fine, tangent, &c. of every arc must consist of some number of these equal parts, and by computing them in parts of the radius, we have tables of fines, tangents, &c. to every arc in the quadrant, called natural fines, tangents, &c. and the logarithms of these give us tables of logarithmic fines, tangents, &c. See Logarithms.

71. In any triangle, \(ABC\) (fig. 1.) if one of its plate sides, as \(AC\), be bisected in \(E\) (and consequently \(AC = 2 \times AE\)), and through \(E\) be drawn \(ED\), parallel to \(BC\), and meeting \(AB\) in \(D\), then \(BC\) will be double of \(ED\), and \(AB\) double of \(AD\). Through \(D\) draw \(DF\), parallel to \(AC\), meeting \(BC\) in \(F\): for since, by construction, \(DF\) is parallel to \(AC\), and \(DE\) parallel to \(BC\); therefore (by art. 36.) the angle \(BFD\) will be equal to the angle \(BCA\), and (by the same article) the angle \(BCA\) will be equal to the angle \(DEA\), consequently the angle \(BFD\) will be equal to the angle \(DEA\); also (by art. 36.) the angle \(BDF\) will be equal to the angle \(DAE\); and since \(DF\) is parallel to \(EC\), and \(DE\) parallel to \(FC\), the quadrilateral \(DFCE\) will be a parallelogram; and therefore (by art. 59. cor. 4.) \(DF\) will be equal to \(EC\), which, by construction, is equal to \(AE\); so in the two triangles \(BDF DAE\), the two angles \(BFD\) and \(BDF\) in the one, are equal to the two angles \(DEA\) and \(DAE\) in the other, each to each respectively; and the included side \(DF\) is equal to the included side \(AE\); therefore (by art. 61. cor. 1.) \(AD\) will be equal to \(DB\), and consequently \(AB\) double of \(AD\); also (by the same) \(DE\) will be equal to \(BF\); but \(DE\) is also (by art. 67. cor. 4.) equal to \(FC\); therefore \(BF\) and \(ED\) together, or \(BC\), will be double of \(DE\).

After the same manner it may be proved, that if in the triangle \(AKG\), (fig. 2.) \(AE\) be taken equal to a third part of \(AK\), and through \(E\) be drawn \(ED\), parallel to \(KG\), and meeting \(AG\) in \(D\); then \(ED\) will be equal to a third part of \(GK\), and \(AD\) equal to a third part of \(AG\).

Likewise if in any triangle \(ABC\), (fig. 3.) upon the side \(AB\), be taken \(AE\), equal to one fourth, one-fifth, one-sixth, &c. of \(AB\), and through \(E\) be drawn \(ED\) parallel to \(BC\) and meeting \(AC\) in \(D\); then \(D\) will be one-fourth, one-fifth, one-sixth, &c. of \(BC\), and \(AD\) the like part of \(AC\); and, in general, if in any triangle, General angle ABC, there be assumed a point E on one of its sides AB, and through that point be drawn a line ED, parallel to one of its sides BC, and meeting the other side AC in D; then whatever part AE is of AB, the same part will ED be of BC, and AD of AC.

Cor. Hence it follows, that if in any triangle ABC, there be drawn ED, parallel to one of its sides BC, and meeting the other two in the points E and D, then AE : AB :: ED : BC :: AD : AC; that is, AE is to AB, as ED is to BC, and that as AD to AC.

72. If any two triangles ABC, fig. 4. a b c, fig. 5. are similar, or have all the angles of the one equal to all the angles of the other, each to each respectively; that is, the angle CAB equal to the angle a c b, and the angle ABC equal to the angle a c b, and the angle ACB equal to the angle a c b; then the legs opposite to the equal angles are proportioned, viz. AB : a b :: AC : a c :: and AB : a b :: BC : b c :: and AC : a c :: BC : b c. On AB of the largest triangle let off AE equal to a b, and through E draw ED parallel to BC, meeting AC in D; then since DE and BC are parallel, and AB crossing them, the angle AED will (by art. 36.) be equal to the angle ABC, which (by supposition) is equal to the angle a b c, also the angle DAE is (by supposition) equal to the angle a c b; so in the two triangles AED, a b c, the two angles DAE AED of the one, are equal to two angles a b a b c of the other, each to each respectively, and the included side AE is (by construction) equal to the included side a b; therefore, (by art. 61. cor. 1.) AD is equal to a c, and DE equal to b c; but since, in the triangle ABC, there is drawn DE parallel to BC one of its sides, and meeting the two other sides in the points D and E, therefore (by cor. art. 71.) AB : AE :: AC : AD, and AB : AE :: BC : DE, and AC : AD :: BC : DE; and in the three last proportions, instead of the lines AE, DE, and AD, putting in their equals a b, b c, and a c, we shall have AB : a b :: AC : a c, and AB : a b :: BC : b c, and lastly, AC : a c :: BC : b c.

73. The chord, fine, tangent, &c. of any arc in one circle, is to the chord, fine, tangent, &c. of the same arc in another, as the radius of the one is to the radius of the other, fig. 6. 6. Let ABD a b d be two circles, BD b d two arcs of these circles, equal to one another, or consisting of the same number of degrees; FD f d the tangents, BD b d the chords, BE b e the fines, &c. of these two arcs BD b d, and CD c d the radii of the circles; then lay, CD : c d :: FD : f d, and CD : c d :: BD : b d, and CD : c d :: BE : b e, &c. For since the arcs BD b d are equal, the angles BCD b c d will be equal; and FD f d, being tangents to the points D and d, the angles CDF c d f will be equal, being each a right angle (art. 22.) so in the two triangles CDF c d f, the two angles FCD CDF of the one, being equal to the two angles f c d c d f of the other, each to each, the remaining angle CFD will be equal to the remaining angle c f d (by art. 60.); therefore the triangles CFD c f d are similar, and consequently (by art. 73.) CD : c d :: FD : f d. In the same manner it may be demonstrated, that CD : c d :: BD : b d, and CD : c d :: BE b e, &c.

74. Let ABD (fig. 7.) be a quadrant of a circle described by the radius CD; BD any arc of it, and BA its complement; BG or CF the fine, CG or BF the cosine; DE the tangent, and CE the secant of that arc BD. Then since the triangles CDE CBG are similar or equiangular, it will be (by art. 72.) DE : EC :: GB : BC, i.e. the tangent of any arc, is to the secant of the same, as the fine of it is to the radius. Also since DE : EC :: GB : BC; therefore, by inverting that proportion, we have EC : DE :: BC : GB, i.e. the secant is to the tangent, as the radius is to the fine of any arc.

Again, since the triangles CDE CGB are similar, therefore (by art. 72.) it will be CD : CE :: CG : CB, i.e. as the radius is to the secant of any arc, so is the co-fine of that arc to the radius. And by inverting the proportion we have this, viz. as the secant of any arc is to the radius, so is the radius to the co-fine of that arc.

75. In all circles the chord of 60° is always equal in length to the radius. Thus in the circle AEBD, (fig. 8.) if the arc AEB be an arc of 60° degrees, then drawing the chord AB, I say AB shall be equal to the radius CB or AC; for in the triangle ACB, the angle ACB is 60° degrees, being measured by the arc AEB; therefore the sum of the other two angles is 120° degrees, (by cor. 1. of 60°.) but since AC and CB are equal, the two angles CAB, CBA will also be equal; consequently each of them half their sum 120°, viz. 60° degrees; therefore, all the three angles are equal to one another, consequently all the legs, therefore AB is equal to CB.

Cor. Hence the radius from which the lines on any scale are formed, is the chord of 60° on the line of chords.

Geometrical Problems.

Prob. 1. From a point C (fig. 9.) in a given line AB to raise a perpendicular to that line.

Rule. From the point C take the equal distances CB, CA on each side of it. Then stretch the compasses to any greater distance than CB or CA, and with one foot of them in B, sweep the arc EF with the other; again, with the same opening, and one foot in A, sweep the arc GH with the other, and these two arcs will intersect one another in the point D; then join the given points C and D with the line CD, and that shall be the perpendicular required.

2. To divide a given right line AB (fig. 10.) into two equal parts; that is, to bisect them.

Rule. Take any distance with your compasses that you are sure is greater than half the given line; then setting one foot of them in B, with the other sweep the arc DFG; and with the same distance, and one foot in A, with the other sweep the arc CED; these two arcs will intersect one another in the points CD, which joined by the right line CD will bisect AB in G.

3. From a given point D, (fig. 11.) to let fall a perpendicular on a given line AB.

Rule. Set one foot of the compasses in the point D, and extend the other to any greater distance than the least distance between the given point and the line, and with that extent sweep the arc AEB, cutting the line in the two points A and B; then (by the last prob.) bisect the line AB in the point C; lastly, join C and D; and that line CD is the perpendicular required. 4. (Fig. 12.) Upon the end B of a given right line BA, to raise a perpendicular.

**Rule.** Take any extent in your compasses, and with one foot in B fix the other in any point C without the given line; then with one point of the compasses in C, describe with the other the circle EBD, and through E and C draw the diameter ECD meeting the circle in D; join D and B, and the right line DB is that required; for EBD is a right angle (by cor. 4. of 63.)

5. (Fig. 13.) To draw one line parallel to another given line AB, that shall be distant from one another by any given distance D.

**Rule.** Extend your compasses to the given distance D; then setting one foot of them in any point of the given line (suppose A,) with the other sweep the arc FCG; again, at the same extent, and one foot in any other point of the given line B, sweep the arc HDK, and draw the line CD touching them, and that will be parallel to the given line AB, and distant from it by the line D as was required.

6. (Fig. 14.) To divide a given line AB into any number of equal parts, suppose 7.

**Rule.** From the point A draw any line AD, making an angle with the line AB, then through the point B, draw a line BC parallel to AD; and from A, with any small opening of the compasses, set off a number of equal parts (on the line AD) less by one than the proposed number (here 6); then from B set off the same number of the same parts (on the line BC); lastly, join 1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1, and these lines will cut the given line as required.

7. (Fig. 15.) To quarter a given circle, or to divide it into four equal parts.

**Rule.** Through the centre C of the given circle, draw a diameter AB, then upon the point C raise a perpendicular DCE to the line AB; and these two diameters AB and DE shall quarter the circle.

8. (Fig. 16.) Through three given points A, B, and D, to draw a circle.

**Note.** The three points must not lie in the same straight line.

**Rule.** Join A and B, also B and D, with the straight lines AB BD; then (by prob. 2.) bisect AB with the line EC, also BD with the line FC, which two lines will cut one another in some point C; that is the centre of the circle required: then fixing one point of our compasses in D, and stretching the other to A, describe the circle ABDG, which will pass through the three points given. The reason of this is plain from cor. 1. of art. 64.

9. (Fig. 17.) From the point A of the given line AB, to draw another line (suppose AC) that shall make with AB an angle of any number of degrees, suppose 45.

**Rule.** Let the given line AB be produced, then take off your scale the length of the chord of 60 degrees, which is equal to the radius of the circle the scale was made for (by art. 75.) and setting the foot in A, with the other sweep off the arc BC; then with your compasses take from your scale the chord of 45 degrees, and let off that distance from B to C. Lastly, join A and C, and the line AC is that required. For the angle CAB, which is measured by the arc BC, is an angle of 45 degrees, as was required.

10. An angle BAC (fig. 18.) being given, to find how many degrees it contains.

**Rule.** With your compasses take the length of your chord of 60 from your scale. Then, setting one foot of them in A, with the other sweep the arc BC, which is the arc comprehended between the two legs AB, AC produced if needful. Lastly, take with your compasses the distance BC, and applying it to your line of chords on the scale, you will find how many degrees the arc BC contains, and consequently the degrees of the angle BAC which was required.

11. Three lines x, y, and z being given. (fig. 19. 19.) to form a triangle of them; but any two of these lines taken together must always be greater than the third.

**Rule.** Make any one of them, as x, the base; then with your compasses take another of them, as z, and setting one foot in one end of the line x, as B, with the other sweep the arc DE; and taking with your compasses the length of the other y, set one foot of them in A, the other end of the line x, and with the other sweep the arc FG, which will cut the other in C; lastly, join CA and CB, and the triangle CAB is that required.

12. To make a triangle, having one of its legs of any number of equal parts (suppose 160,) and one of the angles at that leg 50 degrees, and the other 44 degrees.

**Rule.** Draw an indefinite line ED, (fig. 20.) then take off the line of equal parts with your compasses, 160 of them, and set them on the indefinite line, as BC; then (by prob. 9.) draw BA, making the angle ABC of 50 degrees, and (by the same) draw from C the line AC, making the angle ABC of 44 degrees; which two lines will meet one another in A, and the triangle ABC is that required. See Trigonometry.

13. Upon a given line AB (fig. 21.) to make a square.

**Rule.** Upon the extremity A of the given line AB, raise a perpendicular AC (by prob. 4.) then take AC equal to AB, and with that extent, setting one foot of the compasses in C, sweep with the other foot the arc GH; then with the same extent, and one foot in B, with the other sweep the arc EF, which will meet the former in some point D; lastly, join C and D, D and B, and the figure ABDC will be the square required.

14. On a given line AB (fig. 22.) to draw a rhomb that shall have one of its angles equal to any number of degrees, suppose 60 degrees.

**Rule.** From the point A of the given line AB, draw the line AC, making the angle CAB of 60 degrees, (by prob. 9.) then take AC equal to AB, and with that extent, fixing one foot of the compasses in B, with the other describe the arc GH; and at the same extent, fixing one foot of the compasses in C, with the other describe the arc EF cutting the former in D; lastly, join CD and DB, and the figure ACDB is that required.

15. Given two lines x and z, of these two to make a rectangle.

**Rule.** Draw a line, as AB, (fig. 23. 23.) equal in length to one of the given lines x; and on the extremity A of that line raise a perpendicular AC, on which Part II. GEOMETRY.

Lines and Angles.

Rule. Draw a line AB (fig. 24.) equal in length to one of the lines, as x; then draw the line AC, making with the former the angle BAC equal to the proposed, suppose 50 degrees, and on that line take AC equal to the given line x; then with your compasses take the length of AB, and fixing one foot in C, sweep the arc EF; also taking the length of AC, and setting one foot in B, with the other sweep the arc GH, which will cut the former in D; then join CD and DB, so the figure ACDB will be that required.

16. Two lines x and z being given, of these to form a rhomboides that shall have one of its angles any number of degrees, suppose 50.

Part II. The Application of the Foregoing Principles to the Mensuration of Surfaces, Solids, &c.

Chapter I. Of the Mensuration of Lines and Angles.

A line or length to be measured, whether it be distance, height, or depth, is measured by a line less than it. With us the least measure of length is an inch: not that we measure no line less than it, but because we do not use the name of any measure below that of an inch; expressing lesser measures by the fractions of an inch: and in this treatise we use decimal fractions as the easiest. Twelve inches make a foot; three feet and an inch make the Scots ell; six ells make a fall; forty falls make a furlong; eight furlongs make a mile: so that the Scots mile is 1184 paces, accounting every pace to be five feet. These things are according to the statutes of Scotland; notwithstanding which, the glaziers use a foot of only eight inches; and other artists for the most part use an English foot, on account of the several scales marked on the English foot-measure for their use. But the English foot is somewhat less than the Scots; so that 185 of these make 186 of those.

Lines, to the extremities and any intermediate point of which you have easy access, are measured by applying to them the common measure a number of times. But lines, to which you cannot have such access, are measured by methods taken from geometry; the chief whereof we shall here endeavour to explain.

The first is by the help of the geometrical square.

"As for the English measures, the yard is 3 feet, or 36 inches. A pole is fifteen feet and a half, or five yards and a half. The chain, commonly called Gunter's chain, is four poles, or 22 yards, that is, 66 feet. An English statute-mile is four-score chains, or 1760 yards, that is, 5280 feet.

"The chain (which is now much in use, because it is very convenient for surveying) is divided into 100 links, each of which is \( \frac{9}{10} \) of an inch: whence it is easy to reduce any number of those links to feet, or any number of feet to links.

"A chain that may have the same advantages in surveying Scotland, as Gunter's chain has in England, ought to be in length 74 feet, or 24 Scots ells, if no regard is had to the difference of the Scots and English foot above mentioned. But if regard is had to that difference, the Scots chain ought to consist of 74\(\frac{2}{3}\) English feet, or 74 feet 4 inches and \(\frac{4}{3}\) of an inch. This chain being divided into 100 links, each of those links is 8 inches and \(\frac{4}{3}\) of an inch. In the following table, the most noted measures are expressed in English inches and decimals of an inch."

| Measure | English Inches | Dec. | |--------------------------------|----------------|------| | The English foot | 12 | 000 | | The Paris foot | 12 | 788 | | The Rhindland foot measured by Mr Picart | 12 | 362 | | The Scots foot | 12 | 065 | | The Amsterdam foot, by Snellius and Picart | 11 | 172 | | The Dantzig foot, by Hevelius | 11 | 297 | | The Danish foot, by Mr Picart | 12 | 465 | | The Swedish foot, by the same | 11 | 692 | | The Brussel's foot, by the same | 10 | 828 | | The Lyons foot, by Mr Auzout | 13 | 458 | | The Bononian foot, by Mr Cassini | 14 | 938 | | The Milan foot, by Mr Auzout | 15 | 631 | | The Roman palm used by merchants, according to the same | 9 | 791 | | The Roman palm used by architects | 8 | 779 | | The palm of Naples, according to Mr Auzout | 10 | 314 | | The English yard | 36 | 000 | | The English ell | 45 | 000 | | The Scots ell | 37 | 200 | | The Paris aune used by mercers, according to Mr Picart | 46 | 786 | | The Paris aune used by drapers, according to the same | 46 | 680 | | The Lyons aune, by Mr Auzout | 46 | 570 | | The Geneva aune | 44 | 760 | | The Amsterdam ell | 26 | 800 | | The Danish ell, by Mr Picart | 24 | 930 | | The Swedish ell | 23 | 380 | | The Norway ell | 24 | 510 | | The Brabant or Antwerp ell | 27 | 170 | | The Brussel's ell | 27 | 260 | | The Bruges ell | 27 | 550 | | The brace of Bononia, according to Auzout | 25 | 200 | | The brace used by architects in Rome | 30 | 730 | | The brace used in Rome by merchants | 34 | 270 | | The Florence brace used by merchants, according to Picart | 22 | 910 | | The Florence geographical brace | 21 | 570 | | The vara of Seville | 33 | 127 | | The vara of Madrid | 39 | 166 | | The vara of Portugal | 44 | 031 | | The cavedo of Portugal | 27 | 354 | | The ancient Roman foot | 11 | 632 | | The Persian arish, according to Mr Graves | 38 | 364 | | The shorter pike of Constantinople, according to the same | 25 | 576 | | Another pike of Constantinople, according to Messrs Mallet and De la Porte | 27 | 020 |

PRO. PROPOSITION I.

PROP. To describe the structure of the geometrical square.—The geometrical square is made of any solid matter, as brass or wood, or of any four plain rulers joined together at right angles (as in fig. 1.), where A is the centre, from which hangs a thread with a small weight at the end, so as to be directed always to the centre. Each of the sides BE and DE is divided into an hundred equal parts, or (if the sides be long enough to admit of it) into a thousand parts; C and F at two sights, fixed on the side AD. There is moreover an index GH, which, when there is occasion, is joined to the centre A, in such manner as that it can move round, and remain in any given situation. On this index are two sights perpendicular to the right line going from the centre of the instrument: these are K and L. The side DE of the instrument is called the upright side; E the reclining side.

PROPOSITION II.

FIG. 2. To measure an accessible height AB by the help of a geometrical square, its distance being known.—Let BR be a horizontal plane, on which there stands perpendicularly any line AB: let BD, the given distance of the observator from the height, be 96 feet: let the height of the observator's eye be supposed 6 feet; and let the instrument, held by a steady hand, or rather leaning on a support, be directed towards the summit A, so that one eye (the other being shut) may see it clearly through the sights; the perpendicular or plumb-line meanwhile hanging free, and touching the surface of the instrument; let now the perpendicular be supposed to cut off on the right side KN 80 equal parts. It is clear that LKN, ACK, are similar triangles; for the angles LKN, ACK are right angles, and therefore equal; moreover, LN and AC are parallel, as being both perpendicular to the horizon; consequently (by art. 60. cor. 1. Part I), the angles KLN, KAC, are equal; wherefore (by art. 60. cor. 2. of Part I), the angles LNK and CAK are likewise equal: so that in the triangles NKL, KAC, (by art. 72. of Part I.) as NK : KL :: KC (i.e. BD) : CA; that is, as 80 to 100, so is 96 feet to CA. Therefore, by the rule of three, CA will be found to be 120 feet; and CB, which is 6 feet, being added, the whole height is 126 feet.

But if the distance of the observator from the height, as BE, be such, that when the instrument is directed as formerly toward the summit A, the perpendicular falls on the angle P, opposite to H, the centre of the instrument, and BE or CG be given of 120 feet; CA will also be 120 feet. For in the triangles HGP, ACG, equiangular, as in the preceding case, as DG : GH :: GC : CA. But PG is equal to GH; therefore GC is likewise equal to CA: that is, CA will be 120 feet, and the whole height 126 feet as before.

Let the distance BF be 300 feet, and the perpendicular or plumb-line cut off 40 equal parts from the reclining side: Now, in this case, the angles QAC, QZI, are equal, and the angles QZI, ZIS, are equal: therefore the angle ZIS is equal to the angle QAC. But the angles ZSI, QCA are equal, being right angles; therefore, in the equiangular triangles ACQ, SZI, it will be, as ZS : SI :: CQ : CA; that is, as 100 to 40, so is 300 to CA. Therefore, by the rule of three, CA will be found to be of 102 feet. And, by adding

N° 137.

the height of the observator, the whole BA will be Lines and Angles.

Plate CCXVII.

126 feet. Note, that the height is greater than the distance, when the perpendicular cuts the right side, and less if it cut the reclined side; and that the height and distance are equal, if the perpendicular fall on the opposite angle.

SCHOLIUM.

If the height of a tower to be measured as above, end in a point (as in fig. 3.), the distance of the observator opposite to it, is not CD, but is to be accounted from the perpendicular to the point A; that is, to CD must be added the half of the thickness of the tower, viz. BD: which must likewise be understood in the following propositions, when the case is similar.

PROPOSITION III.

FIG. 4. From the height of a tower AB given, to find a distance on the horizontal plane BC, by the geometrical square.—Let the instrument be so placed, as that the mark C in the opposite plane may be seen through the sights; and let it be observed how many parts are cut off by the perpendicular. Now, by what hath been already demonstrated, the triangles AEF, ABC, are similar; therefore, it will be as EF to AE, to AB (composed of the height of the tower BG, and of the height of the centre of the instrument A, above the tower BG) to the distance BC. Wherefore, if, by the rule of three, you say, as EF to AE, so is AB to BC, it will be the distance sought.

PROPOSITION IV.

FIG. 5. To measure any distance at land or sea, by the geometrical square.—In this operation, the index is to be applied to the instrument, as was shown in the description; and, by the help of a support, the instrument is to be placed horizontally at the point A; then let it be turned till the remote point F, whose distance is to be measured, be seen through the fixed sights; and bringing the index to be parallel with the other side of the instrument, observe by the sights upon it any accessible mark B, at a sensible distance: then, carrying the instrument to the point B, let the moveable sights be directed to the first station A, and the sights of the index to the point F. If the index cut the right side of the square, as in K, in the two triangles BRK, and BAF, which are equiangular, it will be as BR to RK, so BA (the distance of the stations to be measured with a chain) to AF; and the distance AF sought will be found by the rule of three. But if the index cut the reclined side of the square in any point L, where the distance of a more remote point is sought: in the triangles BLS, BAG, the side LS shall be to SB, as BA to AG, the distance sought; which accordingly will be found by the rule of three.

PROPOSITION V.

FIG. 6. To measure an accessible height by means of a plain mirror.—Let AB be the height to be measured; let the mirror be placed at C, in the horizontal plane BD, at a known distance BC; let the observer go back to D, till he see the image of the summit in the mirror, at a certain point of it, which he must diligently mark; and let DE be the height of the observator's eye. The triangles ABC and EDC are equiangular; for the angles at D and B are right angles; and ACB, ECD, are equal, being the angles of Part II.