The compleat surveyor containing the whole art of surveying of land by the plain table, theodolite, circumferentor, and peractor ... : together with the taking of all manner of heights and distances, either by William Leybourn.

THE COMPLEAT SURVEYOR. The First Book. (Book 1)

THE ARGƲMENT.

THis first Book consisteth of divers Definitions & Pro∣blemes Geometricall, ex∣tracted out of the Wri∣tings of divers ancient and modern Geometrici∣ans, as Euclid, Ramus, Clavius, &c. and are here so methodically disposed, that any man may gradually proceed from Probleme to Probleme without interruption, or being referred to any other Author for the Practicall performance of any of them. Onely the Demonstration is whol∣ly omitted; partly, because those Books, out of

which they were extracted, are very large in that particular, and also for the avoiding of many o∣ther Propositions and Theoremes, which (had the ensuing Problemes been demonstrated) must of ne∣cessity have been inserted. Also, the figures would have been so incumbred with multiplici∣ty of lines, that the intended Problemes would have been thereby much darkened. And besides it was not my intent in this place to make an ab∣solute or entire Treatise of Geometry, and there∣fore I have onely made choice of such Problems as I conceived most usefull for my present pur∣pose, and come most in use in the practice of Sur∣veying, and ought of necessity to be known by every man that intendeth to exercise himselfe in the Practice thereof, and those are chiefly such as concern the reducing of Plots from one forme to another, and to inlarge or diminish them according to any assigned Proportion, also divers of the Problemes in this Book will abun∣dantly help the Surveyor in the division and se∣peration of Land, and in the laying out of any assigned quantity, whereby large parcels may be readily divided into divers severals; and those again sub-divided if need be. Also for the better satisfaction of the Reader, I have performed di∣vers of the following Problemes both Arithme∣tically and Geometrically.

GEOMETRICALL DEFINITIONS.

1. A Point is that which cannot be divided.

[illustration]

A Point or Signe is that which is void of all Magni∣tude, and is the least thing that by minde and understanding can be imagined and conceived, than which there can be nothing lesse, as the Point or Prick noted with the letter A, which is neither quantity nor part of quantity, but only the terms or ends of quantity, and herein a Point in Geometry differeth from Unity in Number.

2. A Line is a length without breadth or thicknesse.

[illustration]

A Line is created or made by the moving or drawing out of a Point from one place to another, so the Line AB, is made by mo∣ving of a Point from A to B, and according as this motion is, so is the Line thereby created, whe∣ther streight or crooked. And of the three kindes of Magnitudes in Ge∣ometry, viz. Length, Breadth, and Thicknesse, a Line is the first, consist∣ing of Length only, and therefore the Line AB, is capable of divisi∣on in length only, and may be divided equally in the point C, or un∣equally in D, and the like, but will admit of no other dimension.

3. The ends or bounds of a Line are Points.

[illustration]

This is to be understood of a finite Line only, as is the line AB, the ends or bounds whereof are the points A and B: But in a Circular Line it is other∣wise,

for there, the Point in its motion returneth again to the place where it first began, and so maketh the Line infinite, and the ends or bounds thereof undeterminate.

4. A Right line is that which lieth equally between his points.

[illustration]

As the Right line AB lyeth streight and equall between the points A and B (which are the bounds thereof) without bowing, and is the shortest of all other lines that can be drawn between those two points.

5. A Superficies is that which hath only length and breadth.

[illustration]

As the motion of a point produceth a Line, the first kinde of Magnitude, so the motion of a Line produceth a Superficies, which is the se∣cond kinde of Magnitude, and is capable of two dimensions, namely, length and breadth, and so the Superficies ABCD may be divided in length from A to B, and also in breadth from A to C.

6. The extreams of a Superficies are Lines.

As the extreams or ends of a Line are points, so the extreams or bounds of a Superficies are Lines, and so the extreams or ends of the Superficies ABCD, are the lines AB, BD, DC, and CA, which are the terms or limits thereof.

7. A plain Superficies is that which lieth equally between his lines.

So the Superficies ABCD lieth direct and equally between his lines: and whatsoever is said of a right line, the same is also to be understood of a plain Superficies.

8. A plain Angle is the inclination or bowing of two lines the one to the other, the one touching the other, & not being directly joyned together.

As the two lines AB and BC incline the one to the other, and touch one another in the point B, in which point, by reason of the

inclination of the said lines, is made the Angle ABC. But if the two lines which touch each other be without in∣clination, and be drawn directly one to the other, then they make no angle at all, as the lines CD and DE, touch each other in the point D, and yet they make no angle, but one continued right line.

[illustration]

¶ And here note, that an Angle commonly is signed by three Letters, the middlemost whereof sheweth the angular point: As in this figure, when we say the angle ABC, you are to understand the very point at B: And note also, that the length of the sides con∣taining any angle, as the sides AB and BC, do not make the angle ABC either greater or lesser, but the angle still retaineth the same quantity be the containing sides thereof either longer or shorter.

9. And if the lines which contain the angle be right lines, then is it called a right lined angle.

So the angle ABC is a right lined angle, because the lines AB and BC, which contain the said angle, are right lines. And of right lined Angles there are three sorts, whose Definitions follow.

10. When a right line standing upon a right line maketh the angles on either side equall, then either of those angles is a right angle: and the right line which standeth erected, is called a perpendicular line to that whereon it standeth.

[illustration]

As upon the right line CD, suppose there do stand another right line AB, in such sort that it maketh the angles on either side thereof equall, namely, the angle ABD on the one side, equall to the angle ABC on the other side: then are either of the two angles ABC, and ABD right angles, and the right line AB, which standeth erected upon the right line CD, without inclining to either part thereof, is a perpendicular to the line CD.

11. An Obtuse angle is that which is greater than a right angle.

[illustration]

So the angle CBE is an obtuse angle, because it is greater than the angle ABC, which is a right angle; for it doth not only con∣tain that right angle, but the angle ABE also, and therefore is obtuse.

12. An Acute angle is lesse than a right angle.

So the angle EBD is an acute angle, for it is lesse than the right angle ABD (in which it is con∣tained) by the other acute angle ABE.

13. A limit or term is the end of every thing.

As a point is the limit or term of a Line, because it is the end there∣of, so a Line likewise is the limit and term of a Superficies; and a Superficies is the limit and term of a Body.

14. A Figure is that which is contained under one limit or term or many.

[illustration]

As the Figure A is contained under one limit or term, which is the round line. Al∣so the Figure B is con∣tained under three right lines, which are the limits or terms thereof. Likewise, the Figure C is contained under four right lines, the Figure E under five right lines, and so of all other figures.

¶ And here note, that in the following work we call any plain Superficies whose sides are unequall, (as the Figure E) a Plot, as of a Field, Wood, Park, Forrest, and the like.

15. A Circle is a plain Figure contained under one line, which is called a Circumference, unto which all lines drawn from one point within the Figure, and falling upon the Cir∣cumference thereof are equall one to the other.

[illustration]

As the Figure ABCDE is a Circle, contained under the crook∣ed line BCDE, which line is called the Circumference: In the middle of this Figure is a point A, from which point all lines drawn to the Circumference thereof are equall, as the lines AB, AC, AF, AD: and this point A is called the center of the Circle.

16. A Diameter of a Circle is a right line drawn by the Center thereof, and ending at the Cir∣cumference, on either side dividing the Circle into two equall parts.

So the line BAD (in the former Figure) is the Diameter there∣of, because it passeth from the point B on the one side of the Cir∣cumference, to the point D on the other side of the Circumference, and passeth also by the point A, which is the center of the Circle. And moreover it divideth the Circle into two equall parts, namely, BCD being on one side of the Diameter, equall to BED on the other side of the Diameter. And this observation was first made by Thales Miletius, for, saith he. If a line drawn by the center of any Circle do not divide it equally, all the lines drawn from the center of that Circle to the Circumference cannot be equall.

17. A Semicircle is a figure contained under the Diameter, and that part of the Circumference cut off by the Diameter.

As in the former Circle, the figure BED is a Semicircle, because it is contained of the right line BAD, which is the Diameter, and of the crooked line BED, being that part of the circumference which is cut off by the Diameter: also the part BCD is a Semicircle.

18. A Section or portion of a Circle, is a Figure contained under a right line, and a part of the circumference, greater or lesse then a semicircle.

[illustration]

So the Figure ABC, which consisteth of the part of the Cir∣cumference ABC, and the right line AC is a Section or portion of a Circle greater than a Semi∣circle.

Also the other figure ACD, which is contained under the right line AC, and the part of the cir∣cumference ADC, is a Section of a Circle lesse than a Semicircle.

¶ And here note, that by a Section, Segment, Portion, or Part of a Circle, is meant the same thing, and signifieth such a part as is either greater or lesser then a Semicircle, so that a Semicircle cannot properly be called a Section, Segment, or part of a Circle.

19. Right lined figures are such as are contain∣ed under right lines.

 

20. Three sided figures are such as are contained under three right lines.

 

21. Four sided figures are such as are contained under four right lines.

 

22. Many sided figures are such as have more sides than four.

 

23. All three sided figures are called Triangles.

[illustration]

And such are the Triangles BCD.

24. Of four sided Fi∣gures, a Quadrat or Square is that whose sides are equal and his angles right. [As the Figure A.]

[illustration]

25. A Long square is that which hath right an∣gles but unequal sides. [As the Figure B]

[illustration]

26. A Rhombus is a Figure having four equall sides but not right angles. [As the Figure C.]

[illustration]

27. A Rhomboides is a Figure whose opposite sides are equall, and whose opposite angles are also e∣quall, but it hath neither e∣quall sides nor equal angles. [As the Figure D.]

[illustration]

28. All other Figures of four sides (besides these) are called Trapezias.

[illustration]

Such are all Figures of four sides in which is observed no equality of sides or angles, as the figures A and B, which have neither equall sides nor equall angles, but are described by all adventures without the ob∣servation of any order.

29. Parallel, or equidistant right lines are such which being in one and the same Superficies and produced infinitely on both sides, do ne∣ver in any part concur.

[illustration]

As the right lines AB and CD are parallel one to the other, and if they were infinitely extended on either side would never meet or concur toge∣ther, but still retain the same distance.

Geometricall Theoremes.

• 1. ANy two right lines crossing one another, make the contrary or verticall angles equall.
• 2. If any right line fall upon two parallel right lines, it maketh the outward angles on the one, equall to the inward angles on the other, and the two inward opposite angles on contrary sides of the falling line also equall.
• 3. If any side of a Triangle be produced, the outward angle is equall to the two inward opposite angles, and all the three angles of any Trian∣gle are equall to two right angles.
• 4. In equiangled Triangles, all their sides are proportionall, as well such as contain the equall angles, as also the subtendent sides.
• 5. If any four Quantities be proportionall, the first multiplied in the fourth, produceth a Quantity equall to that which is made by multi∣plication of the second in the third.
• 6. In all right angled Triangles, the square of the side subtending the right angle, is equall to both the squares of the containing sides.
• 7. All parallelograms are double to the triangles that are described upon their bases, their altitudes being equall.
• 8. All triangles that have one and the same Base, and lie between two parallel lines, are equall one to the other.

GEOMETRICALL PROBLEMES.

PROBLEME I. Ʋpon a right line given, how to erect another right line, which shall be perpendicular to the right line given.

THe right line given is AB, upon which from the point E it is required to erect the perpendicular EH.

[illustration]

Opening your Compasses at pleasure to any conveni∣ent distance, place one foot in the assigned point E, and with the other make the marks C and D, equidistant on each side the given point E. Then open∣ing your Compasses again to any other convenient distance wider then the former, place one foot in C, and with the other describe the arch GG; also (the Compasses remaining at the same distance) place one foot in the point D, and with the other describe the arch FF, then from the point where these two arches intersect or cut each other (which is at H) draw the right line HE which shall be perpendicular to the given right line AB, which was the thing required to be done.

PROB. II. How to erect a Perpendicular on the end of a right line given.

[illustration]

LEt OR be a line given, and let it be required to erect the perpendicular RS. First, upon the line OR, with your Compasses opened to any small distance, make five small di∣visions beginning at R, noted with 1, 2, 3, 4, 5. Then take with your Compasses the di∣stance from R to 4, and placing one foot in R, with the other describe the arch PP. Then take the distance R 5, and placing one foot of the Compasses in 3, with the other foot describe the arch BB, cut∣ting the former arch in the point S. Lastly, from the point S, draw the line RS, which shall be perpendicular to the given line OR.

PROB. III. How to let fall a perpendicular, from any point as∣signed, upon a right line given.

[illustration]

THE point given is C, from which point it is required to draw a right line which shall be perpendicular to the given right line AB.

First, from the given point C, to the line AB, draw a line by chance, as CE, which divide into two equall parts in the point D, then placing one foot of the Compasses in the point D, with the di∣stance DC, describe the Semicircle CFE, cutting the given line AB in the point F. Lastly, if from the point C you draw the right line CF, it shall be a perpendicular to the given line AB, which was required.

PROB. IV. How to make an angle equall to an angle given.

[illustration]

LEt the angle given be ACB, and let it be required to make another angle equall thereunto.

First, draw the line EF at pleasure, then upon the given an∣gle at C, (the Com∣passes opened to any di∣stance) describe the ark AB, also, upon the point F (the Compasses un-altered) describe the arke DE: then take with your Compasses the distance AB, and set the same distance from E to D. Lastly, draw the line DF, so shall the angle DFE be equall to the given angle ACB.

PROB. V. A right line being given, how to draw another right line which shall be parallel to the former, at any distance required.

[illustration]

THe line given is AB, unto which it is required to draw ano∣ther right line parallel thereunto, at the distance AC, or BD.

First, Open your Com∣passes to the distance AC or AD, then placing one foot in A, with the other describe the arke C; also, place one foot in B, and with the other describe the arch D. Lastly, Draw the line CD so that it may only touch the arks C and D, so shall the line CD be parallel to the line AB, and at the distance required.

PROB. VI. To divide a right line given into any number of equall parts.

LEt AB be a line given, and let it be required to divide the same into four equall parts.

[illustration]

First, From the end of the given line A, draw the line AC, making any angle, then from the other end of the given line, which is at the point B, draw the line BD parallel to AC, or make the angle ABD e∣quall to the angle CAB; then upon the lines AC and BD set off any three equall parts (which is one lesse then the number of parts into which the line AB is to be divided) on ••ace line, as 1 2 3, then draw lines from 1 to 3, from 2 to 2, and from 3 to 1, which lines, crossing the given line AB, shall divide it into four equall parts as was required.

PROB. VII. A right line being given, how to draw another right line parallel thereunto, which shall also passe through a point assigned.

[illustration]

LEt AB be a line given, and let it be required to draw another line parallel thereunto which shall passe through the given point C.

First, Take with your com∣passes the di∣stance from A to C, and place∣ing one foote thereof in B, with the other describe the ark DE; then take in your compas∣ses the whole line AB, and placing one foot in the point C, with the other de∣scribe the arke FG, crossing the former arke DE in the point H. Lastly, if you draw the line CH it shall be parallel to AB.

PROB. VIII. Having any three points given, which are not situate in a right line, how to finde the center of an arch of a Circle which shall passe directly through the three given points.

[illustration]

THe three points given are A B and C, now it is required to finde the center of a Circle, whose circumference shall passe through the three points given.

First, Opening your Compasses to any di∣stance greater then halfe BC, place one foot in the point B, and with the other de∣scribe the arch FG, then, the Compasses remaining at the same distance, place one foot in C, and with the o∣ther turned about make the marks F and G in the former arch, and draw the line FG at length if need be.

Again, opening the Compasses to any di∣stance greater then halfe AB, place one foot in the point A, and with the other describe the arch HK, then, the Compasses re∣maining at the same distance, place one foot in the point B, and turning the other about make the marks HK in the former arch.

Lastly, draw the right line HK cutting the line FG in O, so shall O be the center upon which if you describe a Circle at the di∣stance of OA, it shall passe directly through the three given points A B C, which was required.

PROB. IX. Any three right lines being given, so that the two shortest together be longer then the third, to make thereof a Triangle.

LEt it be required to make a Triangle of the three lines A B and C, the two shortest whereof, viz. A and B together, are longer then the third line C.

[illustration]

First, Draw the line DE equall to the given line B, then take with your Compasses the line C, and setting one foot in E, with the other describe the arch HG, also, take the given line A in your Compasses, and placing one foot in D, with the other describe the arch HF, cutting the for∣mer arch HG in the point H. Lastly, if from the point H you draw the lines HE and HD, you shall constitute the Triangle HDE, whose sides shall be equall to the three given lines A B C.

PROB. X. Having a right line given, how to make a Geome∣tricall Square, whose side shall be equall thereunto.

[illustration]

THe line given is QR, and it is required to make a Geome∣tricall Square whose side shall be equall to the line QR.

First, Draw the line AB, making it equall to the given line QR, then (by the first or second Probleme) upon the point B raise the perpendicular BC, making the line BC equall to the given line QR also. Then taking the line QR in your Compasses, place one foot in C, and with the other describe the arch D, also the Compas∣ses so resting, place one foot in A, and with the other de∣scribe another arch crossing the former in the point D. Lastly, draw the lines DC and DA, which shall include the Geome∣tricall Square ABCD.

PROB. XI. Two right lines being given, how to finde a third right line which shall be in proportion unto them.

[illustration]

LEt the two given lines be A and B, and let it be re∣quired to finde a third line which shall be in proportion unto them.

First, Draw two right lines making any angle at pleasure, as the lines OP and ON, making the angle PON; then taking the line A in your Com∣passes, set the length thereof from O to S, also, take the line B in your Compasses, and set the length thereof from O to R, and also from O to D, then draw the right line SD, and from the point R draw the line RC parallel to SD, so shall OC be the third proportionall required, for,

• As OS to OD ∷ so OR to OC.
• As 8 to 12 ∷ so 12 to 18.

PROB. XII. Three right lines being given, to finde a fourth in proportion to them.

[illustration]

THe three lines given are A B C, unto which it is re∣quired to finde a fourth proportion∣all line. This is to perform the rule of three in lines.

As in the last Problem, you must draw two lines ma∣king any angle, as the angle DEF. Then take the line A in your Compasses, and set it from E to G, then take the line B in your Compasses and set

that length from E to H. Then take the third given line in your Compasses, and set that from E to K, and through that point K draw the line KL parallel to GH, so shall the line EL be the third pro∣portionall required; for,

• As EG to EH ∷ so EK to EL.
• As 24 to 28 ∷ so 36 to 42.

¶ Here note that in the performance of this Probleme, that the first and the third termes (namely the lines A and C) must be set upon one and the same line, as here upon the line ED, and the second term (namely the line B) must be set upon the other line EF, upon which line also the fourth proportional EL will be found.

PROB. XIII. To divide a right line given into two parts, which shall have such proportion one to the other as two given right lines.

[illustration]

THe line given is AB, and it is required to divide the same into two parts, which shall have such proportion one to the other, as the line C hath to the line D.

First, from the point A, draw the line AE, at pleasure, making the angle EAB; then take in your Compas∣ses the line C, and set it from A to F, also take the line D, and set it from F to E, and draw the line EB, then from the point F, draw the line FG parallel to EB, cutting the given line AB in the point G; 〈◊〉〈◊〉 is the line AB divi∣ded into two parts in the point G, being in proportion one to the other, as the line C is to the line D; for,

• As AE to AB ∷ so AF to AG.

Arithmetically.

LEt the line AB contain 40 Perches, and let the line C be 20, and the line D 30; and let it be required to divide the line AB into two parts, being in proportion one to the other, as the line C is to the line D.

First, Adde the lines C and D together, their summe is 50, then say by the Rule of Proportion: If 50 (which is the summe of the two given terms) give 40 the whole line AB, what shall 30, the greater given term give? Multiply and divide, and you shall have in the quotient 24 for the greater part of the line AB, which being taken from 40 the whole line, there remains 16 for the other part AG; for,

• As AE to AB ∷ so FE to GB.
• As 50 to 40 ∷ so 30 to 24.

PROB. XIV. How to divide a Triangle into two parts, according to any proportion assigned, by a line drawn from any angle thereof, and to lay the lesser part towards any side assigned.

[illustration]

LEt ABC be a Triangle given, and let it be required to divide the same, by a line drawn from the angle A, into two parts, the one bearing proportion to the other, as the line F doth to the line G, and that the lesser part may be towards the side AB.

By the last Probleme divide the base of the Triangle BC in the point D, in proportion as the line F is to the line G, (the lesser part being set from B to D.) Lastly, draw the line AD, which shall divide the Triangle ABC in proportion as F to G; for,

• As the line F, is to the line G;
• So is the Triangle ADC; to the Triangle ABD.

PROB. XV. The Base of the Triangle being known, to perform the foregoing Probleme by Arithmetick.

[illustration]

SUppose the Base of the Triangle BC to be 40, and let the proportion into which the Triangle ABC is to be divided, be as 2 to 3.

First, Adde the two proportionall terms together, 2 and 3, which makes 5, then say by the rule of proportion: If 5 (the sum of the proportionall terms,) give 40 (the whole base BC,) what shall 3 (the greater term) give? Multiply and divide, and the quotient will give you 24 for the greater segment of the Base DC, which being deducted from the whole base 40, there will remain 16 for the lesser segment BD.

PROB. XVI. How to divide a Triangle, whose area or content is known, into two parts, by a line drawn from an angle assigned, according to any proportion re∣quired.

LEt the Triangle ABC contain 8 Acres, and let it be required to divide the same into two parts, by a line drawn from the angle A, the one to contain 5 Acres, and the other 3 Acres.

First, Measure the whole length of the Base, which supppse 40, then say, If 8 Acres (the quantity of the whole Triangle) give 40, (the whole Base,) what parts of the Base shall 5 Acres give? Mul∣tiply and divide, the Quotient will be 25 for the greater segment of the base CD, which being deducted from 40 (the whole Base,) there will remain 15 for the lesser segment of the Base BD, then

draw the line AD, which shall divide the Triangle ABC according to the proportion required.

PROB. XVII. How to divide a Triangle given into two parts, ac∣cording to any proportion assigned, by a line drawn from a point limited in any of the sides thereof: and to lay the greater or lesser part towards an angle assigned.

[illustration]

THe Triangle given is ABC and it is required from the point E, to draw a line that shall divide the Triangle into two parts, being in proportion one to the other as the line I is to the line K, and to lay the lesser part towards B.

First, From the limited point E, draw a line to the opposite angle at A; then divide the base BC in proportion as I to K, which point of division will be at D, then draw DF parallel to AE. Lastly, from F, draw the line FE, which will divide the Triangle into two parts being in proportion one to the other as the line I is to the line K.

PROB. XVIII. To perform the foregoing Probleme Arithmetically.

IT is required to divide the Triangle ABC, from the point E, in∣to two parts in proportion as 5 to 2.

First, Divide the base BC according to the given proportion, then (because the lesser part is to be laid towards B) measure the di∣stance from E to B, which admit 30, then say by the rule of Propor∣tion; If EB 30, give DB 15, what shall AG 29 (the perpendi∣cular of the Triangle) give? Multiply and divide, the Quotient will

be 14½, at which distance draw a parallel line to BC, namely F, then from F draw the line FE, which shall divide the Triangle accor∣ding to the required proportion.

PROB. XIX. How to divide a Triangle, whose area or content is known, into two parts, by a line drawn from a point limited in any side thereof, according to any number of Acres, Roods and Perches.

IN the foregoing Triangle ABC, whose area or content is 5 Acres, 1 Rood, let the limited point be E in the base thereof, and let it be required from the point E to draw a right line which shall divide the Triangle into two parts between M and N, so that M may have 3 Acres, 3 Roods thereof, and N may have 1 Acre and 2 Roods thereof.

First, reduce the quantity of N (being the lesser) into perches, which makes 240, then (considering on which side of the limited point E this part is to be laid, as towards B) measure that part of the base from E to B 30 Perches, whereof take the halfe, which is 15, and thereby divide 240, the part belonging to N; the quotient will be 16, the length of the perpendicular FH, at which parallel distance from the base BC cut the side AB in F, from whence draw the line FE which shall cut off the Triangle FBE, containing 1 Acre, 2 Roods, the part belonging to N, then will the Trapezia AFEC (which is the part belonging to M) contain the residue, namely, 3 Acres, 3 Roods.

PROB. XX. How to divide a Triangle according to any propor∣tion given, by a line drawn parallel to one of the sides.

THe following Triangle ABC is given, and it is required to divide the same into two parts by a line drawn parallel to the side AC, which shall be in proportion one to the other, as the line I is to the line K.

First (by the 13 Probleme) divide the line BC in E, in propor∣tion as I to K, then (by the 24 Probleme following) finde a mean proportionall between BE and BC, which let be BF, from which point F, draw the line FH parallel to AC, which line shall divide the Triangle into two parts, viz. the Trapezia AHFC, and the

Triangle HFB, which are in proportion one to the other as the line I is to the line K.

PROB. XXI. To perform the foregoing Probleme Arithmetically.

[illustration]

LEt the Triangle be ABC, and let it be required to divide the same into two parts, which shall be in proportion one to the other, as 4 to 5, by a line drawn parallel to one of the sides.

First, Let the base BC containing 54 be divided according to the proportion given, so shall the lesser segment BE contain 24, and the greater EC 30; then finde out a mean proportionall line between BE 24, and the whole base BC 54, by multiplying 54 by 24, whose product will be 1296, the square root whereof is 36, the mean proportionall sought, which is BF, then, by the rule of pro∣portion say: If BF 36 give BE 24, what AD 36? the answer is HG 24, at which distance draw a parallel line to the base, to cut the side AB in H, from whence draw the line HF parallel to AC, which shall divide the Triangle as was required.

PROB. XXII. To divide a Triangle of any known quantity, into two parts, by a line drawn parallel to one of the sides, according to any number of Acres, Roods, and Perches.

THe Triangle given is ABC, whose quantity is 8 Acres, 0 Roods, 16 Perches, and it is required to divide the same (by a line drawn parallel to the side AC) into two parts, viz. 4 Acres, 2 Roods, 0 Perches, and 3 Acres, 2 Roods, 16 Perches.

First, Reduce both quantities into perches (as is hereafter taught) and they will be 720, and 576, then reduce both those numbers, by abbreviation, into the least proportionall terms, viz. 5 and 4, and according to that proportion, divide the base BC 54 of the given Triangle in E, then seeke the mean proportionall between BE and BC, which proportionall is BF 36, of which 36 take the halfe, and thereby divide 576, the lesser quantity of Perches, the Quotient will be HG 32, at which parallel distance from the base, cut off the line AB in H, from whence draw the line HF parallel to the side AC, which shall divide the Triangle given according as was required.

PROB. XXIII. From a line given, to cut off any parts required.

[illustration]

THe line given is AB, from which it is required to cut off 3/7 parts. First, draw the line AC, making any angle, as CAB, then from A, set off any seven equall parts, as 1 2 3 4 5 6 7, and from 7 draw the line 7B. Now because 3/7 is to be cut off from the line AB, therefore from the point 3, draw the line 3D parallel to 7B, cut∣ting the line AB in D, so shall AD be 3/7 of the line AB, and DB shall be 4/7 of the same line; for,

• As A7, is to AB ∷ so is A3, to AD.

PROB. XXIV. To finde a mean proportionall between two lines given.

IN the following figure, let the two lines given be A and B, be∣tween which it is required to finde a mean proportionall.

Let the two given lines A and B, be joyned together in the point E, making one right line, as CD, which divide into two e∣quall parts in the point G, upon which point G, with the distance GC or GD, describe the Semicircle CFD; then, from the point E, (where the two lines are joyned together) raise the perpendicular EF, cutting the Periferie of the Semicircle in F, so shall the line

EF be a mean proportionall between the two given lines A and B; for,

• As ED to EF ∷ so EF to CF.
• As 9 to 12 ∷ so 12 to 16.

PROB. XXV. How to divide a line in power according to any proportion given.

[illustration]

IN this figure let CD be a line given to be divided in power as the line A is to the line B.

First, divide the line CD in the point E, in pro∣portion as A to B, (by the 13 Probleme:) then di∣vide the line CD into two equall parts in the point G, and on G, at the di∣stance GC or GD, de∣scribe the Semicircle CFD, and upon the point E, raise the perpendicular EF, cutting the Semicircle in F: Lastly, draw the lines CF and DF, which together in power shall be equall to the power of the given line CD, and yet in power one to the other as A to B.

PROB. XXVI. How to inlarge a line in power, according to any proportion assigned.

IN the former figure, Let CE be a line given, to be inlarged in power as the line B to the line G.

First, (by the 13 Probleme) finde a line in proportion to the given line CE, as B is to G, which will be CD, upon which line describe the Semicircle CFD, and on the point E, erect the perpendicular EF; then draw the line CF, which shall be in power to CE, as G to B.

PROB. XXVII. To inlarge or diminish a Plot given, according to any proportion required.

LEt ABCDE be a Plot given, to be diminished in power as L to K.

Divide one of the sides (as AB) in power as L to K, in such sort, that the power of AF, may be to the power of AB, as L to K. Then from the angle A, draw lines to the points C and D, that done, by F draw a parallel to BC, to cut AC in G, as FG. Again, from G, draw a parallel to CD to cut AD in H. Lastly, from H, draw a parallel to DE, to cut AE in I, so shall the plot AFGHI be like ABCDE, and in proportion to it, as the line L, to the line K, which was required.

[illustration]

Also, if the lesser Plot were given, and it were re∣quired to make a greater in proportion to it as K to L. Then from the point A, draw the lines AC and AD, at length, also increase AF and AI: that done, inlarge AF in power as K to L, which set from A to B, then by B draw a paral∣lel to FG to cut AC in C, as BC. Likewise from C draw a parallel to GH, to cut AD in D, as CD. Lastly, a parallel from D to HI, as DE, to cut AI being increased in E, so shall you include the Plot ABCDE, like AFGHI, and in proportion thereunto, as the line K is to the line L, which was required.

PROB. XXVIII. How to make a Triangle which shall contain any number of Acres, Roods and Perches, and whose base shall be equal to any (possible) number given.

IF it be required to make a Triangle which shall contain 5 Acres, 2 Roods, 30 Perches, whose base shall contain 50 Perches, you must first reduce your 5 Acres, 2 Roods, 30 Perches, all into Perches in this manner.

First, (because 4 Roods make one Acre) multiply your 5 Acres by 4 which makes 20, to which adde the two odde Roods, so have

you 22 Roods in your 5 Acres 2 Roods. Then (because 40 Perch∣es make one Rood) multiply your 22 Roods by 40, which makes 880 Perches, to which adde the 30 odde Perches, and you shall have 910, and so many Perches are contained in 5 Acres, 2 Roods, 30 Perches.

[illustration]

Now to make a Triangle which shall contain 910 perches, & whose base shall be 50 Perches, do thus, Double the num∣ber of perches gi∣ven, namely 910, and they make 1820, then be∣cause the base of the triangle must contain 50 Perches, divide 1820 by 50, the quotient will be 36⅖, which will be the length of the perpendicular of your Triangle. This done, From any equall Scale lay down the line AB equall to 50 Perches, then upon B, raise the perpendicular BD equal to 36⅖ perch∣es, and draw the line CD parallel to AB then, from any point in the line CD (as from E) draw the lines EA and EB, including the Triangle AEB, which shall contain 5 Acres, 2 Roods, 30 Perches, which was required.

PROB. XXIX. How to reduce a Trapezia into a Triangle, by a line drawn from any angle thereof.

[illustration]

THe Trapezia gi∣ven is ABCD, and it is required to reduce the same into a Triangle.

First, Extend the line DC, and draw the Di∣agonall BD, then from the point A, draw the line AE parallel to BD, extending it till it cut the side CD in the point E. Lastly, from B, draw the line BE, constituting the Tri∣angle EBC, which shall be equall to the Trapezia ABCD.

PROB. XXX. How to reduce a Trapezia into a Triangle, by lines drawn from any point in any of the sides thereof.

[illustration]

LEt ABCD be a Trapezia given, and let H be a point in one of the sides thereof, from which point H let it be requi∣red to draw lines which shall reduce the Trapezia into a Triangle.

First, Extend the side which is opposite to the given point, namely, the side CD, both wayes to E and F, and then from the point H, draw lines to the angles C and D, as the lines HC and HD; also, draw the lines AE and BF parallel to HC and HD, cutting the extend∣ed line CD in the points E and F. Lastly, If from the point H you draw the lines HE and HF, you shall constitute the Triangle HEF, which shall be equall to the Trapezia ABCD.

PROB. XXXI. How to reduce an irregular Plot of five sides into a Triangle.

[illustration]

THe irregular Plot given is ABCDE, and it is required to reduce the same into a Triangle.

First, extend the side AE both wayes to F and G, and from the angle C, draw the lines CA and CE, to the angles A and E. Then from the point B, draw the line BF parallel to CA

cutting the extended side AE, in F; also, from the point D, draw the line DG parallel to CE, cutting also the extended side in G. Last∣ly, from the angle C, draw the lines CF and CG, constituting the Triangle CFG which is equall to the Plot ABCDE.

PROB. XXXII. A Trapezia being given, how from any angle there∣of of to divide the same into two parts being in pro∣portion one to the other as two given right lines, and to set the part cut off towards an assigned side.

[illustration]

LEt the Trapezia given be ABCD, and let it be required to draw a line from the angle B, which shall divide the Tra∣pezia into two parts, being in proportion one to the other, as the line G is to the line H, and that the lesser part of the Figure cut off, may be towards the side AB.

First (by the 29 Probleme) reduce the Trapezia ABCD into a Triangle, by drawing the line BF from the assigned angle, thereby constituting the Triangle ABF, equall to the Trapezia ABCD: this done, divide the base of the Triangle AF in proportion as G to H, which will be in the point E. Lastly, draw the line BE, which shall divide the Trapezia in proportion as G to H. Now because the lesser part of the Trapezia was to be set towards the side AB, therefore the lesser part of the line must be set from A to E. Here note that the same manner of working is to be observed, if it had been required to divide the Trapezia by a line drawn from any of the other angles.

PROB. XXXIII. A Trapezia being given, how, from a point limited in any side thereof, to draw a line which shall di∣vide the same into two parts in proportion as two given lines.

[illustration]

THe Trapezia given is ABCD, and it is required from the point H, to draw a line which shall divide the Trapezia in proportion as O to Q.

First, Prolong the side CD, and reduce the whole Trapezia into the Triangle HEF by the 30 Pro∣bleme, then divide the line EF in proportion as O to Q, which will fall in the point G, therefore draw the line HG which shall divide the Trapezia into two parts in proportion as O to Q, which was required,

PROB. XXXIV. A Trapezia being given, how to divide the same into two parts in proportion as two lines given, and so that the line of partition may be parallel to any side thereof.

THe Trapezia given is ABCD, and it is required to divide the same into two parts, which shall be in proportion one to the other as the line K is to the line L, and that the line of partition may be parallel to the side BD.

Consider first, through which sides of the Trapezia the line of partition will passe, as in this Figure it will passe through the sides AB and CD (because parallel to BD,) therefore, extend the sides AB and CD, till they concur in E, then (by the 32 Probleme) re∣duce the Trapezia ABCD into the Triangle BGD, whose base is GD, which line GD, divide in the point H in proportion as K to L; so that,

• As K to L ∷ So DH to HG.

[illustration]

This done, finde a mean proportionall between ED and EH (by the 24 Probleme) as ER. Lastly, through this point R, draw the line RF parallel to BD, which shall divide the Trapezia into two parts being in proportion one to the other, as the line K is to the line L, and with a line parallel to the side BD, which was required.

But if it had been required to divide the Trapezia by a line drawn parallel to the side CD, then the lines CA and DB must have been extended, but the rest of the work must be performed as is before taught.

PROB. XXXV. The figure of a Plot being given, how to divide the same into two parts, being in proportion one to the other as two given lines are, with a line drawn from an angle assigned.

LEt the following Figure ABCDE represent the Plot of a Field or such like, and let it be required to divide the same into two parts, being in proportion one to the other as the line R is to the line S, by a line drawn from the angle B.

First, Reduce the Plot ABCDE into the Triangle BFG, (by the 31 Probleme) so shall the line FG be the base of a Triangle equall to the given Plot, then (by the 13 Probleme) divide this line FG into two parts in the point H, in proportion one to the other, as the line R is to the line S; so that,

• As R to S ∷ so GH to HF.

Lastly, draw the line BH, which shall divide your given Plot into two parts which shall have such proportion one to the other, as the line R hath to the line S.

[illustration]

PROB. XXXVI. How to divide a Triangle into any number of equall parts, by lines drawn from a point given in any side thereof.

[illustration]

LEt it be required to divide the Triangle ABC into five equall parts, by lines drawn from the point D.

First, From the given point D, to the opposite angle B, draw the line DB, then divide the side AC of the Triangle into five equall parts, at E F G and H, and through each of those points draw lines parallel to DB, as EM, FL, GK, and HI: then from the point D, draw the lines DI, DK, DL, and DM, which shall divide the Triangle ABC into five equall parts from the point D, as was required.

PROB. XXXVII. How to divide an irregular Plot of six sides, into two parts, according to any assigned proportion, by a right line drawn from a point limited in any of the sides thereof.

[illustration]

THe irregular Plot given is ABCDEF, and it is required to divide the same into two parts, being in proportion one to the other, as the line R is to the line S.

First, Draw the right line HK, and (by the 30 Probleme) re∣duce the Trapezia ABFG into the Triangle HGK, then divide ••he base thereof, namely HK, into two parts in proportion as R to •…•…, which will be in the point O, then draw the line GO, which will divide the Trapezia ABFC into two parts in proportion one ••o the other, as the line R is to the line S.

Secondly, From the point O (by the 31 Probleme) reduce the Trapezia FCED into the Triangle OLM, and divide the base thereof, namely LM, in the point N, in proportion as R to S, and draw the line ON, which will divide the Trapezia FCED into two parts in proportion as R to S: and by this means is the whole Plot ABCDEF divided into two parts in proportion as R to S, by the lines GO and ON. But it is required to resolve the Pro∣bleme by one right line only drawn from the point G, therefore, from the point G, draw the line GN, and through the point O, draw the line OP parallel to GN: and lastly, from G, draw the right line GP, which shall divide the whole Plot ABCDEF into two parts, being in proportion one to the other as the line T is to the line S.

PROB. XXXVIII. How to divide an irregular Plot according to any proportion, by a line drawn from any angle thereof.

[illustration]

LEt ABCDEFG be an irregular Plot, and let it be required to divide the same into two equall parts, by a line drawn from the angle A.

First, draw the line HK, dividing the Plot into two parts, namely, into the five sided figure ABCFG, and into the Trapezia FCED, then (by the 31 Probleme) reduce the five sided figure ABCFG into the Triangle HAK, the base whereof HK divide into two equall parts in O, and draw the line OA, which shall divide the five sided figure ABCFG into two equall parts. Then (by the 30 Probleme) reduce the Trapezia FCDE into the Triangle OLM, and divide the base thereof LM into two equall parts in the point P, and draw the line OP, which will divide the Trapezia FCDE into two equall parts, and so is the whole Plot divided into two equall parts by the lines AO and OP, but to performe the Probleme by one right line only, do thus, from the point A, draw the line AP, and parallel thereunto, through the point O, draw the line ON. Lastly, if you draw a right line from A to N, it shall divide the whole Plot into two equall parts.

The end of the First Book.