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For by similar triangles ABC, ACD, ADE, ..........

we have

AB AC AC: AD :: AD : AE :: AE : AF ::

This construction is adapted to the antecedent AB, being the less of the given lines: when AC the greater is the antecedent, the construction will be reversed, as is exemplified in the unlettered lines of the figure *.

PROBLEM XIII.

To divide a given line AB in extreme and mean ratio.

Ar one extremity B draw a perpendicular, and take in it BC = AB: with centre C and radius CB describe a circle, and join AC meeting it in D. With centre A and radius AD describe a circle cutting AB in E. Then AB is divided in E in extreme and mean ratio; or such that BA AE :: AE: EB.

:

F

C

For, produce AC to meet the circle again at F. Then, since AB is at right angles to BC, therefore it is a tangent to the circle at B, and BA : AD :: AF: AB; or since AD = AE, and AB = 2BC = DF, we have AB: AE :: AB + AE: AB, and, dividendo, AB: AE :: AE : EB.

PROBLEM XIV.

(See preceding figure.)

To extend a given line AB so that AB shall be the greater segment of the whole line divided in extreme and mean ratio; or, to extend a given line in extreme and mean ratio.

CONSTRUCT as in the last problem, but set off AF in prolongation of AB to G. Then GB: BA :: BA: AG, and the line AB is produced in extreme and mean ratio.

For, since BA : AE :: AE : EB, by comp. BA + AE : BA :: AE + EB : AE; that is, GA: BA :: BA : EA.

PROBLEM XV.

To divide a given line AB harmonically in a given ratio, M: N.

DRAW any two lines through A and B parallel to each other, and in them take AG, AH, each equal to M, and BK equal to N. Join GK, HK, meeting AB in D and C: then the line AB will beharmonically divided in C and D in the given ratio of M to N. For AC

H

G

K

B

M

N

CB:: AH: BK:: AG: BK :: AD: DB.

* An instrument founded on this principle was proposed by Descartes for the construction of a series of mean proportionals, but it has never received any improvement, and is mechanically inadequate to the purpose, though theoretically correct in principle.

PROBLEM XVI.

Through a given point P to draw a line which shall make equal angles with two given lines AB, CD, whose intersection K would be beyond the limits of the picture.

THROUGH P draw PD, PE, parallel to AB, CD, respectively, and take in them equal distances, PF, PG, PH, as in the figure, of which the dark lines represent the edges of the paper. Through P draw PL parallel to GH, and MN parallel to FH. These will be the lines sought.

G

E

B

H

K

For, since PE is parallel to CD, the angle EPL is equal to PLD, and since PD is parallel to AK, the angle DPL is equal to EBP. But by problem iv. the angle EPL is equal to DPL; and hence EPB is equal to PLK.

In the same way we may prove that EMP is equal to DNP.

PROBLEM XVII.

To draw through a given point C a straight line which shall tend to the intersection of two given lines BD, AE, but whose point of intersection falls beyond the limits of the drawing.

First method. Through C draw any line whatever meeting the given lines BD, AE, in B and A, and any other line EG parallel to it meeting AE in E. B Draw any two parallels AD, EF, meeting BF in D and F. Then if FG be drawn parallel to CD meet

D

H

ing EG in G, and CG be drawn, it will tend to the intersection of BD, AE. For, by parallels, DE: BA :: DF: DB :: DG; BC, and hence AB BC :: ED: DG, and the lines AE, BD, CG, will, if produced, meet in the same point H *.

Scholium. The usual process for constructing this problem, hitherto adopted in England, is as follows:

Draw any line AC through the given point C and any other line EG parallel to it. Find DG a fourth proportional to AB, BC, ED. The line CG being drawn is that sought. It is the method of finding DG adopted above, that constitutes the improvement of the process.

Second method. Draw any two parallels CA, GE, cutting the given lines as before. Join AD, BE, meeting in F, and through F draw QF parallel to AB or DE; and join CE meeting it in K, and AK meeting ED in G. Then CG will tend to the intersection of BD, AE.

:

For, AB BC:: QF : FK :: ED: DG, and hence, as before, the three lines AE, BD, CG, tend to one point, H.

B

H

* This solution was first given, so far as he is aware, by the present editor of the Course, in the Monthly Magazine for August 1825. It is here divested of the technicalities of perspective, the original one having been published as the solution of a problem of frequent occurrence in drawings, and therefore given in a form exclusively adapted to that particular art. That is, however, almost the only case in which it particularly occurs in practice. It may be remarked, that the point D, to which the arbitrary line from A is drawn, may be taken anywhere in BH; and it is only taken in EG for the convenience of having as few marks as possible in the line BH.

Third method. Through C draw any line AP, and from any point P in it draw any other line PE, and let the intersections be as in the figure. Join AD, BE meeting in F, and draw PFQ; and through the intersection K of CE and PQ draw AK meeting PE in G. Then CG being joined, will tend to the point H, in which AE, BD, intersect.

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The truth of this construction follows from the demonstration of th. 97, p. 339 *.

PROBLEM XIX.

On a given line AB to describe a square.

First method. At A draw AC perpendicular to AB, and take AD equal to AB. Draw DE, BE, parallel to AB, AD: then ABED is the square required (def. 39) †.

G

Second method. Produce AB till AG is equal to AB. With any radius greater than AG or AB describe arcs intersecting in C; join AC, and take AD equal to AB. With centres B and D describe arcs intersecting in E, and join DE, BE; then ADEB is the square required ‡.

PROBLEM XX.

To make a rectangle whose length and breadth shall be given lines AB, CD.

AT A make AE perpendicular to AB and equal to CD, and draw the parallels as in the last problem.

Or a course of constructions analogous to the second construction of the last proposition may be employed.

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*These methods all effect the purpose very completely, and without any bye-work beyond what appears in the figures. The latter methods are much used by the continental draughtsmen; and the last especially, being independent of parallels, is much valued by them. As, however, these problems are never performed but under advantageous circumstances for drawing, it seems to partake of affectation to reject the use of instruments so simple as the parallel ruler. Besides, in comparison of the work which must be done in executing the process by the different methods, it will appear that each of the two last has more than the first; and what is of greater consequence, it is work of that kind to injure the picture itself.

Mr. Nicholson invented two instruments for the purpose of practically solving this problem, which he called centrolineads, and for which he received rewards from the Society of Arts. The parallel ruler does the work as effectively and rapidly by the first method, as the centrolinead does; and being a simple instrument, is less liable to derangement.

This method is convenient when instruments for perpendiculars and parallels are available. The most simple is the Marquois.

This method does not differ in principle from the last; the additional work prescribed being only that necessary for finding the perpendicular AC and the parallels DE, BE, when the ruler and compasses only are used.

PROBLEM XXI.

To make a square which shall be equal to any number of given squares, viz. those whose sides are A, B, C, D, E, ...

DRAW two lines HG, GK, at right angles to each other;

M

in GH take GL equal to A, and in GK take GM equal to B, and join ML. Then in GH take GN equal to ML, and GP equal to C, and join PN. Again, in GH take GQ equal to NP, and in GK take GR equal to D, and join RQ. Proceed thus till all the given lines have been employed in the HSQN Í GABCDE construction. The square described upon the last, as ST,

is equal to the sum of the squares on A, B, C, D, E: as is evident from successive applications of th. 34.

PROBLEM XXII.

To describe a square equal to the difference of two squares, viz. of those on

A and B.

DRAW two lines HG, GK, at right angles to each other; in GK take GC equal to the less of the given sides, B; and with centre C and radius equal to A, describe an arc EF cutting GH in D. Then (cor. th. 34) the square of IG is the difference of the squares on A and B.

I

PROBLEM XXIII.

To describe a triangle equal to a given rectilineal figure ABCDEFG.

HAVING fixed upon the line, as AG, which is to be the base of the triangle required, find the highest point, D, in reference to the base (by sliding the parallel ruler is the most ready method), and draw line DA or DG, separating the polygon into two polygons of equal altitude.

P

C

B

Η

E

M

Produce AG both ways. Through B draw BB' parallel to AC, meeting AG in B', and join B C; through C draw CC' parallel to B'D meeting AG in C', and join C'D; and proceed thus till the highest point of the figure is attained. Then, commencing at the other end G of AG, draw FF' parallel to GE meeting AG in F', and join EF'; parallel to DF' draw EE' meeting AG in E', and join DE'. Proceed thus till the highest point is attained, as at D. Then the triangle C'DE' is that required.

For the triangle CB'A is equal to the triangle CBA (th. 25), and hence the quadrilateral DCB'A is equal to the quadrilateral DCBA. Again, the triangle DC'B' is equal to the triangle DCB', and hence the triangle DCA' is equal to the quadrilateral DCB'A, or to the quadrilateral DCBA. In like manner, the triangle DAE' is equal to the pentagon ADEFG: and hence, the triangle C'DE is equal to the given polygon.

VOL. I.

CC

PROBLEM XXIV.

To construct a parallelogram equal to a given polygon ABCDEFG, and which shall have an angle equal to a given angle HKG (see figure to last problem). CONSTRUCT the triangle C'DE' as in the last problem. Through D draw LM parallel to AG, and make the angle GC'L equal to the given angle HKG, and draw E'M parallel to C'L. Draw the diagonals LE', MC', meeting at N, and through N draw PQ parallel to AG. Then PC'E'Q is the parallelogram required, as is obvious by th. 26, cor. 2.

PROBLEM XXV.

To find a square equal to a given parallelogram.

THE side of a square is a mean proportional between the two sides of the rectangle (th. 87); and constructions for the mean proportional have been given in problem 11.

When the parallelogram is not a rectangle, a rectangle equal to it stands upon the same base and between the same parallels (th. 25), into which, therefore, the parallelogram may be converted, and the square equal to it found as above.

PROBLEM XXVI.

To describe a parallelogram one of whose angles BAD, and one of whose sides AE are given, and which shall itself be equal to a given polygon.

CONSTRUCT a parallelogram ABCD, (prob. 24,) equal to the given polygon, and having an angle BAD equal to the given angle: and let AE be the given side of the parallelogram to be constructed. Join DE, and draw BF parallel to DE meeting AD in F; and complete the figure FGEA. It is that sought.

D

F

For, (th. 82,) BA : AF :: AE : AD, or the sides about the angle A are reciprocally proportional; and hence, (th. 81,) the parallelograms AG, AC, are equal *.

PROBLEM XXVII.

To find the centre and radius with which a given circle or segment of a circle was described.

First method. Take any point B in the circumference, and with any radius BA describe a circle AEG; and from the points A and C in which it cuts the given circumference, and the same radius as before, describe arcs cutting AEG in E and F, G and H. Then GH, LF, being drawn to meet in O, will give O the centre of the circle; and the distance at which either of them, as GH, cuts the given circle from O, is the required radius.

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* This problem most frequently occurs where the given and required figures are rectangles.

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