PROBLEM XXIII.

To describe a circle that shall touch a given straight line (AB) in a given point (E), and shall also pass through another given point (F) not in the same straight line (AB).

From the given point E (Fig. 45) draw EG perpendicular to AB; join EF, and bisect it by a perpendicular CD, cutting EG in the point 0, with which as a centre, and the radius OE, describe the circle EDF, which will touch the line AB in the given point E, and also pass through the given point F.

C

PROBLEM XXIV.

To describe a part of a circle that shall touch a given straight line (AB) in a given point (E), and shall also touch a given arc (DGH).

Draw EC (Fig. 46) perpendicular to AB; and through the centre K of the given arc DGH, draw KL parallel to EC, and meeting the arc DGH continued in the point L; join LE, and produce it to meet the given arc in the point D, and join DK, cutting EC in 0; then with the centre O, and radius OE, describe a part of a circle which will touch the given line AB in the given point E, and also the given arc DGH in the point D.

PROBLEM XXV.

To describe two arcs that shall meet each other in the line of their centres, and shall touch two given straight lines (AB, CD) at the given points (E, F) in those lines, the radius (EG) of the lesser are being also given.

F

I

From the point E (Fig. 47) draw EG perpendicular to AB, and equal to the given radius, and from the point F draw FH perpendicular to CD, and make it equal to EG ; join GH; bisect GH by a perpendicular KO, cutting FH produced in the point O; join OG, and produce it, then with O as a centre, and radius OF, describe an arc meeting OG produced in L. With the centre G, and radius GE or GL, describe an arc meeting the former arc in L; then the arcs FL and EL are those required, meeting each other at the point L in the line LO, passing through both their centres G and O, and also touching AB and CD in E and F.

A

"Fig. 47.

PROBLEM XXVI.

To find a straight line nearly equal to the length of a given arc (ABC).

Bisect the given arc in B (Fig. 48), and draw the lines AB, AC; with the centre A, and radius AB, describe an are cutting AC in D; make DE in AC produced equal to AD, and divide CE into three equal parts, and make EF equal to one of the parts; then AF will be nearly equal to the arc ABC.

PROBLEM XXVII.

To draw a straight line nearly equal to the circumference of a given circle (ADBC).

Draw the diameters AB and CD (Fig. 49) at right angles to each other; produce CD until DE is equal to three-fourths of DO; through C draw FG parallel to AB; join EA, EB, and produce them until they meet FG in F and G; then FG will be nearly equal to the semi-circumference of the circle AD BC.

B

D

Fig. 48.

PROBLEM XXVIII.

To inscribe a circle in a given triangle (ABC) that shall touch all its sides. (Euclid,

Book IV., Prop. IV.)

Produce AC (Fig. 50) both ways; with the centre A, and any radius, describe an arc

cutting AC produced in D, and AB in E;

also with the centre C, and the same radius, describe an arc cutting AC produced in F, and BC in G; join DE and FG, and draw through A and C lines parallel to DE and FG respectively, and cutting each other in the point 0; then O is the centre of the required circle (the lines AO and OC will bisect the angles BAC and ACB, which might be done by Problem XII). From O draw OH per

B

A

H

C

Fig. 50.

pendicular to AC; then with the centre O, and radius OH, describe a circle which will touch all the sides of the given triangle ABC.

PROBLEM XXIX.

To describe a circle about a given triangle (ABC). (Euclid, Book IV., Prop. V.)

This Problem is simply to describe a circle passing through three given points, ABC, as given before in Problem XVIII.

PROBLEM XXX.

To draw a square within or about a given circle (ACBD). (Euclid, Book IV.,
Props. VI. and VII.)

Draw through the centre O (Fig. 51) any two diameters, AB and CD, at right angles to each other, meeting the circle in the points ACBD. To inscribe a square in the circle, join those four c points; then ACBD is the square required; and to describe a square about the circle, draw tangents or lines parallel to the diameters at those four points, then EFHG is the square required.

E

F

B

II

Fig. 51.

PROBLEM XXXI.

To circumscribe and inscribe circles to a given square (ABCD). (Euclid, Book IV., Props. VIII. and IX.)

Draw the diagonals, AC, BD (Fig. 52), intersecting in the point O, the centre of the required circles. With the centre O, and radius OA, describe a circle which will pass through A,B,C and D, and be circumscribed about the given square ABCD. From 0, draw OE perpendicular to DC; and again with the centre O, and radius OE, describe a circle which will touch all the four sides of the square, and be inscribed in the given square ABCD.

E

B

Fig. 52.

PROBLEM XXXII.

To describe a pentagon on a given straight line (AB). (Euclid, Book IV.

Props. XI. and XII.)

1ST METHOD.-Bisect AB in C (Fig. 53), and from B draw BD perpendicular to AB, and equal to AC or BC; join AD and produce it; with the centre D and radius DB describe a circle cutting AD produced, in E; with the centres A and B and radii, each equal to AE, describe arcs intersecting in F; lastly, with the centres A and F, and radii, each equal to AB, describe arcs intersecting in G, and with B and F as centres, and the same radii, describe arcs intersecting in H; join AG, FG, FH, and BH; then ABHFG is the pentagon required.

F

B

Fig. 53

2ND METHOD.-Bisect AB in C (Fig. 54), and from B draw BD perpendicular to AB, and equal to AC or BC; join AD and produce it; with the centre D, and radius DB, describe a circle cutting AD produced, in E; join BE with the centres A and B and radii, each equal to BE, describe arcs intersecting in O, with which as a centre and radius OA or OB, describe a circle. With the centres A and B and radii, each

F

Fig. 55.

3RD METHOD.-With the centres A and B (Fig. 55), and radii, each equal to AB, describe circles intersecting in C and D; join CD, cutting AB in E, and produce it.

Make

EK equal to AB; join AK, and produce it, till KL is equal to AE or EB; with the centre A, and radius AL, describe an arc cutting DC produced in F; and with the centre F and radius AB describe a circle cutting the circles DAC and DBC in H and G respectively; join AG, FG, FH, and BH; then ABHFG is the pentagon required.

PROBLEM XXXIII.

E

To describe a regular hexagon upon a given straight line. (Euclid, Book IV., Prop. XV.) With the centres A and B (Fig. 56), and radii, each equal to AB, describe circles intersecting in O; with the centre O, and the same 'radius, describe another circle, cutting the two former in F and C respectively; join AO and BO, and produce them till they meet the latter circle E in D and E respectively; join BC, CD, DE, EF, and FA; then ABCDEF is the hexagon required. From this it will be seen that the side of a hexagon is equal to the radius of the circumscribed circle; and by merely finding the centre O as above, and setting off the length of the given line six times round the circumference, the hexagon will be constructed.

PROBLEM XXXIV.

To describe a regular octagon on a given straight line (AB). With the centres A and B (Fig. 57), and radii equal to AB, describe circles intersecting in K; with the centre K, and the same radius, describe a circle cutting the two former in L and M; with the centres L and M, and still the same radius, describe arcs cutting the last circle in N and O; join AN and BO, by lines cutting the two first circles in P and Q, and produce them; join AQ and BP, and produce them till PG and QD are each equal to AB; and lastly, with the

H

Fig. 56.

centres G and D, and the same radii, equal to AB, describe circles cutting the two first circles in H and C, and the lines AN and BO produced, in F and E. Join BC, CD, DE, EF, FG, GH and HA; then ABCDEFGH is the octagon required.

2ND METHOD.-With the centres A and B (Fig. 58), and radii, each equal to AB, describe circles cutting each other in K and L; join KL, and through A and B draw AF and BE parallel to KL, cutting the circles in M and N; join AN and BM, and parallel to them through A and B draw AH and BC, meeting the circles in H and C; through H and C draw HG and CD parallel to KL, and meeting BM and AN produced in G and D; and lastly, through G and D draw GF and DE parallel to AD and BG, cutting AM and BN produced in F and E; join FE; then ABCDEFGH is the octagon required.

H

F

M K

PROBLEM XXXV.

Fig. 58.

To describe a regular polygon (containing any number of given sides) upon a given straight

line (AB).

With the centre B (Fig. 59), and radius AB, describe a semicircle, meeting AB produced in H; divide the semicircle into as many equal parts as in the proposed polygon; join B and the second point of division C; bisect each of the sides AB and BC by perpendiculars intersecting in O; then with the centre 0, and radius OA, OB, or OC, describe a circle, which will pass through the points A, B, and C; this circle will be that described about the required polygon; and therefore, by setting off AB as many times as required round the circle thus formed, the polygon will be described. The example given in the figure is a nonagon, or ninesided figure.

Fig. 59.

ON THE CONSTRUCTION OF THE CONIC SECTIONS.

The preceding propositions depend entirely on the properties of straight lines and circle, and therefore admit of construction by rule and compasses. Now, beside the circle there are several curves which are used more or less by draughtsmen; these are for the most part drawn, when required, by determining accurately several points in them which are then neatly joined by the hand. Of these curves the ellipse is the one oftenest needed, and is used, in fact, almost as much as the circle, in consequence of the perspective representation of a circle being an ellipse. Besides the ellipse, the hyperbola and parabola are often needed in the delineation of shadows. These three curves are generally called conic sections, and for this reason-if a right cone is cut by a plane, the curve which bounds the section is one of these three, except in the special cases when the elliptic section degenerates into a circle, and the hyperbolic into two