For, CB, BA, being drawn, GH, EF, bisect them at right angles, and hence pass through the centre; dem. of th. 41, and prob. 7 *.

Second method. Draw any line AB cutting the circle, and set off AE, AD, BF, BC, each equal to AB, the two former in the line, and the two latter in the circle. Parallel to DE and CF draw AO, BO, intersecting at O. Then O is the centre of the circle.

For this is only bisecting the equal angles at A and
B, (prob. 4, second method,) and the bisecting lines pass through the centre.
Third method. Take any four equal distances AB, BC,
CD, DE, in succession in the circumference of the circle:
draw AB and CD meeting in G, BC and DE meeting in
H, AC and BD meeting in K, and BD, CE meeting in
L: then GK, HL, being drawn to meet in O, give this
point as the centre of the circle.

For it is easily shown that GK, HL, bisect BC, CD, at right angles, and hence they pass through the centre.

H

C

D

L

E

Scholium. Any of these methods enable us to find the centre and radii of the inscribed and circumscribed circles to a regular polygon. The chief value of the latter two consists in their ready adaptation to this purpose; especially the last, which requires the use of the ruler only.

PROBLEM XXVIII.

To draw a tangent to a given circle, from a given point A in the circumference.

First method. Find the centre O, and join OA; draw PQ through A perpendicular to AO: it will be the tangent sought, th. 46.

Second method. Draw any line BD through A, and with radius BA describe the circle BCD from centre A, cutting the circle in C, and BA produced in D; with centres D and C, and with equal radii describe

B

arcs intersecting in E; then AE being drawn, it is the tangent required.

For, (prob. iii. fourth method,) AE is parallel to BC; and BC is perpendicular

to the diameter through A, and hence it is a tangent at A.

Third method. Describe the arc BC from centre A, as in the last, and draw AQ parallel to it: then AQ is the tangent ↑.

Fourth method. Take any other point E in the given circumference, with which as centre, and with radius EA, describe a circle cutting AE produced in F, and the given circumference in B: make the arc FG equal to FB, and join GA, which will be the tangent required.

For, since BF is equal to FG, and E is the centre of the circle BFG, the angles BEF, FEG, are equal; and because BEA, GEA, are isosceles triangles, the angle

It is only from convenience of working, and not from mathematical necessity, that AB is taken equal to BC.

In the case where the parallel ruler is available, this is by far the neatest method.

FEB is double of EBA, and FEG is double of GAE: the angle GAE is equal to the angle EBA in the alternate segment. Whence AG touches the circle in A *.

PROBLEM XXIX.

To draw a tangent to a given circle BCD from a given point A without it.

First method. Find the centre O of the given circle, join AO, and on it as diameter describe a circle ACOC', meeting the given circle in C and C': then if AC, AC', be drawn, they will be tangents to the given circle.

For since ACO is a semicircle, the angle ACO is a right angle; and hence AC is perpendicular to the radius OC, and therefore a tangent.

proved.

E

In a similar manner the other case is

Second method. With A as centre, and any radius, describe an arc cutting the given circle in B and D ; draw AB, AD, meeting the circle again in E and F, and draw BF, DE, meeting in G, and through G a line CC' parallel to BD; it will cut the given circle in C, C' the points of contact.

Third method. Through the given point A, draw any three lines cutting the circle in E, F, in B, D, and in G, H, respectively: draw the diagonals BF, ED, intersecting in L, and GF, EH, intersecting in K: then a line drawn through KL to cut the given circle in C and C' determines the points of contact †.

*This, or some modification of this, is the only method available when the centre is unknown or inaccessible, and the point A at or near one of the extremities of the given portion of the circumference.

The first of these methods is the most usually employed in this country; but the others are more convenient when the centre of the given circle is not given. The last requires the use of the ruler only, and can be employed when only a part of the circle is given, provided it be not almost wholly one side of the point of contact. If, however, this last condition should not be fulfilled, we have no alternative but to find the centre, or at least in some way find other points in the circumference in the required region for completing the operation. ·

Respecting the two latter methods it may be generally remarked that they depend upon the principles intimated in theorems 70 and 71 of the Miscellaneous Exercises, pp. 346, 7, of this volume. They form a branch of a system of inquiries much cultivated by the continental writers under the name of Geometry of the Rule, and sometimes under the name of transversals. It forms one of the most interesting of all the departments of geometry in reference to practical utility. The same construction that is here used (the third) is also applicable to all the conic sections. For some further notice of these subjects the reader is referred to the second volume. The writings of Carnot, Garnier, Brianchon, Chasles, Servois, and De Gelder, may also be consulted with considerable advantage by the inquiring student.

PROBLEM XXX.

From a given point A in the circumference of a given circle ABC, to cut off a segment to contain a given angle EDF.

DRAW any line AC from the given point to meet the circle in C with any radius DE describe the arcs EF and GK, from centres D and C; and take GH equal to FE, and join CH cutting ABC in B: then AB being drawn will be the line, or the arc BCA will be that required.

For this construction makes the angles ACB, EDF, equal, and hence fulfils the condition.

E

H

K

PROBLEM XXXI.

In a given circle ABC to inscribe a triangle similar to a given triangle EDH, and having the angle which is equal to EHD at a given point A in the given circle.

FROM A draw a line AB to cut off a segment containing

one of the other given angles, as EDH, by the last problem; and make the angle BAC equal to the angle EHD. Then joining BC, the triangle ABC is that sought.

For it has by construction two of its angles BAC, ACB, equal to two of the angles EHD, EDH; and hence the third angles are equal and the triangles similar.

D

PROBLEM XXXII.

On a given line AB to describe a segment of a circle to contain a given angle

CEF.

HG

M

First method. If CEF be less than a right angle, make CED equal to it, and take ED equal to EF, (the particular length of ED being any whatever,) and join FD. In FD take FG equal to the given line AB, and draw GH parallel to EF to meet ED in H. Then with radii equal to EH, and centres A, B, describe arcs meeting in M; and M will be the centre of the circle sought, and the segment ANB may be described.

A

B

If the given angle PEC be greater than a right angle, construct as before with its adjacent acute angle CEF; and the segment AN'B will be that required. For, suppose HK drawn parallel to GF: then HK = GF

=

AB; and hence

the sides of HEK are equal each to each to those of AMB, and consequently the angle AMB is equal HEK. But HEK is double of CEF by construction, and AMB is double of the angle ANB in the segment. The angle in the segment is, therefore, equal to the given angle CEF.

Second method. Take any point O in EF and describe a semicircle from O as centre, and with OE as radius, cutting EC in C: join FC, and produce it till FG is equal to the given line AB: through G and F draw parallels to EF, EG, meeting in S, and join SE cutting

FG in T: draw TV parallel to EF meeting EC in V, and VK parallel to GF meeting EF in K. Then EK is the radius of the segment, with which proceed as in the last construction *.

Third method. Take any line AL making any angle with AB, and at any point G in it make the angle AGH equal to the given angle DEF, and through B draw BK parallel to HG meeting AL in K. Then a circle described through the three points A, K, B will evidently be that which has the required segment cut off by AB ↑.

Fourth method. Draw the perpendicular GH to bisect the given line AB, and make the angle HGK equal to the given angle DEF, and through B draw BM parallel to GK, meeting GH in M. Then M is the centre from which the segment is to be described.

For join MA: then, since M bisects the base AB. of the triangle AMB, it bisects the angle AMB. Whence the angle AMB is double of

G

K

/H B

F

D

P

K

BMP, that is of KGP, that is of DEF; and AMB is likewise double of ANB: whence ANB is equal to DEF 1.

Fifth method. With radius equal to the given line AB, and with centres A, B, E, describe circles, the two former mutually intersecting in G and H, and the latter cutting CE, FE, in C and F. Make BI equal to CF, draw AI meeting the circle HAL in K, and draw the diameter KBL. Join AL, GH, intersecting at M: then M is the centre from which the segment is to be described.

M

B

E

For, GH is by this construction a perpendicular to the middle AB of the base of the triangle, and hence the centre of any circle through A and B is in this line. Also, the angle BAK being equal to the given one at E, and KAL, being an angle in a semicircle, is a right angle; and therefore, BAK and BAM are together equal to one right angle: but MPA being a right angle, PAM and AMP are together equal to one right angle; and hence AMP is equal to BAK or to CEF. Again, AMB is double of AMP, and therefore double of CEF; whence M is the centre of the circle required, in the same manner as in the former constructions §.

* This is but an obvious variation of the last construction. With the parallel ruler, however, it is a convenient one.

+ This method anticipates the construction for a circle through three given points; but this is immaterial, as that problem is quite independent of this. It has, however, when carried out, more actual work than either of the former two; and has, moreover, the disadvantage of performing the work in the important part of the drawing, and thereby rendering the paper liable to injury.

A few of the lines viz. AM, AN, NB, not requisite in the construction, are omitted, intentionally, in the figure.

§ This is essentially the same construction as that given by Euclid, but having all the implied operations detached, so as to suit the particular circumstances of the problem.

PROBLEM XXXIII.

Through three given points A, B, C, to describe a circle.

FROM B describe a circle with a radius greater than half the distance AB or BC, and from A and C circles with the same radii, cutting the former in D, E, and F, G: then DE, FG, being drawn to intersect in O, will give the centre required, and hence the circle may be described.

B

For if AB, BC, be joined, DE and FG would bisect them at right angles, and hence pass through the centre.

1. Scholium. The construction is the same when a circle is required to be described about a given triangle ABC.

2. Scholium. Sometimes a particular case of this problem arises in practice under the following form :

Given the span AB and rise SR of a circular arch to describe it. Join AR, RB, and bisect them by the perpendiculars MO, NO: then, as in the problem, O is

the centre.

The joints between the stones, or voussoirs, are only continuations of the radii from the centre O.

M

PROBLEM XXXIV.

To inscribe a circle in a given triangle ABC.

PRODUCE AB both ways, and take AE, AD, BG, BF, all equal: parallel to ED and FG draw AO, BO, intersecting in O: then O is the centre of the inscribed circle. With centre O and radius AO describe an arc cutting AB in P, and with centres A and P and equal radii describe

arcs intersecting in H. Join HO cutting AB in K: then OK is the radius of the inscribed circle, whence the circle may be drawn.

For, draw OL, OM, perpendicular to AC and BC: then, since AO bisects the angle CAB, and those at K and L are right angles, the two triangles AOK, AOL, are equiangular, and have the side AO opposite two equal angles common, they are equal in all respects, and KO is equal to OL *.

Scholium. When the given triangle is equilateral, and a parallel ruler not available, the following method is useful.

Find by intersections the vertices D and L of equilateral triangles on the sides AC, BC. Draw AL, DB, intersecting in F, and let one of them

D

This method implies the use of the parallel ruler for bisecting the angle. When this is not The rest available, bisect the angles at A and B by means of the first method of prob. iv. of the construction is the same.