sally, one similar to any regular polygon whatever, the method of inscribing which in a circle is known.

In practice, the construction may often be effected more easily without describing a similar figure in a circle. Thus, if it were required to describe a regular pentagon on CD, (see the figure for the preceding proposition) CD may be divided in extreme and mean ratio, and on it the triangle CDA may be described having CA, DA, each equal to the sum of CD and its greater segment: then, if the triangles ABC, AED be described, having the sides AB, BC, AE, ED, each equal to CD, ABCDE will be the required figure, as is evident from the preceding proposition, and from the scholium to the eleventh proposition of the second book.

In like manner, to describe a regular decagon on a given straight line BF, (see the figure for the next proposition) describe the triangle BAF, having the sides BA, FA each equal to the sum of BF and its greater segment, when it is cut in extreme and mean ratio: then A is the centre of the circle in which the required figure is to be inscribed, and the construction will be completed by inscribing nine other chords in the circle each equal to BF.

PROP. XXI. THEOR.

THE Square of the side of a regular pentagon inscribed in a circle, is equal to the squares of the sides of a regular hexagon and regular decagon, inscribed in the same circle.

G

K

L

Let AF, a radius of the circle BCG, be divided (II. 11.) in D, so that the rectangle AF.FD = AD2: inscribe the chord FB equal to AD: make the arc FC equal to FB, and join BC: then (IV. 11. and 15.) BC, BF, AF are respectively the sides of the inscribed pentagon, decagon, and hexagon; and it is now to be proved that the square of BC is equal to the squares of BF, FA. Produce FA to G, and join AB, BD. Then (by the demonstration of IV. 10.) AD, BD, BF, are equal, and therefore (III. 3. cor. 1.) DE, EF are equal. Now, (III. 35.) the square of BE is equal to the rectangle GE.EF; H and four times the square of BE, or (II. 4. cor. 2.) the square of BC, is equal to the rectangle under GE, and four times EF; that is, since AF is equal to AG, and four times EF equal to twice DF, BC2 = 2AF.DF + 2AD.DF+ 2DE.DF. For 2AF.DF substitute AD + AD2, which (const.) is equal to it; and for 2DE.DF put its equal DF2: then BC2 = AD2 + AD2 + 2AD.DF + DF2. But (II. 4.) the last three terms are equal to the square of AF, and the square of AD to the square of BF: therefore BC2 = BF2 + AF2.

M

B

Cor. Hence we have the following easy and elegant method of inscribing a regular pentagon and decagon in a circle. Draw the two perpendicular diameters GF, HK: bisect AK, one of the radii, in L, and join GL: make LM equal to LG, and join GM. Then it is plain (II. 11.) that the square of AM is equal to the rectangle AH.HM; and therefore AM is the side of the decagon; while MG is that of the pentagon, its square being (I. 47.) equal to the squares of GA, AM, the sides of the hexagon and decagon.

PROP. XXII. PROB.

To divide a given straight line into two parts, such that the square of one of them may be equal to the rectangle contained by the other, and a given straight line.

Let AB, AC be two given straight lines: it is required to divide AB into two parts, such that the square of one of them may be equal to the rectangle under AC and the other.

On CB, the sum of the given lines, describe the semicircle CDB, and draw AD perpendicular to CB; bisect CA in E, and join DE; and make EF equal to ED: then the square of AF is equal to the rectangle AC.FB.

C E A

F B

Describe the semicircle CGA, cutting ED in G, and join GF. Then, because FE, EG are respectively equal to DE, EA, and the angle FED common, GF is equal to AD, and the angle EGF, to EAD, which is a right angle; and therefore GF touches the circle CGA. Hence (III. 35.) the rectangle CA.AB is equal to the square of AD, or of FG, or (III. 36.) to the rectangle CF.FA. But (II. 1.) the rectangle CA.AB is equal to the two CA.AF, CA.FB, and (II. 3.) the rectangle CF.FA is equal to CA.AF and the square of AF. From these equals take away the rectangle CA.AF, and there remains the rectangle CA.FB equal to the square of AF.

Schol. The eleventh proposition of the second book is the particular case of this problem in which the given lines are equal.

PROP. XXIII. PROB.

To divide a given straight line into two parts, such that the rectangle contained by the whole and one of the parts may have a given ratio to the square of the other.

Let AB (see the last figure) be the given line; it is required

to divide AB into two parts, such that the square of one of them may have a given ratio to the rectangle under AB and the other.

Let the given ratio of the rectangle to the square be that of AB to AC: then, the construction being made as in the last proposition, F is the point of section required. For, as was there proved, the rectangle CA.FB is equal to the square of AF. But (VI. 1.) the rectangle CA.FB has to the rectangle AB.FB the same ratio that CA has to AB. Therefore the square of AF has to the rectangle AB.FB, the same ratio that CA has to AB.

Schol. The eleventh proposition of the second book is also a case of this; being the same as this proposition, when the given ratio is a ratio of equality.

PROP. XXIV. PROB.

To draw a common tangent to two given circles.

Let BDC, FHG be given circles; it is required to draw a common tangent to them.

Join their centres A, E, and make BK equal to FE: from A as centre, with AK as radius, describe a circle cutting another, described on AE as diameter, in

M: draw AM, meeting the circle BCD in D; and draw EH paral· lel to AM: join DH: it touches both the given circles.

For MD, EH, which (const.) are parallel, are equal to one another, because each of them is equal to KB: therefore (I. 33.) DH is parallel to ME. Now, (III. 31.) the angle AME in a semicircle is a right angle; and therefore (I. 29.) the angles ADH, DHE are right angles, and (III. 16.) DH touches both the circles, since it is perpendicular to the radii AD, EH.

H

E

B

G

F

B

K

Schol. In the first figure the circles lie on the same side of the tangent, which is therefore exterior to them; but in the second the tangent is transverse, or lies between the circles. It is plain also, that in these figures, by using the point L instead of M, another exterior and another transverse tangent would be obtained: and this will always be so, when each of the circles lies wholly without the other, and does not touch it. But if the circles touch one another externally, the two transverse tangents coalesce into a single line passing through their point of contact: if they cut

one another, there will be two exterior tangents, but no transverse one: if one of the circles touch the other internally, they can have only one common tangent, and this passes through their point of contact: and, lastly, if one of them lie wholly within the other without touching it, they can have no common tangent.

If the circles be equal, the points of contact of the exterior common tangents are the extremities of the diameters perpendicular to the line joining the centres: for (I. 33. and 29.) the lines con. necting the extremities of these towards the same parts are perpendicular to the diameters, and therefore (III. 16. cor.) they touch the circles.

PROP. XXV. THEOR.

IF a common tangent to two circles cut the straight line passing through their centres, the segments between the point of intersection and the centres, are proportional to the radii.

Let A, B be the centres of two circles, and CD an exterior common tangent meeting AB produced in E; AE: BE:: AC: BD.

For the triangles CAE, DBE are equiangular, and therefore (VI. 4.) AE: AC:: BE: BD,

and alternately, AE: BE:: AC: BD.

In like manner, if the transverse common tangent FG cut AB in H, the triangles AFH, BGH are equiangular, and therefore (VI.

C

D

K

M

A

N H

B L

E

4. and alternately) AH: HB:: AF: BG.

Cor. 1. Since AC: BD:: AE: BE, we have, by division, AC-BD: BD::AB: BE. It would be shown in a similar way, that the distance from B to the point in which the other exterior common tangent cuts AB produced would also be a fourth proportional to AC-BD, BD, and AB. The two exterior tangents, therefore, both pass through the same point E. It would appear, in like manner, by composition, from the analogy, AF: BG:: AH: HB, that the two transverse tangents both pass through the same point H.

Cor. 2. Since AE: BE:: AC: BD, and AH: HB:: AC: BD; therefore (V. 11.) AE: BE:: AH: HB; and (VI. def. 5.) AE is divided harmonically in H and B.

PROP. XXVI. THEOR

IF on the three sides of any triangle, equilateral triangles be described, either all externally, or all inter

nally; straight lines joining the centres of the circles inscribed in those three triangles, form an equilateral triangle.

On the three sides of any triangle ABC, let the equilateral triangles ABD, BCF, CAE be described externally, and find G, H, K, the centres of the circles described

in those triangles; draw GH, HK, KG: GHK is an equilateral triangle.

D

F

C

E

Join GA, GB, HB, HC, HF, KC, AF. Then the angle FBC is two-thirds of a right angle, and the angles GAB, GBA, FBH, BFH, each one-third. The triangles FBH, ABG are therefore similar, and (VI. 4.) FB: BH:: BA: BG; whence, alternately, FB: BA :: HB: BG; that is, in the triangles FBA, HBG, the sides about the angles FBA, HBG are proportional: and these angles are equal, each of them being equal to the sum of the angle ABC and two-thirds of a right angle. Hence (VI. 6.) these triangles are equiangular; and therefore (VI. 4.) FB or BC: FA:: BH: HG; and it would be shown in the same manner, by means of the triangles ACF, KCH, that FC or BC: FA:: CH: HK: therefore (V. 11.) BH: HG :: CH: HK. But BH is equal to CH, and therefore (V. 9.) HG is equal to HK: and it would be demonstrated in a similar manner, that HG, HK are each equal to GK.

If the equilateral triangles were described on the other sides of the lines AB, BC, CA, the angles ABF, GBH would be the difference between ABC and two-thirds of a right angle; but the rest of the proof is the same.

* If ABC exceed a right angle and a third, the sum of it and two-thirds of a right angle is greater than two right angles. In that case the angles ABF, GBH, understood in the ordinary sense, are each the difference between that sum and four right angles. If ABC be a right angle and a third, the sum becomes two right angles, and FB, BA are in the same straight line, as are also HB, BG. In this case it is proved as before, that FB: BA:: HB: BG; and then (V. E. cor.) FB or BC: FA:: HB: HG. The rest of the proof would proceed as above.

It may be remarked, that if an equilateral triangle be described on a straight line, and if on the two parts into which it is divided at any point, other equilateral triangles be described, lying in the opposite direction, the lines joining the centres of the three equilateral triangles will also form an equilateral triangle. The connexion of this and the proposition in the text will be perceived by supposing two angles of the triangle continually to diminish, till they vanish, as the triangle may thus be conceived to become a straight line.