four lines divide harmonically any straight line that falls upon them, or their continuations.

E

Let AB be a straight line divided harmonically in C and D, and E a point without it; join EA, EC, ED, EB: any straight line FGHK, falling on these, is divided harmonically in G and H.

F

A

K

H

M

N

C

D

B

Through D draw LM (APP. I. 7, cor. 2) so that LD may be equal to DM, and through H draw NO parallel to LM. Then, in the similar triangles LDE, NHE, and MDE, OHE, LD : NH :: DE: HE, and DM: HO :: DE: HE, whence (V. 11) LD : NH :: DM: HO; and therefore (V. 14) because LD, DM are equal, NH is equal to HO. Again, since LD, DM are equal, EF (APP. I. 29, cor.) is parallel to LM, and therefore (I. 30) to NO: and since FGHK is drawn through the middle point of NO, the base of the triangle ENO, it is divided harmonically (APP. I. 29) in G and H.

PROP. XXXI. THEOR.-If through any point in the diameter of a circle produced, two straight lines be drawn making equal angles with the diameter, and cutting the circumference in four points, the straight lines joining the opposite points intersect one another in the diameter: and their point of intersection and the circumference divide the produced diameter harmonically.

Let A be a point in the diameter BC produced, and let the straight lines ADE, AFG, making equal angles with AB, cut the circumference in D, E, F, G: the straight lines joining DG, EF, cut BC in the same point H, and AB is divided harmonically in H and C.

Jein BE, BG, DB, DC. Then (III. 8) AD is equal to AF, and AE to EG; and the angles ADG, AFE are equal, being supplements of the angles EDG, EFG in the same segment. Hence the triangles in which are

these two angles and the equal angles at A, and which have the equal sides AD, AF, have (I. 26) their sides which are opposite to ADG, AFE, equal, and consequently DG, FE must cut AB in the same point H. Now, since AE, AG are equal, AB common, and the angles at A equal, the

G

H

chords BE, BG are (I. 4) equal, and therefore (III. 28) the arcs themselves. Hence (III. 27) the angles BDE, BDG are equal, and BD bisects the exterior angle EDH of the triangle ADH. Again, BDC is (III. 31) a right angle, and is therefore cqual to

the two BDE, ADC: from these equals take the equals BDG, BDE, and the remaining angles CDH, CDA are equal: and therefore, since DC, DB bisect the interior and exterior angles at the vertex of the triangle ADH, BA is divided harmonically (VI. A, cor.) in C and H.

Schol. 1. Since BA: AC:: BH: HC, the ratio of BH to HC is always the same, so long as A is a fixed point: and therefore, however the positions of the points D, E, F, & may vary, provided the angles at A are equal, the transverse lines DG, FE will always cut the diameter in the same point H.

Schol. 2. Conversely, were a point H taken in the diameter, and transverse chords DG, EF drawn through it, making equal angles with the diameter, it would be shown in a similar manner, that the straight lines ÉDA, GFA, each passing through the extremities that lie on the same side of the diameter, would always meet in the same point in the continuation of the diameter, and that BACA :: BH : HC.

PROP. XXXII. THEOR.-If in the diameter of a circle and its continuation, two points be taken on the opposite sides of the centre, such that the rectangle under their distances from the centre may be equal to the square of the radius, any circle whatever described through these points, bisects the circumference of the other circle.

Let ABC be a circle, D a point in any diameter, and E a point in the same diameter produced on the opposite side of the centre F: then, if the rectangle DF.FE be equal to the square of the radius, any circle GDHE described through D, E, bisects the circumference of ABC.

G

P

D

E

Through G, one of the points of intersection of the circles, draw GFH a chord of A GHE. Then (III. 35) the rectangle GF.FH is equal to the rectangle DF.FÉ, that is, (hyp.) to the square of the radius GF. Hence GF, FH are equal, and H is a point in the circumference of the circle ABC. But GH is drawn through its centre, and is therefore a diameter; wherefore its circumference is bisected in the points G, H.

PROP. XXXIII. THEOR.-A straight line drawn from the point of intersection of two tangents of a circle to the concave circumference, is divided harmonically by the convex circumference and the chord joining the points of

contact.

Through the point A let the tangents AB, AC be drawn to the circle BFC, and let BC join the points of contact; any straight line ADEF drawn from A to the concave circumference, is divided harmonically in F D and E.

For (III. 17, schol.) AB, AC are equal; and therefore (II. 5, cor. 5) AB2 or (III. 36) FA.AD=AE2+BE.EC; or (II. 1, and III. 35) FE.AD+EA.AD=AE+ FE.ED (II. 2) AE.ED+EA.AD÷

=

FE.ED. Take away EA.AD; then, since (II. 1) AE.ED+ FE.ED-AF.ED, the remainder becomes FE.AD-AF.ED; whence (VI. 16) AF: AD :: EF : ED.

PROP. XXXIV. THEOR.-If four circles be described, either all outside, or all inside, of any quadrilateral, each of them touching three of its sides, produced or not produced; the circumference of a circle will pass through all their centres.

Let ABCD be the quadrilateral, and E, F, G, H the centres of the external circles, each touching three sides; a circle may be described through E, F, G, H.

K

A

Join AE, BE, BF, &c. Then, since AE, AH bisect two vertical opposite angles, HAD is equal to EAK, and (I. 14, cor.) EAH is a straight line. For the same reason, EBF, FCG, GDH are straight lines. Now (I. 13) the four exterior angles KAB, ABL, NCD, CDM, and the four interior ones, A, B, C, D, are equal to eight right angles; but the interior are equal to four right angles; therefore also the exterior are equal to the same. Again, the angles EAB, EBA are the halves of one pair of those exterior angles, and GCD, GDC the halves of

E

B

the other pair: these four halves, therefore, are equal to two right angles; and if they be taken from four right angles, the sum of the angles of the triangles ABE, CDG, there remain the angles E, G equal to two right angles; and therefore (APP. I. 4) a circle may be described about the quadrilateral EFGH.

When the circles are described internally, the proof proceeds on the same general principle, but it is rather more simple.*

*If, in this case, the circle which touches three of the sides, touches also the fourth, the four circles, as also their centres, may be regarded as coinciding; and the circle passing through the four centres is reduced to a point.

PROP. XXXV. PROB.-To divide a given straight line into two parts, such that their rectangle may be equal to a given space, not exceeding (II. 5, cor. 2) the square of half the given line.

Let AB be a given straight line: it is required to divide it into two parts, such that their rectangle may be equal to a given space, not greater than the square of the half of AB. Bisect AB in C; then if the given space be equal to the square of AC or CB, AB is divided in C in the manner required. But if not, on AB describe a semicircle; draw ED parallel to AB, and at a distance equal to the side of a square containing the given space; and draw DG perpendicular to AB; G is the point of section required.

A

G

B

For (VI. 13) DG is a mean proportional between AG, GB, and (VI. 17) the rectangle AG.GB is equal to the square of GD; that is (const.) to the given space.

PROP. XXXVI. PROB.-To produce a given straight line, so that the rectangle under the whole produced line, and the part produced, may be equal to a given space.

Let it be required to produce a given straight line AB, so that the rectangle under the whole produced line, and the part added, may be equal to the square of a given

straight line C.

Bisect AB in D, and draw BE perpendicular to AB, and equal to C: from Das centre, at the distance DE, describe a semicircle cutting AB produced in F and G: either of these is the extremity of the produced part.

E

For (VI. 13) BE is a mean proportional between GB, BF, or AF, BF, since AF, GB are obviously equal. Therefore (VI. 17) the rectangle AF.FB is equal to the square of BE, that is, of C.

Schol. This proposition and the foregoing are the most useful cases of the 28th and 29th of the sixth book.

PROP. XXXVII. THEOR.—A line ABC, whether curved or polygonal, which is convex throughout, that is, which can be cut by a straight line in only two points, is less than any line, whether curved or polygonal, which envelops it, from one end to the other.

For, if ABC be not shorter than any of the lines that envelop it, some one of those lines will be less than any of the others, and

F

G

H

D

will be either less than ABC, or, at most, equal to it. Let AHFGEC be that line, and draw the straight line DG cutting it in D, G, and either touching ABC, or not meeting it. Then DG is less than DHFG, and consequently (I. ax. 4) ADGEC is less than AHFGEC: but, by hypothesis, AHFGEC is less than any other enveloping line; and yet the enveloping line ADGEU is less than it, which is absurd; therefore every enveloping line is greater than ABC.

Schol. In exactly the same manner it may be shown, that a convex line returning into itself, is shorter than any line enveloping it on all sides, and either touching it, or encompassing it without touching it.

A

O

PROP. XXXVIII. THEOR.-The circumferences of circles are to each other as their radii.

Let AB, CD be the radii of two circles; their circumferences are proportional to these.

For, if the circumference of the circle OR be not to that of FK, as AB to CD, AB will be to CD, as the circumference of OR to the circumference of a circle having a radius either less or greater than CD: and first, suppose its radius to be CE, less than CD. In the circle FK inscribe a regular polygon, the sides of which will not meet the circle EN, and in OR describe a similar polygon.*

Then, if AT, AO, CM, CF were joined, we should have (VI. 4) TO: MF :: AT or AB CM or CD. But (VI. 20, cor. 4) the perimeter of the polygon in OR is to that of the polygon in FK, as TO to MF, that is, as AB to CD; and, by hypothesis, AB is to CD, as the circumference of OR to the circumference of EN;

therefore (V. 11) the perimeter of the polygon in OR is to that of the polygon in FK, as the circumference of OR to the circumference of EN, which (V. 14) is absurd; as, by the preceding proposition, the first term of this analogy is less than the third, and the second greater than the fourth. Hence, AB cannot be to CD, as the circumference of OR to a circumference less than that of FK: and it would be proved in a similar manner, that CD cannot be to AB, as the circumference of FK to a circumference less than that of OR.

The method of effecting the construction here pointed out, is evident from the 16th proposition of the twelfth book.