D

H

K, the centre of the circle, with E, B, C, G; draw also KF perpendicular to BC. Then KF is equal to the radius of EG. For since the hypotenuse KB is common to the two right-angled triangles KEB, KFB, and KC to KGC, KFC, if KF were less than KE or KG, BF and FC, that is BC, would (I. 47. cor. 6.) be greater than BE and CG; while if KF were greater than KE or KG, BC would be less than BE and CG: but BC is equal to BE, CG, because AE, DG are equal to AD. Therefore F is on the circumference; and BC touches the circle, because the angles at F are right angles.

E

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K

It is plain from this scholium, and from the fourth proposition of the first book of this Appendix, that a quadrilateral which has two opposite sides equal to the other two, and the opposite angles equal to two right angles, may have one circle inscribed in it, and another about it. It also follows, that any quadrilateral having two opposite sides equal to the other two, may have a circle described touching all its sides, however the angles may vary in their magnitude, and even should one of them become re-entrant.

PROP. XXXVIII. PROB.

THE circumference of a circle being divided into any number of equal parts, it is required to describe as many circles as there are parts, each touching the given circle and two of the others.

This will be effected simply by drawing tangents to the circle through the points of division; by drawing lines from their points of intersection to the centre; and by inscribing circles in the triangles thus formed. Other circles, touching the given one externally, and each touching two others, will be found by describing circles touching the sides of the circumscribed figure, and the radii produced through the intersections of the tangents. The proof is easy in both cases.

PROP. XXXIX. THEOR.

In the figure for the forty-seventh proposition of the first book of Euclid, the straight lines BK, CF, and the perpendicular from the right angle to the hypotenuse, pass through the same point.

Produce BC both ways, and draw AM, FN, KO perpendicular

N

B

M C

K

to it. Let AM, FC intersect each other in P, and join PN, FM. Then, since ABF is a right angle, ABM, NBF are together equal to a right angle, and therefore to NBF, BFN; whence ABM, BFN are equal; and (I. 26.) since BF, BA are equal, BN is equal to AM, and FN to BM. In the same manner it would be proved, that CO, OK are respectively equal to AM, MC; whence NB is equal to CO, and NC to BO. Now (I. 37.) the triangles NMP, FMP are equal: to each add P MC, and the wholes NPC, FMC are equal. So likewise are their doubles, which are the rectangles NC.PM, and MC.FN or MC.BM. would be proved in a similar manner, that the rectangle under BO or NC, and the part of MA intercepted between M and the straight line joining BK, is likewise equal to BM.MC: and therefore FC, BK cut AM in the same point.

PROP. XL. PROB.

It

LET a given straight line FE be cut in extreme and mean ratio, and in its continuation let ED be taken equal to its greater segment; in like manner let DA be equal to the greater segment of ED, AC equal to the greater segment of DA, and so on: it is required to find the length of the entire line that would be obtained by the addition of an infinite number of these successive lines, FE, ED, DA, &c.

Let S be put to denote the required sum. Then, by the corollary to the 19th proposition of the Supplement to the fifth book, FE-ED: FE:: FE: S. But (II. 11. schol.) FE-ED:ED:: ED: EF, or as EF: FD; whence, alternately, FE-ED: EF ::

F

E

D

A

C B

ED: FD; and therefore (V. 11.) ED: FD :: EF: S, or, alternately, ED: EF:: FD: S. But ED: EF::FD: FD + FE, and therefore FD: FD + FE :: FD: S, and (V. 9.) S = FD +FE = 2FE+ED; that is, the required sum is equal to twice

This is one out of many curious propositions that may be founded on the figure for Euclid's proof of the 47th proposition of the first book. Another very simple one is, that if GH, FD, EK be joined, each of the triangles thus formed is equal to ABC. This follows immediately from the second corollary to the 38th proposition of the first book. Thus, the triangles FBD, ABC, are equal, because FB, BD are respectively equal to AB, BC, and the contained angles supplemental.

the given line together with its greater segment; or to a straight line, of which, when cut in extreme and mean ratio, the given line is the less segment.

PROP. XLI. PROB.

In a given semicircle ACB let another be inscribed lying in the opposite direction; in the second let another be inscribed in a similar manner; in it another, and so on; their diameters being all parallel: it is required to find the position of the point to which the centres approach, when the process is continued without limit.

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H

C

G

M

L

P

B

Through the centre F draw FC perpendicular to AB; then, if radii be drawn bisecting the right angles at F, and if their extremities D, E be joined by DGE, it is evident that the triangles DFE, FGD, FGE are all right-angled and isosceles; and therefore G is the centre of the first inscribed semicircle. Let this be described; and then by bisecting the angles at G, the next semicircle HGK and its centre L may be determined; and so on. Join CD, HF. Then the triangles CGD, HLF, have each a right angle, and the angles GDC, LHF are equal, because they stand on like parts of the circumferences of the two circles, of which ACB, DFE are halves. Hence (VI. 4. and alternately) CG: LF:: GD or GH: LH; that is, as the diagonal of a square to its side. In exactly the same manner, if M and N, the centres of the next semicircles were found, it would be shown, that GM is to LN, as the diagonal of a square to its side; and the same must hold in regard to each successive pair of semicircles. Now, the square of HG or DG is double of the square of GL, and the square of DF or CF is double of the square of DG, and therefore quadruple of the square of GL; wherefore (II. 4. cor. 2.) CF is double of GL, and consequently of the sum of CG, LF: and in the same manner GL would be shown to be double of the sum of GM, LN, and so on. Hence (XII. lem. 1.) if from CF there be taken CG, LF, and if from GL there be taken GM, NL, and so on; a remainder will at length be obtained less than any line however small and therefore the two lines made up of CG, GM, &c., and of FL, LN, &c., may be made to approach a point in FC, so that the distances between it and their extremities will be as small as we please. Now (V. 12.) the sum of CG, GM, &c., is to the sum of FL, LN, &c., as CG to FL, that is, as the diagonal of a square

to the side; and therefore, if CF be divided in O, so that CO is to OF as the diagonal of a square to the side, O will be the required point.

PROP. XLII. PROB.

A SEMICIRCLE ACB being given, and other semicircles being described as in the preceding proposition; it is required to find the sum of the areas of all those inscribed semicircles.

Circles (XII. 2.), and consequently semicircles, are as the squares of their diameters or of their radii. Now the square of GD is half the square of DF or CF, and therefore the semicircle DFE is half of ACB. For the same reason HGK is half of DFE; and universally, each semicircle is half of the one in which it is inscribed. Hence the entire amount will be the sum of the infinite series ACB+ACB+ ACB +ACB+ &c. and therefore (V. SUP. 19.) ACB-ACB: ACB:: ACB ACB, the required sum; and it thus appears that the sum of all the inscribed semicircles is equal to the given semicircle.

:

PROP. XLIII. THEOR.

IN any triangle, the centre of the circumscribed circle, the point in which the three perpendiculars intersect one another, and the point of intersection of the straight lines drawn from the angles to bisect the opposite sides, lie all in the same straight line.

Let ABC be a triangle, and let the two perpendiculars AD, CE intersect in F; bisect AB, BC in H, G, and draw AG, CH intersecting in K; draw also GI, HI perpendicular to BC, BA, and intersecting in I. Then (VI. H.) F is the intersection of the three perpendiculars, K (APP. I.7.) the intersection of the three lines drawn from the angles to bisect the opposite sides, and (IV. 5. cor. 1.) I is the centre of the circumscribed circle. Join FK, KI: FKI is a straight line.

Join HG: it is (VI. 2. and 4.) parallel to AG, and is half of it. Also the trian

H

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F

D

gles ACF, GHI are (1.34. cor. 4.) equiangular, and therefore GI is half of AF. So likewise (APP. I. 7. schol.) is KG of KA. Hence

the two triangles AKF, GKI have the alternate angles KAF, KGI equal, and the sides about them proportional: therefore (VI. 6.) the angles AKF, GKI are equal, and (I. 14. cor.) since AKG is a straight line, FKI is also a straight line.

Schol. It is plain (VI. 4.) that FK is double of KI. We have also seen that AF is double of GI. Hence it appears, that the distance between any of the angles and the point of intersection of the three perpendiculars, is double of the perpendicular drawn from the centre of the circumscribed circle to the side opposite to that angle.

PROP. XLIV. THEOR.

STRAIGHT lines drawn from the angles of a triangle to the points in which the opposite sides touch the inscribed circle, all pass through the same point.

Let ABC be a triangle, and D, E the points in which the sides AB, AC touch the inscribed circle; draw BFE, CFD; draw also AFG cutting BC in G: G is the point in which BC touches the inscribed circle.

B

H

AG C

E

If possible, let another point K be the point of contact, and draw DH, DI parallel to BC, CA. Then in the similar triangles FDI, FCE, FD: DI:: FC: CE, or CK; and in the similar triangles FDH, FGC, DH: DF:: GC: FC; from which and from the preceding analogy we get, ex æquo, DH: DI:: CG: CK. Again BD: DI BA: AE or AD:: BG: DH. Hence, alternately, and by inversion, BG: BD :: DH: DI; whence (V. 11.) BG: BD or BK: CG: CK, or alternately, BG: CG :: BK: CK; and by composition, BC: CG:: BC: CK; and therefore CG, CK are equal; that is, G and K coincide, and AFG passes through the point in which BC touches the circle.

PROP. XLV. THEOR.

In a circle, the sum of the perpendiculars drawn from the centre of the circumscribed circle to the three sides, is equal to the sum of the radii of the inscribed and circumscribed circles.

Let ABC be a triangle, having its sides bisected in D, E, F, by perpendiculars meeting in G, the centre of the circumscribed circle: the sum of GD, GE, GF is equal to the sum of the radii of the inscribed and circumscribed circles.