D

points E and F with the distance BG, describe other two arcs intersecting in H: This is the point of medial section. For it is evident that this point H lies in the straight line AB. And because the triangles AGB, CGB have their 'sides respectively equal, the angle ABG (I. 2.) is a right angle, and consequently (II. 10.) AGAB2+BG"; but AG=AE, and AE-3AB2 (IV. 17. cor. 2.); wherefore 3AB AB2 + BG2, and BG2AB2. Now since BE EC, it follows, (II. 20.)

=

A

B

C

that HE-BE-CH.HB; but HE-BE2 = BG*— BEAB2, and therefore ABCH.HB. Whence CH is cut by a medial section at B, and consequently (II. 19. cor. 1.) its greater segment BC or AB is likewise divided medially at H by the remaining portion BH.

PROP. VIII. PROB.

To bisect a given arc of a circle.

Let it be required to bisect the arc AB of a circle whose centre is C.

From the extremities A and B with the radius AC, describe opposite arcs, and from the centre C inflect the chord AB to D and E; from these points, with the distance DB describe arcs intersecting in F; and from D or E, with the distance CF, cut the given arc AB in G : AB is bisected in that point.

For the figures ABCD and ABEC being evidently rhomboids, DC and CE are parallel to AB, and hence con

in the rhomboid ABCD, DB2+CA2=2DC*+2CB2 (II. 22.), or BD2=2DC2+CB2; and since DB=DF, 2DC2+CB2=DC2+CF", whence DC+CB2=CF2, or DC2+CG2=DG2, and therefore (II. 11.) DCG is a right angle. And because CG is perpendicular to DC, it is likewise (I. 22.) perpendicular to AB, and the triangles . CAP and CBP are equal (I. 21.), and the angle ACG equal to BCG; whence (III. 12.) the arc AG-BG.

PROP. IX. PROB.

To find the centre of a circle,

Assume an arc AB greater than a quadrant, and from one extremity B, with the distance BA, describe a semicircle ADC, cutting the given circumference in D; from the points B and C, with the distance CD, describe arcs intersecting in E, and, from that point with the same distance, describe an arc cutting ADC in F; and lastly, from

the points A and B, with the distance AF, describe arcs intersecting in G: This point is the centre of the circle ADB.

E

For the isosceles triangles BEC, BEF, being evidently equal, the angle FBC is equal to both the angles at the base; but FBC is (I. 32. El.) equal to the interior angles BAF and BFA of the isosceles triangle ABF, and hence that triangle is similar to BEF. Wherefore BE: BF:: BA AF, or CD: BD :: BA : AG; consequently the isosceles triangles CBD and BGA (VI. 12. cor.) are similar, and the angle BCD is e

D

B

qual to GBA; BG is, therefore, parallel to CD, and hence (I. 30. El.) the angle BDC, or BCD, is equal to GBD. The triangles BGA and BGD, having thus the side BA equal to BD, BG common, and equal contained angles GBA and GBD, are (I. 3. El.) equal, and therefore the side GA is equal to GD. The point G being thus equidistant from three points, A, D, and B in the circumference, is hence (III. 8. cor.) the centre of the circle.

PROP. X. PROB.

To divide the circumference of a given circle successively into four, eight, twelve, and twentyfour equal parts.

1. Insert the radius AB three times from A to D, E, and C; from the extremities of the diameter AC, and with a distance equal to the chord AE, describe arcs intersecting in the point F; and from A, with the distance BF, cut the circumference on opposite sides at G and H : AG, GC, CH, and HA are quadrants.

For, as before, AF2=AE2=3AB2; and the triangle ABF being right-angled, 3AB2=AF2=AB2+BF*, and therefore BFAG=2AB*; whence (II. 12.) ABG is a right angle, and AG a quadrant.

2. From the point F with the radius AB, cut the circle in I and K, and from A and C inflect the chord AI to L, H and M; the circumference is divided into eight equal portions by the points A, I, G, K, C, M, H, and L.

For BF, being equal to 2AB2, is equal to the squares

H

of BI and IF, and consequently BIF is a right angle; but the triangle BIF is also isosceles, and therefore the angle IBF at the base is half a right angle; whence the arc IG is an octant.

3. The arc DG, on being repeated, will form twelve equal sections of the circumference.

For the arc AD is the sixth or two-twelfth parts of the circumference, and AG is the fourth or three-twelfths; consequently the difference DG is one-twelfth.

4. The arc ID is the twenty-fourth part of the circumference.

For the octant AI is equal to three twenty-fourths, and

the sextant AD is equal to four twenty-fourths; their difference ID is hence one twenty-fourth part of the circumference.

PROP. XI. PROB.

To divide the circumference of a given circle successively into five, ten, and twenty equal parts.

Mark out the semicircumference ADEC, by the triple insertion of the radius, from A and C, with the double chord AE, describe arcs intersecting in F, from A, with the distance BF, cut the circle in G and K, inflect the chords GH and GI equal to the radius AB, and, from the points H and I, with the distance

BL to that of the decagon inscribed in the given circle. Hence AL may be inflected five times in the circumference, and BL ten times; and consequently the arc MK, or the excess of the fourth above the fifth, is equal to the twentieth part of the whole circumference.

Scholium. This proposition, and the preceding, include the happiest application of the circle to the solution of such problems.