PROP. VI. PROB.

To cut off any aliquot part of a given distance.

Suppose it were required to cut off the fifth part of the distance between the points A and B.

Repeat (App. II. 1.) the distance AB four times, to F; from F, with the radius FA, describe the arc GAH; inflect the chords AG

is the fifth part of the line of communication AB.

For, as before, the point I is situate in AB. But since AGI is evidently an isosceles triangle, and AF is equal to FG, it follows (II. 23. cor.) that AG2=AF.AI, and consequently AB5AB.AI; whence AB=5AI.

PROP. VII. PROB.

To divide a given distance by medial section.

Let it be required to cut the distance AB, such that BH-BA.AH.

From B describe a circle with the radius BA, which insert successively from A to D, E, C, and F; from the extremities of the diameter AC and with the chord AE, describe two arcs intersecting in G; and, from the

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 trian

gles AGB, CGB have their sides respectively equal, the angle ABG (I. 2.) is a right angle, and consequently (II. 10.) AG2=AB+BG"; but AG=AE, and AE=3AB2 (IV. 17. cor. 2.); wherefore 3AB AB2 + BG2, and BG22AB". Now since BE EC, it follows, (II. 20.)

A

B

that HE-BE2=CH.HB; but HE-BE2 — BG2. BE2=AB2, and therefore AB2=CH.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=2DC2+2CB* (II. 22.), or BD2=2DC2+CB2; and since DB=DF, 2DC*+CB2=DC+CF2, whence DC+CB=CF", 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.

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

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, 3AB AF2=AB2+BF", and therefore BF AG2=2AB2; whence (II. 12.) ABG is a right angle, and AG a quadrant.

=

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