PROP. XIV, PROB.

The side of a regular octagon being given, to mark out the figure.

Let the side of an octagon terminate in the points A and B; to find the remaining corners of the figure.

From the centres A and B, with the radius AB, describe the two semicircles AEFC and BEGD; with the double chord AF, and from A, C and B, D describe arcs intersecting in H, I; from these points, with the radius AB, cut the semicircles in K, L: on HI de scribe the square HMNI, by making the diagonals HN,

D

A

the octagon.

C

M

B

S

: For (by App. II.Pond AS Prop. 10.) BH, AI are both of them perpendicular to BA, and BKH, ALI are right-angled isosceles triangles; HI is therefore parallel to BA, and HMNI, consisting of triangles equal to BKH, is a square; whence all the sides AB, BK, KO, OM, MN, NP, PL, and LA of the octagon are equal: But they likewise contain equal angles; for ABK, composed of ABH and HBK, is equal to three half right angles, and BKO, by reason of the parallels BH and KO, being the supplement of HBK, is also equal to

three half right angles. In the same manner, the other angles of the figure may be proved to be equal.

PROP. XV. PROB.

On a given diagonal to describe a square.

Let the points A and B be the opposite corners of a square which it is required to trace.

From B as a centre describe the semicircle ADEC, from A and C with the double chord AE describe arcs intersecting in F, from C with the distance BF describe an arc and cut this from A with the radius AD in G, and lastly from B and A with the distance BG describe arcs inter

secting in H and I: ABHI is the required square.

For, in the triangle AGC, the straight line GB bisects the base, and consequently (II. 22.) AG2 + CG2 = 2 AB2 + 2BG* ; but, (by App. II. Prop. 10.).

==

D

B

CG BF 2AB'; whence AG AB'= 2BG*, and (II. 11.) AHB is a right angle; and the sides AH, HB, BI, and IA, being all equal, the figure is therefore a

square.

PROP. XVI. PROB.

Two distances being given, to find a third proportional to them.

Let it be required to find a third proportional to the

distances AB and CD.

From any point F with the distance AB, describe a portion of a circle, in which insert FG equal to CD, and from G, with that distance, describe the

semicircle FHI; HI

is the third propor

tional required.

For the angles GEH

and IGH are each of

them double the angle GFH or IFH at the circumference (III. 17. El.); whence the triangles GEH and IGH must also have the angles at the base equal, and are consequently similar: Wherefore (VI. 12. El.) EG: GH:: GH: HI.

If the first term AB be less than half the second term CD, this construction, without some help, would evidently not succeed. But AB may be previously doubled, or assumed 4, 8, or 16 times greater, so that the circle FGH shall always cut FHI; and in that case, HI, being likewise doubled, or taken 4, 8, or 16 times greater, will give the true result.

PROP. XVII. PROB.

To find a fourth proportional to three given distances.

Let it be required to find a fourth proportional to the

distances AB, CD, and EF.

From any point G, describe two concentric circles HI and KL with the distances AB and EF; in the circumference of the first insert HI equal to CD, assume any point K in the second circumference, and cut this in L by an arc described from I with the distance HK; the chord LK is the fourth proportional required.

I

A

B

CD

EF

H

L

K

For the triangles ILG and HKG are equal, since their corresponding sides are evidently equal; whence the angle IGL is equal to HGK, and taking away HGL, the angle IGH remains equal to LGK; consequently the isoceles triangles GIH and GLK are similar, and GI: IH:: GL: LK, that is, AB: CD:: EF: LK.

If the third term EF be more than double the first AB, this construction, it is obvious, will not answer without some modification. It may, however, be made to suit all the variety of cases, by multiplying equally AB and the chord LK, as in the last proposition.

PROP XVIII. PROB.

To find the linear expressions for the square roots of the natural numbers, from one to ten Inclusive.

This problem is evidently the same as, to find the sides of squares which are equivalent to the successive multiples of the square constructed on the straight line representing the unit. Let AB, therefore, be that measure: And from B as a centre, describe a circle, in which inflect the radius four times, from A to C, D, E, and F; from the opposite points A and E, with the double chord AD, describe arcs intersecting in G and H,-with the same distance, and from the points D, F, describe arcs intersecting in I,and, with still the same distance and from E, cut the circumference in K;

and IL 10 AB'.

For, in the isosceles triangles ACB and BDE, the perpendiculars CO and DP must bisect the bases AB and BE; and the triangle ADI being likewise isosceles, IP= But from what AP, and consequently IB=AE=2AB. has been formerly shown, it is evident that AK' = 2 AB* and AD 3 AB'; and since AE= 2 AB, AE' = 4AB'.