GHKLM.

B

form a polygon GHKLM, which has as many sides as there are plane angles constituting the solid angle at A. Also each of the sides of the polygon as GH is the base of a triangle in the plane of one of the plane angles, and having its opposite angular point at A. Now since the solid angle at a is constituted by the three plane angles AGM, AGH, MGH, the two AGM, AGH are greater (xi. 20) than the third мGH. For the same reason the two plane angles at each of the points H, K, L, M, which are at the bases of the triangles having the common angular point at A, are each greater than the third angle at the same point, which is one of the angles of the polygon Therefore all the angles at the bases of the triangles are together greater than all the angles of the polygon GHKLM. To each of these unequals add four right angles; therefore all the angles at the bases of the triangles together with four right angles are greater (Ax. 4) than all the angles of the polygon together with four right angles. But all the angles of the polygon GHKLM together with four right angles are equal to twice as many right angles as the polygon has sides (i. 32. Cor. 1), that is, as there are plane angles constituting the solid angle at A; therefore likewise all the angles at the bases of the triangles together with four right angles are greater than twice as many right angles as there are plane angles constituting the solid angle at A. Now because the three angles of each of the triangles are equal to two right angles (i. 32), all the angles of the triangles are

equal to twice as many right angles as there are triangles, that is, as there are plane angles constituting the solid angle at A: hence all the angles at the bases of the triangles together with four right angles are greater than all the angles of the triangles, that is, than all the angles at the bases of the triangles together with all the angles at A. From each of these unequals take away the common angles at the bases of the triangles; then the remaining four right angles are greater (Ax. 5) than the angles of the triangles at a. That is, the plane angles BAC, CAD, DAE, EAF, FAB, which constitute the solid angle at A, are together less than four right angles. Which was to be proved.

OBS. The proof (I.) is but a particular case of the general demonstration (II.), and may be included in it.

THE

ELEMENTS OF EUCLID.

BOOK XII.

PROP. A.

If from the greater of two unequal magnitudes there be taken away a part greater than its half, and from the remainder a part greater than its half; and so on: then this operation can be repeated until there is left a remainder less than the less of the two unequal magnitudes.

OBS. This propn is required for the proof of Bk. xii. Prop. 2. It is the first prop" in the Tenth Book of the Elements.

Let AB and c be two unequal magnitudes of which AB is the greater; and from AB let there be taken away a part greater than its half, from the remainder a part greater than its half, and so on. Then this operation can be repeated until there is left a magnitude less than c.

Since the less magnitude c may be multiplied a sufficient number of times for its multiple to exceed the greater AB, let it be so multiplied, and let its multiple be DE, which is greater than AB. From AB take AH greater than its half; from the remainder FB take FG greater than its half; and repeat the operation till the number of divisions in AB is equal to the number of times c was multiplied. Let these divisions be AF, FG, GH, HB; and let DE be divided into magnitudes each equal to c, viz.

сс

DK, KL, LM, ME, the number of these divisions being equal to that of the divisions of AB.

taken a part DK not greater than its half, D while from AB is taken

a part AF greater than

its half: therefore the remainder KE is greater than the remainder FB. Again, because KE is greater than FB; and from KE is taken KL not greater than its half, and from FB is taken FG greater than its half; therefore the remainder LE is greater than the remainder GB. And proceeding in like manner for the corresponding divisions of DE and AE, we should have at last the remainder ME greater than the remainder HB. But ME is equal to c by const; therefore c is greater than HB. Hence the operation has been repeated until there remains a magnitude HB less than c. Which was to be proved.

COR.-If from the greater of two unequal magnitudes there be taken away its half, and from the remainder its half; and so on: then this operation can be repeated until there is left a remainder less than the less of the two unequal magnitudes.

The proof of this is exactly similar to that of the prop".

PROP. I. THEOR.

If similar polygons be inscribed in circles: then they shall be to one another as the squares of the diameters of the circles in which they are inscribed.

Let ABC, FGH be two circles; and in ABC, FGH let there be inscribed the similar polygons ABCDE, FGHKL. Then the polygon ABCDE shall be to the polygon FGHKL as the square of the diameter of the circle ABC is to the square of that of the circle FGH.

The angles at A, B, C, D, E being supposed those which by the defa (vi. Def. 1) of similar polygons are respectively equal to those at F, G, H, K, L, and the sides AB, BC, CD, DE, EA those which are homologous to FG, GH, HK, KL, LF, find (iii. 1) the centres of the circle ABC. FGH; and through A, F draw their diameters AM, FN. Join MB, NG, AC, FH.

Since the angle ABC is equal to the angle FGH, and AB is to BC as FG to GH (vi. Def. 1); therefore the two triangles ABC, FGH have the angle ABC equal to the angle FGH, and the sides about this pair of equal angles proportional. Therefore these two triangles are similar (vi. 6); and hence the angle ACB is equal to the angle FHG. But the angle AMB is equal to the angle ACB, because they are in the same segment AMCB; and the angle FNG is equal to the angle FHG for like reason: therefore the angle AMB is equal to the angle FNG. Also the angle ABM in the semicircle ABM, and the angle FGN in the semicircle FGN are right angles; and all right angles are equal: therefore the angle ABM is equal to the angle FGN. Hence the two triangles ABM, FGN have the two angles AMB, MBA respectively equal to the two angles FNG, NGF, and consequently (i. 32. Cor. A) the third angle BAM equal to the third angle GFN; therefore these two triangles are equiangular to one another. Therefore they are similar (vi. 4); and hence AB is to AM as FG to FN. Therefore, alternando (v. 16), AB is to FG as AM to FN; and the duplicate ratio of the ratio of AB to FG is the same as the duplicate ratio of the ratio of AM to FN. Now the polygons ABCDE, FGHKL are similar by hyps, and the squares described on AM, FN are similar, since regular polygons of the same number of sides are similar (vi. Def. 1. Obs. 3); and similar polygons have