PROP. XIII. PROB.

In a given regular pentagon (ABCDE), to inscribe a circle.

Because then, in the triangles AFE, AFB, the sides AE, AB are equal [Hyp.], AF common, and the angles FAE, FAB equal [Constr.], the angles AEF, ABF are also equal [4. 1]; but the angles AED, ABC are equal [Hyp.], therefore, since ABF is the half of ABC [Constr.], AEF is half of AED; in like manner, it may be demonstrated, that the angles EDC, DCB are bisected by the right lines drawn to them; therefore, in the triangles FEL, FEK, the angles FEL, FEK are equal, as also the angles at L and K, being right angles (Constr.), and the side FE is common, therefore the sides FL, FK are equal [26. 1]; in like manner, it may be demonstrated, that all the other perpendiculars are equal to each other; therefore a circle described from the centre F, at the distance FG, passes through H, I, K, L, and, because of the right angles at G, H, I, K, L, touches the sides of the pentagon in these points [Cor. 16. s], and is therefore inscribed therein (Def. 6. 4).

Schol.-In like manner, a circle may be inscribed, in any regular polygon.

PROP. XIV. PROB.

About a given regular pentagon (ABCDE), to circumscribe a circle.

Bisect any two adjacent angles EAB, ABC by the right lines AF, BF meeting in F; a circle, described from the centre F, at the distance FA, is circumscribed about the given E pentagon.

For let FB, FC, FD, FE be joined, and in the triangles FAB, FAE, the sides FA, AB and the angle FAB, are severally equal to FA, AE and the angle FAE, therefore the angles

D

ABF, AEF are equal (4. 1); but the angles ABC, AED are equal (4. 1); but the angles ABC, AED are equal (Hyp. and Def. 7. 4), and ABF is the half of ABC, therefore AEF is the half of AED, and so the angle AED is bisected by EF; in like manner it may be shewn, that the other angles of the pentagon at D and C are bisected. Since then, in the triangle FAB, the angles FAB, FBA, being the halves of the equal angles EAB, ABC, are equal (Ax. 7), FA is equal to FB (6. 1); in like manner it may be shewn, that FC, FD, FE are each of them equal to FA or FB, therefore the five right lines FA, FB, FC, FD and FE are equal to each other, and the circle, described from the centre F, at the distance FA, passes through B, C, D and E, and is circumscribed about the given pentagon (Def. 4. 4).

Schol.-In like manner, a circle may be circumscribed, about any regular polygon.

PROP. XV. PROB.

In a given circle (ABCDEF), to inscribe a regular hexagon.

Having found the cen

A

I

tre G of the given circle, and drawn the diameter AGD, inscribe in the cir.. cle AB, DC, AF, DE each equal to AG, join GB, GC, GE, GF, then since the triangles AGB, CGD, DGE, FGA are equilateral ones, and therefore equiangular (Cor. 5. 1), their angles at the point G, are each of them equal to one third of two right angles (32. 1), and the angles AGB, DGC together, equal to two third parts of two right angles, and therefore the angle BGC equal to a third part of two right angles (Cor. 13. 1); in like manner it may be shewn, that the angle EGF is a third part of two right angles, and so each of the six angles, formed at the point G, is equal to a third part of two right angles, and therefore to each other; whence, the chords subtending them, which are the sides of the hexagon ABCDEF, inscribed in the circle (Def. 3. 4), are equal (Cor. 2. 29. 3), which hexagon is therefore regular (Schol. 6. 4. and Def. 7. 4). Cor. 1.-The side of a regular hexagon, inscribed in a circle, is equal to the radius.

For, in the above construction, AB one of the sides of the inscribed hexagon, is equal to the radius AG (Constr.).

Cor. 2.-The square of the side of a regular pentagon, inscribed in a circle, is equal to the squares, of the sides of a regular hexagon, and regular decagon, inscribed in the same circle.

For the square of the side of the pentagon, is equal to the squares of the radius and the side of the decagon (Cor. 2. 10. 4), or, radius being equal to a side of the hexagon (Cor. 1. 15. 4), to the squares of the sides of the hexagon and decagon.

Cor. 5.-The square of the side of an equilateral triangle, is triple the square of the radius of the circumscribing circle.

For FB, BD, DF being joined, are equal, because they subtend equal angles FGB, BGD, DGF at the centre (Cor. 2. 29. 3), and so the triangle FBD is equilateral, and the angle DBA in a semicircle is right (31. 3), therefore the squares of DB, BA together, are equal to the square of DA (47. 1), or, which is equal (Cor. 4. 2), to four times the square of AG, or (Constr.), AB; taking from each the square of AB, there remains the square of DB, a side of the equilateral triangle FBD inscribed in the circle, equal to three times the square of AB, or, of the radius AG.

Cor. 4.-The altitude of an equilateral triangle, is triple the radius of the inscribed, and triple to half the radius of the circumscribed circle.

Part 1.-Let the right lines HAI (see fig. on preced. page), ICK and KEH touching the circle ACE in A, C and E, meet. each other in H, I and K; join DK, BI, aud in the triangles DCK, DEK, the sides DC, DE are equal, being sides of the hexagon, CK is equal to EK (Cor. 2. 36. 3), and DK is common, therefore the angles CDK, EDK are equal (8. 1); whence the angles GDC, GDE being angles of equilateral triangles, and therefore equal, the angles GDC, CDK together are half the four angles GDC, CDK, KDE, EDG, and therefore equal to two right angles (Cor. 2. 15. 1), therefore AD, DK make one right line (14. 1); and since the angle DCK is equal to the angle in the alternate segment DFC (32. 3), or to CDB, which is equal to DFC (27. 3), they standing on the equal arches DC, CB, the right lines CK, BD are parallel (27. 1); whence, the angle DCK being equal to DFC in the alternate segment (32. 3), or to BDA, which is equal to DFC (27. 3), they standing on equal arches DC, AB, or, which is equal (29. 1), the internal remote angle CKD, DK is equal to DC (6. 1), or the radius AG; in like manner, EB, BI may be proved to make one right line, and BI to be equal to AG; and since, in the triangles GCI, GCK, the angles at G are equal, those at C right, and GC common, CI, CK are equal (26. 1), and CK, EK are equal (Cor. 2. 36. 3); in like manner may EH, HA and AI be each of them proved equal to CI, CK or EK, and to each other; therefore HI, IK and KH are equal to each other (Ax. 2 or 6), and the triangle HIK equilateral (Def. 28. 1), and circumscribed about the circle ACE (Def. 5. 4); of which triangle, AK, being at right angles to HI (18. 3), is the altitude, which, DK being equal to AG or GD, is triple the radius AG.

Part 2.-In the triangles ABL, GBL, AB is eqnal to BG, BL common, and the angles ABL, GBL standing on the equal arches AF, FE also equal (27. 3), therefore AL and GL are equal, and the angles ALB, GLB are equal (4. 1), and therefore right angles [Def. 20. 1]; whence, FBD being an equilateral triangle, inscribed in the circle ACE, as appears by the preceding corollary, and, because of the right angles at L, DL being its altitude, that altitude is triple the half radius AL or GL.

Cor. 5.-Of two regular polygons, of an unequal number of sides, an angle of that which has the greater number of sides, is greater, than an angle of the other.

First, let the difference of the number of sides be one, and an angle of that, which has the greater number, is not equal to an angle of the other, for then the total of its angles, would exceed the total of the angles of the other, only by one of the angles of that other, and, therefore, by less than two right angles, which is absurd [Cor. 1. 32. 1]. A like absurdity would follow, if an angle of that, which has the greater number of sides, were supposed to be less; since therefore it can neither be equal to, or less than, an angle of that other, it is greater.

Secondly, let the difference of the number of sides be two; and an angle of that, which has the greater number of sides, is greater, than an angle of that, which has the next greater (Part 1. of this Cor.), and an angle of that, than an angle of the other (Part 1. of this Cor.).

In like manner we might proceed, if the difference of the number of sides were three, four, or any other number.

Cor. 6.-Only three regular figures, namely, equilateral triangles, squares and hexagons, can constitute one continued superficies.

For, in order thereto, it is necessary, that a certain number, of the angles of the regular figure, which is to constitute such a superficies, should be equal to four right angles, all the angles which can be formed about any point being equal to so many right ones [Cor. 2. 15. 1]; but six angles of an equilateral triangle are equal to four right angles, three of them being equal to two right ones (32. 1), the four angles of a square are four right angles [Def. 36. 1), and, since all the six angles of a regular hexagon, are equal to eight right angles (Cor. 1. 32. 1), three of them are equal to four right ones.

But the angles, of no other regular figure, can constitute a continued superficies, for no fewer angles, of a regular polygon, than three, can be equal to four right angles, since no such an