Cor. 1. Hence it is evident that every equilateral figure inscribed in a circle is equiangular.

Cor. 2. An equiangular figure, inscribed in a circle, has its alternate sides equal, as is evident from the equality of the arches which contain equal angles.

Schol. Hence if the number of sides be odd, the figure is equilateral.

Cor. 3. An equilateral pentagon can be constructed Fig. 13. upon a given line, by constructing upon this line as a base an isosceles triangle ACE, in which each of the

angles at the base is double of the angle at the vertex (1), (1)Prop.10. and circumscribing a circle about it; the pentagon in- B. 4. scribed in this circle is constructed upon the given line.

PROP. XII. PROB.

About a given circle (ABCDE) to circumscribe an equi- Fig. 14. lateral and equiangular pentagon.

Let the points A, B, C, D and E be the vertices of the angles of an equilateral pentagon inscribed in the circle and draw GH, HK, KL, LM and MG tangents to the circle at these points; GHKLM is an equilateral and equiangular pentagon, circumscribed about the given circle.

B. 3.

Draw FĂ, FG, FE, FM and FD. In the triangles FGA, FGE the sides GA and GE are equal (1), and (1) Cor. 2. also FA and FE, FG is common, therefore the angles Prop. 36. FGA and FGE are equal, and also AFG and EFG (2), (2) Schol. therefore the angle AGE is double of FGE and AFE Prop. 8, double of GFE; in the same manner it can be demon- B. ì. `strated that DME is double of FME, and that DFE is double of MFE; but, since the arches AE and ED are equal (3), the angles AFE and DFE are equal (4), (3) Const. and therefore their halves GFE and MFE are equal, (4) Schol. and the angles FEG, and FEM are also equal, and the Prop. 29. side EF is common, therefore the angles FGE and FME are equal, and also the sides GE and EM (5), (5) Prop.26. and therefore the line GM is double of GE; in the B. 1. manner it can be demonstrated that GH is double of GA, but GE and GA are equal, therefore GM and GH are equal; in the same manner it can be demon

same

B. 3.

Fig. 15.

strated that the other sides are equal, and therefore the pentagon GHKLM is equilateral: and because the angles DME and AGE are double of FME and FGE, and FME and FGE are equal, DME is equal to AGE, and in the same manner it can be demonstrated that the other angles are equal, and therefore GHKLM is also equiangular.

Cor. In the same manner any equilateral and equiangular figure can be circumscribed about a circle, by inscribing a similar figure and drawing tangents through the vertices of its angles.

PROP. XIII. PROB.

In a given equilateral and equiangular pentagon
(ABCDE) to inscribe a circle.

Bisect any two adjacent angles A and E by the right lines AF and EF, and from their point of concourse F draw FG perpendicular to AE; the circle described from the centre F with the radius FG is inscribed in the given pentagon.

Draw FB, FC and FD and from F let fall the perpendiculars FH, FN, FM, FL.

In the triangles AFB, AFE the sides AB and AE (1)Hypoth. are equal (1), AF is common, and the angles FAB and (2) Constr. FAE are equal (2), therefore the angles ABF and (3) Prop. 4. AEF are equal (3), but the angles ABC and AED are

B. 1.

also equal (1), therefore since AEF is half of AED (2), ABF is half of ABC; in the same manner it can be demonstrated that the other angles of the pentagon are bisected by the lines drawn from F: wherefore in the triangles FBH, FBN, the angles FBH and FBN are equal, the angles at H and N are right, and the side FB opposite to the equal angles H and N is common, (4) Prop.26. therefore the sides FH and FN are equal (4), and in the same manner it is proved that all the perpendiculars are equal; therefore the circle described from the centre F with the radius FG passes through the points H, N, M and L, and the sides of the given pentagon are tangents to it, because the angles at G, H, N, M and L are right.

B. 1.

Schol. In the same manner a circle can be inscribed in any equilateral and equiangular figure.

PROP. XIV. PROB.

To circumscribe a circle about a given equilateral and Fig. 16. equiangular pentagon (ABCDE).

Bisect the angles A and E by the right lines AF and EF, the circle described from their point of concourse F as centre with the radius AF passes through the points B, C, D and E.

B. 1.

Draw FB, FC and FD. In the triangles FAE and FAB the sides FA and AE are equal to FA and AB, and the angle FAE is equal to FAB (1), therefore the (1) Constr. angles FBA and FEA are equal (2), but the angles (2) Prop. 4. ABC and AED are also equal (3), therefore since the (3) Hypoth. angle FEA is half of AED (1), FBA is half of ABC and therefore ABC is bisected by FB, and in the same manner it can be demonstrated that the angles C and D are bisected: Hence in the triangle AFE the angles FAE and FEA, being halves of the equal angles BAE and AED, are equal, and therefore the sides FE and FA are equal (4), and in the same manner it is proved (4) Prop, 6. that the remaining lines FB, FC and FD are equal; B. 1. therefore the five lines FA, FB, FC, FD and FE are equal, and therefore the circle described from the centre F with the radius FA passes through the points B, C, D and E, and therefore is circumscribed about the given pentagon.

Schol. In the same manner a circle can be circumscribed about any equilateral and equiangular figure.

PROP. XV. PROB.

In a given circle (ABCDEF) to inscribe an equilateral Fig. 17. and equiangular hexagon.

Let G be the centre of the given circle, draw any diameter AGD, from the centre A with the radius AG describe a circle and from its intersections B and F with

Prop. 32,
B. 1.

the given circle draw the diameters BE and FC, join AB, BC, CD, DE, EF and FA, and the figure ABCDEF is an equilateral and equiangular hexagon inscribed in the given circle.

Since the lines AB and AG are equal as being radii of the same circle BGF, and GA and GB also equal, as being radii of the same circle ABCDEF, the triangle BGA is equilateral, and therefore the angle BGA is (1) Cor. 4. the third part of two right angles (1); in like manner it is proved that the triangle AGF is equilateral and the angle AGF equal to one third part of two right angles, but the angles BGA and AGF together with FGE are (2) Cor. 1. equal to two right angles (2), therefore FGE is one third part of two right angles, and therefore the three angles BGA, AGF and FGE are equal, and also the angles vertically opposite to them EGD, DGC and CGB, hence the six angles at the centre G are equal, therefore the arches on which they stand are equal, and the lines subtending those arches (3), therefore the (4) Cor. 1. hexagon ABCDEF is equilateral, and also, since it is inscribed in a circle, equiangular (4).

Prop. 13.

B. 1.

(3) Schol. Prop. 29. B. 3.

Prop. 11.

B. 4.

Fig. 17.

Fig. 17.

Fig. 17.

(1) Schol. Prop. 29.

B 3. (2) Prop.26. B. i.

Cor. 1. The side of a hexagon is equal to the radius of the circle, in which it is inscribed.

Cor. 2. If the arches AB, BC, &c. were bisected, a figure of twelve sides might be inscribed, and, these arches being bisected, a figure of twenty-four sides, and

so on.

Cor. 3. An equilateral and equiangular hexagon can be constructed upon a given line BA, by describing upon it an equilateral triangle, BGA, and from the centre G with the radius GA describing a circle and inscribing in it a hexagon.

Cor. 4. Draw AC, AE and CE, and an equilateral triangle is inscribed in the circle, whose sides bisect the radii perpendicular to them. For let GK be perpendicular to CE, it passes through D, therefore in the triangles DCK, GCK, the angles at K are right, the angles DCK and GCK standing upon the equal arches DE and EF are also equal, (1) and the side CK between the equal angles common, therefore the sides DK and KG are equal (2), and therefore GD is bisected in K.

Cor. 5. Hence it follows that the square of the side of an equilateral triangle is triple the square of the ra

dius of the circle in which it is inscribed; draw AC and its square is equal to the sum of the squares of AG and GC and to twice the rectangle under AG and GK (1), (1) Prop.12. but KG is the half of GD (2), therefore double the B. 2. rectangle AGK is equal to the square of AG, therefore (2) Cor. the square of AC is equal to twice the square of AG and the square GC, or to triple the square of radius.

PROP. XVI. PROB.

prec.

In a given circle (CAD) to inscribe an equilateral and Fig. 18.

equiangular quindecagon.

Let CD be the side of an equilateral triangle inscribed in the circle CAD, and CA the side of an equilateral pentagon also inscribed in the circle CAD, bisect the arch AD, the right line joining AB is the side of the inscribed quindecagon. For if the whole circumference be divided into fifteen parts, the arch CD, since it is the third part of the whole circumference, contains five of these parts, in like manner the arch CA contains three of them, therefore the arch AD contains two, and therefore the arch AB is the fifteenth part of the whole circumference, and AB is the side of the inscribed equilateral quindecagon.

See N.

N