PROP. X. PROB.

To constitute an isosceles triangle, having each of the angles at the base double the vertical angle.

Take any right line AB, and divide it in the point C, so that the rectangle ABC may be equal to the square of AC [11. 2], and, from the centre A, at the distance AB, describe a circle BDG, in which inscribe BD equal to AC [1. 4], and join AD; the isosceles triangle ABD is such as is required, having each of the angles at the base ABD, ADB, double the vertical angle BAD.

G

B

C

Draw DC, and let the circle DCA be circumscribed about the triangle ACD [5. 4].

Because the rectangle ABC is equal to the square of AC [Constr.], or its equal [Constr.] BD, the right line BD touches the circle ACD [37. 3], and therefore the angle BDC is equal to the angle DAC in the alternate segment [32. 3]; adding to each the angle CDA, the angle BDA is equal to CDA, CAD together; but, because the sides AB, AD are equal, the angles ABD, ADB are equal (5. 1), therefore the angle ABD is equal to the angles CDA, CAD together, or, which is equal (32. 1), the angle BCD, and therefore CD is equal BD (6. 1), or its equal CA, and, of course, the angles CAD, CDA are equal (5. 1); whence the angle BDA, or its equal ABD, being equal to CDA, CAD together, is double to CAD, and so an isosceles triangle BAD is constituted, having each of the angles at the base BD, double the vertical angle BAD.

Cor. 1.-The greater segment AC, of the radius AB, of a circle, so divided into two parts, that the square of the greater part, is equal to the rectangle under the whole and the other part, is equal to the side of a regular decagon, inscribed in a circle.

For, since the angle BAD is half of either of the angles ABD, ADB, it is one fourth of both together, and one fifth of all the angles of the triangle ABD, or (32. 1), of two right angles,

and therefore one tenth part of four right angles, or [Cor. 2. 15. 1], of all the angles which can be formed about the point A; whence, since in equal circles, equal angles at the centre, are subtended by equal chords (Cor. 2. 29. 3), BD, or its equal AC, is equal to the side of an equilateral, and therefore (Schol. 6. 4 and Def. 7. 4), of a regular decagon inscribed in the circle.

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

Inscribe in the circle BDG, the right lines BF, FG each equal to BD (1. 4), draw AF, AG, GB, GC, and let fall the perpendicular GH on AB; and, since all the sides of the triangle AFB, are severally equal to all the sides of the triangle ABD, the angles FAB, BAD are equal (8. 1); in like manner the angles GAF, BAD may be proved equal; therefore the angle GAB is double the angle BAD, and one fifth part of four right angles; whence, in like manner, as in the preceding is shewn of BD, the right line GB may be shewn to be the side of a regular pentagon inscribed in the circle; and since, in the triangles GAC, ADB, the sides GA, AC and the angle GAC, are severally equal to the sides AD, DB and the angle ADB, the base GC is equal to AB (4. 1), or its equal GA, and so the triangle GAC is isosceles, and the perpendicular GH bisects AC (Cor. 26. 1); therefore, in the triangle BGA, the rectangle under the sum and difference of GB, GA, or, which is equal (Schol. 6. 2), the difference of their squares, is equal to the rectangle under the sum and difference of BH, HA (Cor. 1. 5 and 6. 2), or the rectangle ABC, or square of AC or BD; and so the square of GB the side of a regular pentagon inscribed in the circle, is equal to the squares of the radius AG, and of BD the side of a regular decagon inscribed in the same circle.

PROP. XI. PROB.

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

Make an isosceles triangle FGH, having each of the angles F, G double the angle H (10. 4), and inscribe in the given circle, the triangle ABD equiangular to the triangle FGH (2. 4), bisect the angles at the bases DAB, DBA by the right

lines AC, BE, and join AE, ED, DC and CB.

B

Because the angles DAB, DBA, are each of them double to ADB, and bisected by the right lines AC, BE, the five angles ADB, DAC, CAB, ABE, EBD are equal, and there-. fore, the right lines AB, BC, CD, DE and EA are equal (Cor. 2. 29. 3); and so the pentagon ABCDE, which is inscribed in the given circle (Def. 3. 4), is equilateral, and, of course, regular (Schol. 6. 4).

the square of AH is equal to the squares of AF and FH (47. 1', therefore AH is equal to the side of a regular pentagon inscribed in the same circle [Cor. 2. 10. 4]; whence the required pentagon may be easily described, by applying in the given circle right lines AB, BC, CD and DE each equal to AH, and drawing AE.

Scholium.-The division of FG in H, as required above, may be performed, by producing GF to meet the circle in K, bisecting KF in L, and taking LH equal to the distance LA, as is manifest from the construction of Prop. 11. 2.

PROP. XII. PROB.

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

Inscribe in the given circle, the regular pentagon ABCDE [11.4], through the vertices of the angles of which, draw GH, HI, IK, KL and LG, touching the circle in these vertices [17. 3]. The pentagon GHIKL, which is circumscribed about the given circle (Def. 5. 4), is a regular one.

For, drawing FA, FB and FC from the centre F, because the triangles FAB, FBC are mutually

equilateral, the angles FAB, FBA, FBC, FCB are equal (8. 1), which being taken from the angles FAH, FbH, FBI, FCI, which are equal, being right [18. 3], the remaining angles HAB, HBA, IBC, ICB are equal; whence, in the triangles HAB, IBC, the right lines AB, BC being also equal (Constr. and Def. 7. 4), the sides AH, HB, BI, IČ, as also the angles H and I are equal (26. 1); in like manner, AG may be proved equal to AH, HB or BI, therefore GH, HI are equal (Ax. 2). In like manner, may the sides IK, KL, LG be proved equal to GH or HI, and to each other; therefore the circumscribed pentagon is equilateral; it is also equiangular, since, in like manner as the angles H and I have been proved equal, the angles K, L, G may be proved equal to either of them, and to each other; it is therefore a regular one [Def. 7. 4].

Scholium.-In like manner, as is shewn in this proposition, may a regular polygon, be circumscribed about a given circle, of the same number of sides, as any given regular polygon, inscribed in the same circle.

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. 3], and is therefore inscribed therein (Def. 6. 4).

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