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BD, intersecting each other in E. Then the tri. angles ABC, ADC, have two sides AB, AD, equal,
А and AC common; also the bases BC, DC are equal; therefore the angle CAB equals the angles CAD (a).
În like manner, it is shown, that the other angles of the square are bisected by the straight lines AC, BD, (b): but the angles of the square are all equal to each other (c); therefore the halves BDA, BDC, ACB, ACD, DBA, DBC are also equal to each other (d); and their bases EA, EB, EC, ED are equal (e), and E is the centre of a circle that shall pass through the angular points A, B, C, D, (f).
Wherefore, a circle has been described about a given square; which was to be done. Recite (a) p. 8, 1;
(6) p. 9, 1; (c) def. 30, ax. 10, 1; (d) ax. 7, 1;
(e) p. 4, 1; ($ ) p. 9, 3.
10 P. To describe an isosceles triangle (ALK), having each of the angles (L, K) at the base double of the angle (A) contained by the equal sides (AL, AK).
Constr. With any radius AB, and centre A, describe the circle BDLK (a); produce BA to meet the circumference in L (6); upon AB de JE scribe the square ABCD (c); bisect AB in E (d), and join ED; make EF equal to ED (e); place
D AF in the circle from L to K (f); join AK, FK (8); about the triangle AFK describe a circle (h).
Argument. The radius AL is so divided in that the rectangle ALXLF equals the
square of AF (i); but AF is equal to LK (k), which meets the circle AFK in the point K, and AL cuts the same circle in F (1); therefore the square of LK equals the rectangle ALXLF, and LK touches the circle AFK (m). Now ALK is an isosceles triangle, because AL, AK are equal radii (n); therefore the angles ALK, AKL are equal (o): Again, because LK touches the circle AFK, and from the point of contact KF is drawn in the circle, the angle FKL is equal to FAK in the alternate segment (P); but the angle AKL equals FKL and FKA, or FAK and FKA together; and the exterior angle KFL equals the same FAK and FKA (q); therefore KFL equals AKL, or ALK (r), and KFL is an isosceles triangle, in which KL and KF are equal (n); but KL is made equal to AF (6); therefore KF equals AF (r), and AKF is an isosceles triangle (n), having the angles FKA, FAK equal (0); but AKL, or ALK proves equal to both, or double the angle FAK, or LAK.
Wherefore, an isosceles triangle has been described, &c. Q. E. F.
(d) p. 10, 1; (e) p. 3, 1; (f) p. 1, 4;
11 P. To inscribe an equilateral and equiangulat pentagon in a given circle (ABCDE).
Describe an isosceles triangle FGH(a), whose equal sides contain an angle F, half as great as the angle G, or H, at the
E base. Then in the circle inscribe a triangle ACD, equiangular to FGH; so that the angle C, or D, shall be double the angle A (). Bisect the angles ACD, G ADC by the chords CE, DB (c); and join AB, BC, CD, DE, EA. ABCDE is the pentagon required.
Because each of the angles ACD, ADC is double of the angle CAD; and are bisected by the chords EC, BD, the five angles DAC, ACE, ECD, CDB, BDA are equal to one another: but equal angles stand upon equal arcs (d); therefore the five arcs AB, BC, CD, DE, EA are equal to each other. Again, equal arcs subtend equal chords (e); therefore the five chords AB, BC, CD, DE, EA are equal to each other: wherefore the pentagon is equilateral.
It is also equiangular: becanse the arc AB equals the arc DE; add to each BCD; the sum ABCD is equal to the sum BCDE: and the angle AED stands on the arc ABCD; and the angle BAE on the arc BCDE; therefore the angles AED, BAE are equal (f); for the same reason, each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED; wherefore the pentagon is equiangular. And thus an equilate-al and equiangular pentagon has been inscribed in a given circle: which was to be done. Recite (a) p. 10, 4; (b) p. 2, 4; (c) p. 9, 1;
(d) p. 26, 3; (e) p. 29, 3; (f) p. 27, 3
12 P. To describe an equilateral and equiangular pentagon about a given circle (ABCDE).
Constr. Let A, B, C, D, E, be the angular points of an inscribed pentagon, as in the last (n); so that the arcs AB, BC, CD, DE, EA are Hi
M equal : then through the points A, B, C, D, E, draw the tangents GH, HK, KL, LM, MG (b). If these tangents be equal, and their angles equal, the required pentagon is described about the circle. Find the centre F, and draw the radii FB, FC, FD; join FK, FL.
Argument. In the triangles FBK, FCK, the angles at B, C arə right angles (c); and the common side FK subtends them: therefore the squares of FB, BK, and also of FC, CK are equal to the square of FK (d), and therefore equal to each other (e). But the squares of the equal radii FB, FC, are equal; therefore the remaining squares of BK, CK are equal; und BK is equal to CK (f). For the same reason CL and DL are equal: therefore the triangles FBK, FCK, FDL, FCL have two sides in each eqnal to two sides in every other,
and the angles contained by the two sides equal; therefore the other angles are equal, each to each, to which the equal sides are opposite (8); and so, FKB, FKC, FLC, FLD are equal; and any two of them are equal to any other two: therefore FKB, FKC are equal to FLC, FLD; that is, HKL is equal to KLM.
In like manner, it may be shown that each of the angles at H, G, M, is equal to HKL, or KLM: therefore the pentagon is equiangular.
Again, becanse the perpendicular FC bisects KL, the perpendiculars FB, FD bisect HK and LM: for they intercept equal arcs : and since the halves BK, KC, LD are equal, the wholes HK, KL, LM are equal. In like manner, it may be shown, that GH, or GM is equal to HK, KL, or LM. The pentagon is therefore equilateral ; and it proves also to be equiangular, and is inscribed in the given circle ABCDE; which was to be done. Recite (a) p. 11, 4; (6) def. 2, p. 16 and cor., also p. 17, of b. 3;
(c) def. 10, 1; (d) p. 47,1; (e)ax. 1;
13 P. To inscribe a circle in a given equilateral and equiangular pentagon (ABCDE).
Constr. Bisect the angles BCD, CDE, by the A А. straight lines CF, DF; and from the point F, in which they meet, draw FB, FA, FE.
B equal angles, by bisection, the bases FB, FD are equal; and the angle CBF equals the angle H Н CDF (a): but CDF is the half of CDE; therefore CBF is the half of CBA: because CBA equals CDE. Therefore ABC is bisected by the straight line BF. In the eame way,,
it may be shown, that the angles BAE, AED are bisected by the straight lines FA, FE.
From the point F draw FG, FH, FK, FL, FM, perpendicular upon the sides of the pentagon (b). Now, in the triangles FCH, FCK, the side FC is common; and the angles at C are equal, by bisection; and those at H, K are equal, as right angles (c); therefore the sides FH, FK are equal (d). In like manner, it may be shown, that each of the perpendiculars FL, FM, FG is equal to FH, or FK: therefore, these five straight lines are equal to one another; and a circle described upon the centre F, at the distance of any one of them, will pass through the extreme points of the other four; and touch the sides of the pentagon, where they meet the perpendiculars, in the points G, H, K, L, M (e). But a circle is inscribed in a rectilineal figure, when the circumference touches all the sides of the figure (f): wherefore a circle is inscribed, &c., which was to be done. Recite (a) p. 4, 1;
(6) p. 12, 1; (c) ax. 10;
14 P. To describe a circle about a given equilateral and equiangular pentagon (ABCDE).
Bisect two of the angles of the pentagon, as BCD, CDE, by straight lines CF, DF, meeting in the point F (a): join FB, FA, FE.
BA Argument. Because the equal angles BCD, CDE are bisected; the angles FCB, FCD are equal; and they are contained by the equal sides CB, CF; CD, CF: therefore the remaining sides FB, FD are equal, and also the angle CDF to CBF (b): but CDF is the half of CDE, which equals CBA; therefore FB bisects the angle CBA. In like manner it may be shown, that the angles BAE, AED are bisected by the straight lines FA, FE. Now the five triangles, whose vertices are in the point F, bave equal bases; namely, the sides of the pentagon; also equal angles adjacent to the bases (c), as above; therefore the sides FĀ, FB, FC, FD, FE, are equal to one another; and being drawn from the point F to the angular points of the pentagon, a circle described upon F, at the distance of any one of them, will pass through the five points, and be described about the pentagon; which was to be done (d). Recite (a) p. 9, 1;
(6) p. 4, 1; (c) p. 6, 1;
(d) def. 3, 4.
15 P. To inscribe an equilateral and equiangular hex. agon in a given circle (ABCDEF).
Let G be the centre of the given circle, and draw the diameter AGD (a): Again, on the
B point D, where the diameter meets the circumference, describe a circle to pass through G, and cut the circumference in the points E, A
H Н C (b): draw the diameters EGB, CGF; also chords between the points A, B, C, D, E, F, A, in the circumference. If these chords be eq ai, and their angles equal, the required hexagon is inscribed.
Because G and D are centres of equal circles, the radii GE, GD and DE are equal (c): GC, GD and DC are in the same case: therefore, the triangle GED, or GCD is equilateral (d); and being every way isosceles, it is also equiangular (e): therefore, the angle CGD is one third of two right angles (f ), EGD is also one third of the same; and because the straight line EG makes with CF, the adjacent angles equal to two right angles (8), therefore EGF is also one third of two right angles. The chord EF is therefore equal to ED, or DC (h); and these three are placed in the semicircle (i). The opposite, or vertical angles are also equal to these (k): therefore each of the angles AGF, AGB, BGC, is one third of two right angles (f); and the radii GB, GA, GF, being equal (c), tho chords AF, AB, BC are also equal to one another; and they are placed in a semicircle(i). Therefore the six chords AB, BC, CD, DE, EF, FA, divide the circumfer
ence, cut off equal arcs (1), and are therefore sides of an equilateral hexagon inscribed in the given circle ABCDEF.
But, since equal angles stand upon equal arcs (m), the arc AF equals the arc ED; to both add the arc ABCD (n): therefore the whole arc FABCD equals the whole ABCDE; and the angle FED stands upon the former, and the angle AFE upon the latter: therefore the angles IFE, FED are equal. In the same way, it may be proved, that each of the other angles of the hexagon is equal to AFE, or FED. The hexagon (0) is therefore equiangular, and it was shown to be equilateral ; and it is inscribed in the given circle: which was to be done. Recite (a) def. 16, 1; p. 15, 3;
(b) p. 10, 3;
(m) p. 26, 29, 3; (n) ax. 2; (0) Note to def. 36, 1. Cor. The side of the hexagon; that is, the chord of one sixth part of the circumference, is equal to the radius, or semi-diameter of the circle.
Scholium. To describe an equilateral and equiangular hexagon about the circle and about the inscribed hexagon is the same thing: for if through the points A, B, C, D, E, F, tangents be drawn, touching the circle at right angles to the diameters, the angles of the inscribed hexagon will be in the sides of the one described. Def. 2, 4, b. 4.
Tangent literally means they touch. See def. 2, b. 3.
16 P. To inscribe an equilateral and equiangular quindecagon in a given circle (ABCD).
Let AC be the side of an equilateral triangle (a), and AB the side of an equilateral pentagon (6) inscribed in the given circle. Therefore, as
B it is required to cut the circumference into fifteen equal parts, the chord AC cuts off five, AB cuts Ei off three, and the difference BC contains two of those fifteenths. Bisect the arc BC in E (c): therefore, the arc BE, or EC, is one fifteenth part of the circumference. Now, if the chord BE, or EC be drawn, and equal chords be placed all around in the circle (d), an equilateral and equiangular quindecagon shall be inscribed in it; which was to be done.
And, if through the angular points of the inscribed quindecagon, tangents be drawn, an equilateral and equiangular quindecagon shall be described about the circle, and also about the inscribed quindecagon: for the angular points of the inscribed rectilineal figure shall be in the sides of the one described (e). Recite (a) p. 2, 4;
(b) p. 11,4; (c) p. 30,3; (d) p. 1, 4; (e) def. 36, 1, and Note; def. 1, 2, 3, 4, 5, of b. 1.