« ΠροηγούμενηΣυνέχεια »
38 P. To construct a rectangle equal to a given square, having the difference of its adjacent sides equal to a given line.
Let C be the side of a given square, and AB the difference of the sides of an equivalent rectangle.
Describe a circle on the diameter AB; at the extremity. A, of the diameter, draw the tangent AD, equal to C; through the centre 0, draw DF, B cutting the circle in }, F. Then the rectangle DEXDF equals the square of AD (a), or its equal C; and the difference of DE and DF'is EF, which is equal to AB.
Recite (a) p. 36 of b. 3
1. One rectilineal figure is said to be inscribed in another, when all the angles
of the former are upon the sides of the latter. Note.—Regular polygons, which have the same number of equal sides, thus inscribed, form a series, and have a certain ratio to each other.
2. One rectilineal figure is described about another, when all the sides of the former pass through the angular points of the latter.
3. A rectilineal figure is inscribed in a circle, when the angles of the former are all in the circumference of the latter. And in this case, the circle is said to be described about the rectilineal figure.
4. A rectilineal figure is described about a circle, when each side of the former touches the circumference of the latter. And in this case, the circle is said to be inscribed in the rectilineal figure.
5. A straight line is said to be placed in a circle, when its extreme points are in the circumference.
1 P. In a given circle (ABC) to place a chord equal to a given straight line (D), not greater than the diameter.
Draw BC, the diameter of the circle : then if the given straight line D, be equal to BC, there is placed in the circle such a chord as was required.
But if D be not equal to BC, it cannot be greater (a); make CE equal to D (6); and with CE as radius, upon the centre C, describe the circle AEF (c); join CA (d): therefore, in the circle AEF, CA and CE are equal radii (e): but CE is equal D; therefore CA is equal to D; and it is a chord (f), not greater than the diameter, placed in the circle ABC; which was to be done. Recita (a) p. 15, 3; (b) p. 3, 1; (c) pos. 3;
(d) pos. 1; (e) def. 15, 1; ifdef 18, 1.
In a given circle (ABC), to inscribe a triangle equiangular to a given triangle (DEF).
H make the angle HAC equal to the angle E, also the angle GAB equal to the angle F (b); join BC.
Argument. Because the straight line GAH touches the circle ABC, and from the point of contact A, the straight lines AB, AC are drawn, the angle HAC is equal to B, and the angle GAB is equal to C, in the alternate segments of the circle (c). But HAC is made equal to E, and GAB to F, of the given triangle; therefore, the angle B is equal to E, and the angle C is equal to F, (d); so the third angles BAČ and D must be equal (e). Therefore, the triangle ABC has all its angles equal to those of the given triangle DEF, each to each; and it is inscribed in the given circle ABC, which was to be done.
Recite (a) p. 17, 3;
(d) ax. 1;
(c) p. 32, 3;
3 P. About a given circle (ABC), to describe a triangle equiangular to a given triangle (DEF).
Produce EF both ways to the points G, H: find K, the centre of the given circle (a); draw any radius KB; at the point K, in BK, make angles BKA, BKC equal to the exterior angles DEG, DFH, each to each (b);
FH draw tangents through the points A, B, C (c), to meet in the points M, L, N. There fore LM, LN, MN, meet the radii at right M ,B angles in the points A, B, C (d): and because the quadrilateral AMBK may be divided into two right angled triangles, whose angles are equal to four right angles (e); and that two of its angles at A and B, are right angles, the other two, AKB, AMB, are equal to two right angles; but the two angles DEF, DEG are also equal to two right angles (f ), and BKA was made equal to DEG; therefore, the remaining angles DEF, AMB are equal. It may be proved, in this way, that the angles DFE and N are also equal ; and so the third angles D and L must be equal (e). Wherefore, about a given circle a triangle is described equiangular to a given triangle ; which was to be done.
Recite (a) p. 1, 3;
(d) p. 18, 3;
4 P. To inscribe a circle in a given triangle (ABC).
Constr. Bisect the angles B and C (a), by straight lines BD, CD, meeting in D, from which point draw perpendiculars to meet the sides of the triangle in the points E, F, G (6).
Argument. The triangles BDE, BDF are equal, E for the following reasons, viz. The angles EBD, FBD, are halves of the angle EBF (c); BED, BFD are right angles (d), and BD is common to the two triangles: therefore DE is equal to DF (e): for like reason, DG is also equal to DF, or DE; and the circle described upon the centre D, at the distance of any of them, will pass through the points E, F, G; and be inscribed in the given triangle (f); which was to be done. Recite (a) p. 9, 1; (b) p. 12, 1;
(c) ax. 7; (d) ax. 10; (e) p. 26, 1;
y , 4.
5 P. To describe a circle about a given triangle (ABC).
Constr. In points D, E, bisect the sides AB, AC, of the given triangle (a); draw DF, EF, at right angles to the sides (b); and, if the point F be on the side BC, join FA; if within or without the triangle, join also FB, FC.
Argument. Now, on account of the right angles at D (c), and that DA, DF are equal to DB, DF, the bases AF and BF are equal (d). In like manner it may be proved that FC is equal to FA, or FB: therefore F is the centre of a circle passing through the three points of the triangle, as required (e). Recite (a) p. 10, 1; (6) p. 11, 1;
(c) ax. 10, 1; (d) p. 4, 1;
(e) p. 9, 3. Corollary. The point F is within, on the side, or without the triangle, according as it
be acute, right, or obtuse angled.
6 P. To inscribe a square in a given circle (ABCD). Draw, in the circle, the diameters AC, BD, at right angles to each other (a); they will divide the circumference into four equal arcs (b); draw also the chords AB, AD, CB, CD, which all
equal to each other (c). Now each of the an-
(c) p. 29, 3; (d) p. 31, 3;
7 P. To describe a square about a given circle (ABCD).
Draw the diameters AC, BD, at right angles G A to each other (a); also tangents (b) to the circle, through the points A, B, C, D, at right angles to the diameters, meeting in the points F, G, H, K. Now, because of the
qual alternate angles (c) GBD, BDK, the side GH is parallel to FK: for the same reason GF is parallel to HK: therefore FGHK is a parallelogram, whose opposite H
K sides and angles are equal(a): but because any one of the angles at E and an interior angle on the same side of those at the points A, B, C, D, are equal to two right angles (e); therefore each of the angles F, G, H, K, is a right angle. Wherefore, since the equal sides touch the circle and meet at right angles, there is a square described about the circle as was required (f). Recite (a) cor. p. 1, 3; (1) def. 2, 3; (c) p. 27,1;
(d) p. 34, 1; (e) p. 29, 1;
(f) def. 30, 1, and 2, 4.
8 P. To inscribe a circle in a given square (ABCD).
Bisect two sides AB, AD, of the given square, in the points E, F (a); draw EH, FK parallel to AB, AD (6): then the interior angles A, E, and also A, F, are equal to two right angles (c): for
H the same reason all the angles about the point G are right angles; therefore the opposite sides GE, GF, ĜI, GK are equal (d), and G is the centre of a circle that shall touch the sides of the given. square, in E, F, H, K (e), and be inscribed therein, as required (f). Recite (a) p. 10, 1; (b) p. 31, 1; (c) p. 29, 1;
(d) p. 34, 1; (e) def. 3, 3;
9 P. To describe a circle about a given square (ABCD).
Join the opposite angles of the given square by the diameters AC,