to the angle dfe, and join bc: therefore because hag touches the circle abc, and ac is drawn from the point of contact, the angle ha c is equal (iii. 32) to the angle abc in the alternate segment of the circle; but hac is equal to the angle def; therefore also the angle abc is equal to def. For the same reason, the angle a cb is equal to the angle dfe; therefore the remaining angle bac is equal (i. 32) to the remaining angle edf: wherefore the triangle a bc is equiangular to the triangle def, and it is inscribed in the circle abc. Which was to be done.

PROPOSITION III.—PROBLEM.

About a given circle to describe a triangle equiangular to a given triangle. LET abc be the given circle, and def the given triangle; it is required to describe a triangle about the circle a bc equiangular to the triangle def

a

1

d

Produce ef both ways to the points g, h, and find the centre k of the circle a bc, and from it draw any straight line kb; at the point k in the straight line k b, make (i. 23) the angle bka equal to the angle deg, and the angle bkc equal to the angle dfh; and through the points a, b, c, draw the straight lines lam, mbn, ncl, touching (iii. 17) the circle abc: : therefore, because Im, mn, nl touch the circle a bc in the points a, b, c, to which from the centre are drawn ka, kb, kc, the angles at the points a, b, c, are right (iii. 18) angles: and because the four angles of the quadrilateral figure ambk are equal to four right angles, for it can be divided into two triangles; and that two of them kam, kbm are right angles, the other two ak b, amb are equal to two right angles: but the angles deg, def are likewise equal (i. 13) to two right angles; therefore the angles akb, amb are equal to the angles deg, def, of which akbis equal to deg; wherefore the remaining angle amb is equal to the remaining angle def. In like manner, the angle lnm may be demonstrated to be equal to dfe; and therefore the remaining angle mln is equal (i. 32) to the remaining angle edf: wherefore the triangle 1mn is equiangular to the triangle def: and it is described about the circle a b c Which was to be done.

m

b

e 1 g

f h

PROPOSITION IV.-PROBLEM.

To inscribe a circle in a given triangle.

LET the given triangle be a bc; it is required to inscribe a circle in abc. Bisect (i. 9) the angles abc, bca by the straight lines b d, cd meeting one another in the point d, from which draw (i. 12) de, df, dg per

80

pendiculars to a b, bc, ca: and because the angle ebd is equal to the angle fb d, for the angle a bc is bisected by bd, and that the right angle bed is equal to the right angle bfd, the two triangles ebd, fbd have two angles of the one equal to two angles of the other, and the side bd, which is opposite to one of the equal angles in each, is common to both; therefore their other sides shall be equal (i. 26); whereC fore de is equal to df. For the same reason, dg is equal to df; therefore the three straight lines de, df, dg, are equal to one another, and the circle described from the centre d, at the distance of any of them, shall pass through the extremities of the other two, and touch the straight lines a b, bc, ca. Because the angles at the points e, f, g are right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches (iii. 13) the circle: therefore the straight lines ab, bc, ca do each of them touch the circle, and the circle efg is inscribed in the triangle ab c. Which was to be done.

b

f

PROPOSITION V.-PROBLEM.

To describe a circle about a given triangle.

LET the given triangle be abc; it is required to describe a circle about abc.

Bisect (i. 10) a b, a c in the points d, e, and from these points draw df, ef at right angles (i. 11) to ab, ac; df, ef, produced, meet one another. For, if they do not meet, they are parallel, wherefore a b, ac, which are at right angles to them, are parallel; which is absurd. Let

them meet in f, and join fa; also if the point f be not in bc, join bf, cf: then, because a d is equal to db, and d f common, and at right angles

In like manner, it may

to a b, the base af is equal (i. 4) to the base fb. be shewn that cf is equal to fa; and therefore bf is equal to fc; and fa, fb, fc, are equal to one another; wherefore the circle described from the centre f, at the distance of one of them, shall pass through the extremities of the other two, and be described about the triangle abc. Which was to be done.

COR. And it is manifest, that when the centre of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicircle; but, when the centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, is a right angle; and, if the centre falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle: wherefore, if the given triangle be acute-angled, the centre of the circle falls within it; if it be a right-angled triangle, the centre is in the side opposite to the right angle; and if it be an obtuse-angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle.

PROPOSITION VI.—PROBLEM.

To inscribe a square in a given circle.

LET abcd be the given circle; it is required to inscribe a square in abcd.

Draw the diameters a c, bd, at right angles to one another, and join ab, bc, cd, da; because be is equal to ed, for e is the centre, and that ea is common, and at right angles to bd; the base ba is equal (i. 4) to the base ad; and, for the same reason, bc, cd are each of them equal to ba, or a d; therefore the quadrilateral figure abcd is equilateral. It is also rectangular; for the straight line bd, being the dia- b meter of the circle a b c d, bad is a semicircle; wherefore the angle bad is a right (iii. 31) angle; for the same reason, each of the angles abc, bed, cda, is a right angle; therefore the quadrilateral figure abcd is rectangular, and it has been shewn to be equilateral; therefore it is a square; and it is inscribed in the circle abcd. Which was to be done.

d

PROPOSITION VII.—PROBLEM.

To describe a square about a given circle.

LET abcd be the given circle; it is required to describe a square about it. Draw two diameters a c, b d of the circle a bcd, at right angles to one another, and through the points a, b, c, d, draw (iii. 17) fg, gh, hk, kf, touching the circle; and because fg touches the circle abcd, and ea is drawn from the centre e to the point of contact a, the angles at

b

g

a

f

d

a are right angles (iii. 18); for the same reason, the angles at the points b, c, d, are right angles; and because the angle aeb is a right angle, as likewise is ebg, g h is parallel (i. 28) to a c; for the same reason ac is parallel to fk, and in like manner gf, hk may each of them be demonstrated to be parallel to bed; therefore the figures gk, gc, ak, fb, bk are parallelograms; and gf is therefore equal (i. 34) to hk, and gh to fk; and because a c is equal to bd, and that a c is equal to each of the two gh, fk; and bd to each of the two gf, hk: gh, fk are each of them equal to gf or hk; therefore the quadrilateral figure fghk is equilateral. It is also rectangular; for gbe a being a parallelogram, and aeb a right angle, a gb (i. 34) is likewise a right angle. In the same manner it may be shewn that the angles at h, k, f, are right angles; therefore the quadrilateral figure f g h k is rectangular, and it was demonstrated to be equilateral; therefore it is a square; and it is described about the circle a b c d Which was to be done.

h

C

k

PROPOSITION VIII.-PROBLEM.

To inscribe a circle in a given square.

LET a b c d be the given square; it is required to inscribe a circle in abcd. Bisect (i. 10) each of the sides ab, ad in the points f, e, and through e draw (i. 31) eh parallel to ab or d c, and through f draw fk parallel to ad or bc; therefore each of the figures ak, kb, ah, hd, ag, gc, bg gd, is a parallelogram, and their opposite sides are equal (i. 34); and

g

because a d is equal to a b, and that ae is the half of a d, and af the half of ab, ae is equal to af; wherefore the sides opposite to these are equal, viz. fg to ge; in the same manner it may be demonstrated that gh, gk are each of them equal to fg or ge: therefore the four straight lines ge, gf, gh, gk are equal to one another; and the circle described from the centre at the distance of one of them, shall pass through the extremities of the other three, and touch the straight lines ab, bc, cd, da: because the angles at the points e, f, h, k, are right (i. 29) angles, and that the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle (iii. 16); therefore each of the straight lines a b, bc, cd, da, touches the circle, which therefore is inscribed in the square abcd. Which was to be done.

b

h

C

g

PROPOSITION IX.-PROBLEM.

To describe a circle about a given square.

LET abcd be the given square; it is required to describe a circle about it. Join a c, bd, cutting one another in e; and because da is equal to a b, and ac common to the triangles dac, bac, the two sides da, a c, are equal to the two ba, a c, and the base dc is equal to the base bc; wherefore the angle dac is equal (i. 8) to the angle ba c, and the angle da b is bisected by the straight line a c In the same manner it may be demonstrated that the angles a bc, bcd, cda are severally bisected by the straight lines bd, ac; therefore, because the angle dab is equal to the angle a bc, and that the angle e ab is the half of da b, and eba the half of abc: the angle e ab is equal to the angle eba; wherefore the side e a is equal (i. 6) to the side e b. In the same manner it may be demonstrated that the straight lines ec, ed are each of them equal to ea or eb; therefore the four straight lines ea, eb, ec, ed, are equal to one another; and the circle described from the centre e, at the distance of one of them, shall pass through the extremities of the other three, and be described about the square abcd. Which was to be done.

b

PROPOSITION X.-PROBLEM.

To describe an isosceles triangle, having each of the angles at the base double of the third angle.

TAKE any straight line a b, and divide (ii. 11) it in the point c, so that the rectangle a b, bc be equal to the square of ca; and from the centre a, at the distance a b, describe the circle bd e, in which place (iv. 1) the straight line b d equal to a c, which is not greater than the diameter of the circle bde; join da, dc, and about the triangle a dc describe (iv. 5) the circle a cd; the triangle abd is such as is required, that is, each of the angles abd, adb is double of the angle bad.

Because the rectangle a b, bc is equal to the square of a c, and that ac is equal to bd, the rectangle a b, bc is equal to the square of bd; and because from the point b, without the circle acd, two straight lines bca, bd are drawn to the circumference, one of which cuts, and the other meets the circle, and that the rectangle a b, bc, contained by the whole of the cutting line, and the part of it without the circle, is equal to the square of bd, which meets it; the straight line bd touches (iii. 37) the circle a cd; and be

b

d

G