are severally equal to FB, BE, and FE is common to both triangles, therefore the angles FAE, FBE are equal (8. 1); but FBE is a right angle (18. 3), therefore FAE is also a right angle, and, of course AE touches the circle (Cor. 16. 3).

BOOK IV.

DEFINITIONS.

1. A RECTILINEAL figure, is said to be inscribed in another rectilineal figure, when all the angles of the inscribed figure, are in the perimeter of the other.

2. A rectilineal figure, is said to be circumscribed about another rectilineal figure, when the perimeter of the former touches all the angles of the other. 3. A rectilineal figure, is said to be inscribed in a circle, when all its angles, are in the circumference of the circle.

4. A circle, is said to be circumscribed about a rectilineal figure, when every angle of the rectilineal figure, is in the circumference of the circle.

5. A rectilineal figure, is said to be circumscribed about a circle, when every side of it, touches the circle.

6. A circle, is said to be inscribed in a rectilineal figure, when every side of the rectilineal figure, touches the circle.

7. A regular figure, is that, which is equilateral and equiangular.

PROPOSITION I. PROBLEM.

In a given circle (CAB), from a given point (C) in its circumference, to inscribe a right line, equal to a given right line (Ï), not greater than the diameter of the circle.

Draw the diameter of the circle CB, and, if this be equal to D, what was required is done.

If not, take from CB, a part CE equal to D (3. 1), and from the centre C, at the distance CE, describe the cir

cle AEF, and to either of its D

intersections with the given circle, as A, draw CA, this is equal to CE (Def. 10. 1), and therefore to the given right line D (Constr. and Ax. 1. 1).

Cor.-Hence it appears, how, in a given circle, from a given point in its circumference, an arch may be taken, equal to a given arch, of an equal circle; namely, by drawing the chord of the given arch, and inscribing in the given circle, from the given point, a right line equal to that chord (by this prop.), which right line cuts off an arch equal to the given one (by 28. 3).

PROP. II. PROB.

In a given circle, (BAC), to inscribe a triangle, equiangular to a given triangle (EDF).

AD

Draw the right line GH, touching the circle in any point A (17. 3), and at the point A, with the right line AH, make the angle HAC equal to the angle E (23. 1); and at the same point, with the right line AG, make the angle GAB equal to angle F, and join BC.

E

H

The angle E is equal to the angle HAC (Constr.), or, (32. 3), to the angle B in the alternate segment; for a like reason, the angles F and C are equal; therefore the remaining angle D is equal to the remaining BAC (32. 1); therefore the triangle BAC, which is inscribed in the given circle (Def. 3. 4) is equiangular to the given triangle EDF.

PROP. III. PROB.

About a given circle (ABC), to circumscribe a triangle, equiangular to a given triangle (EDF),

Produce any side EF, of the given triangle, both ways as to G and H; find the centre K of the given circle (1. 3), from which draw any radius KA, with which, at K, make the angle BKA equal to DEG (23. 1; and, with BK at K, the angle. BKC equal to DFH; and draw ML, MN and LN, touching the circle in the points A, B and C (17. 3).

Because the four angles, of the quadrangle MAKB, are equal to four right angles (Cor. 1. 32. 1), and the angles KAM, KBM are right angles (18. 3), the remaining angles AKB, AMB are equal to two right angles, and therefore to DEF, DEG, which together are also equal to two right angles [13. 1]; taking away the equal angles AKB, DEĞ, the remaining angles AMB, DEF are equal: in like manner, LNM and DFE may be proved, equal; therefore the remaining angle L is equal to the remaining D [32. 1], and, of course, the triangle LMN, which is circumscribed about the given circle [Def. 5. 4], is equiangular to the given triangle DEF.

PROP. IV. PROB.

In a given triangle (ABC), to inscribe a circle.

Bisect any two angles ABC, BCA of the given triangle by the right lines BD, CD (9. 1), which, because the angles DBC, BCD, are together less than ABC, BCA together [Ax. 9], and therefore than two right angles [17. 1], may be so produced, as to meet (Theor. at 29. 1); let them be so produced, and meet, as in D, from which, let fall the perpendicular

B

DE on AB [12. 1], from the centre D, at the distance De, describe a circle (Post. 3), which is inscribed in the given triangle.

For, the perpendiculars DF and DG being let fall on BC and CA; because, in the triangles DEB, DFB, the angles DEB, DBE are severally equal to DFB, DBF [Constr.], and the side DB common, DE and DF are equal [26. 1]; in like manner, DF, DG may be proved equal; therefore, the three right lines DE, DF, DG are equal [Ax. 1. 1], and the circle, described from the centre D, at the distance DE, passes through F and G; and since the angles formed by AB, BC, CA with the radiuses, at the points E, F, G are right angles, these right lines touch the circle (Cor. 16. 3), which is therefore inscribed in the triangle ABC [Def. 6. 4), as was required,

PROP. V. PROB.

About a given triangle (BAC), to circumscribe a circle.

Bisect any two sides AB, AC of the given triangle, in D and E, and from D and E, draw the perpendiculars DF, EF, and join DE. Because the angles ADF, AEF are right angles [Constr.], the angles EDF, DEF are less than two right angles [Ax. 9], therefore the perpendiculars drawn from D and E may be so produced as to meet [Theor. at 29. 1], let them meet as in F, from whence draw, to any angle A of the triangle ABC, the right line FA; from the centre F, at the distance FA, describe a circle, which is circumscribed about the given triangle,

For, FB, FC being drawn, because, in the triangles FDA, FDB, the sides DA, DB are equal (Constr.), FD common, and the angles at D right [Constr.], FA and FB are equal [4. 11; in like manner, FB and FC may be proved equal, therefore the three right lines FA, FB, FC are equal [Ax. 1. 1], and, of course, a circle, described from the centre F, at the distance FA, passes through B and C, and is therefore circumscribed about the given triangle BAC [Def. 4. 4].

Scholium. This problem is, in effect, the same, as to deseribe a circle through three given points, which are not in the same right line,

Cor. 1.-If the centre (F) of a circle, circumscribing a triangle, be within the triangle (as in fig. 1), all the angles of the triangle are acute; if on any side (BC), of the triangle (as in fig. 2), the angle opposite that side is right; if without the triangle (as in fig. 3), the angle, opposite the side (BC), which is adjacent to the centre, is obtuse.

For, in the case of fig. 1, every angle of the triangle ABC, is in a segment greater than a semicircle, and therefore acute [31. 3]; in the case of fig. 2, the angle BAC opposite BC, is in