FOURTH BOOK.

DEFINITIONS.

1. A rectilineal figure is said to be inscribed in a circle, when each of its angles touches the circumference of the circle.

2. A rectilineal figure is said to be circumscribed about a circle, when each of its sides touches the circle.

3. A circle is said to be inscribed in a rectilineal figure, when each of the sides of the figure touches its circumference.

4. A circle is said to be circumscribed about a rectilineal figure, when each of the angles of the figure touches its circumference.

5. A right line is said to be inscribed in a circle, when its extremities are in the circumference of the circle.

In a given circle (BCA) to inscribe a given right line (D) which is not greater than the diameter of the circle.

Draw a diameter AB of the given circle, and if this be equal to the given line D, that which is proposed is done. If not, assume in it AE, equal to D (by Prop. 3, B. 1), and from the centre A, with the interval AE, describe a circle ECF, and to either of its

B

1

K

intersections, with the given circle, draw the right line AC, this will be equal to AE (by Def. 15, B. 1), and therefore equal to the given line D (by Const. and Ax. 1).

PROPOSITION II. PROBLEM.

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

Draw the right line GH, touching the given circle in any point A; and at the point. A, with the line AH, construct the angle HAC, equal to E; and at the same point, with the right line AG, con

-H

Ds.

E

F

struct GAB equal to F, and draw BC. Because the angle E is equal to the angle HAC (by Const.), and the angle HAC is equal to the angle B in the alternate segment (by Prop. 32, B. 3), the angles E and B are equal; similarly, F and C are equal; therefore the remaining angle D is equal to the remaining angle BAC (by Cor. 2, Prop. 32, B. 1); and therefore the triangle BAC, inscribed in the given circle, is equiangular to the given triangle EDF.

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

KA, an angle BKA equal to the angle EDG; and at the same point K, make, at the other side of KA, an angle AKC equal to the angle EFH, and draw the right lines LM, LN, and MN, touching the circle in the points B, A, and C. Therefore, since the four angles of the quadrilateral figure LBKA are equal to four right angles (by Cor. 6, Prop. 32, B. 1), and the angles KBL and KAL are right angles (by Prop. 18, B. 3), the remaining angles AKB and ALB will be together equal to two right angles, but the angles EDG and EDF together are equal to two right angles (by Prop. 13, B. 1), therefore the angles AKB and ALB together are equal to the angles EDG and EDF taken together; but AKB and EDG are equal (by Const.), and therefore ALB and EDF are equal. Similarly it can be demonstrated that the angles ANC and EFD are equal; therefore the remaining angle M is equal to the remaining angle E (by Cor. 2, Prop. 32, B. 1), therefore the triangle LMN, circumscribed about the given circle ABC, is equiangular to the given triangle.

PROPOSITION IV. PROBLEM.

To inscribe a circle in a given triangle (BAC).

A

Bisect any two angles B and C, by the right lines BD and CD, and from the point in which these right lines meet, draw to any of the sides of the triangle BAC a perpendicular DF; the circle described from the centre D, and with the interval DF, will be inscribed in the given triangle. For draw DE and DG perpendicular to BA and AC; and since in the triangles DEB, DFB, the angles DEB and DBE are equal to the angles DFB and DBF (by Const.), and the side DB is common to both, therefore the sides DE and DF will be equal (by Prop. 26, B. 1); it can be similarly

demonstrated that DG and DF are equal; therefore the three lines DE, DF, and DG, are equal in themselves (by Ax. 1), and therefore the circle described from the centre D and with the interval DF passes through E and G, and since the angles at F, E, and G are right, the right lines BC, BA, and AC touch the circle (by Prop. 16, B. 3), and therefore the circle FEG is inscribed in the given triangle.

To inscribe a circle about a given triangle (BAC).

F

Bisect any two sides BA and AC of the given triangle, in D and E; and through D and E draw DF and EF perpendicular to AB and AC; and from their point of meeting FB draw to any angle A of the triangle BAC, the right line FA; the circle described from the centre F, with the interval FA, shall be circumscribed about the given triangle.

For draw FB and FC; and since in the triangles FDA and FDB, the sides DA and DB are equal (by Const.), but FD is common to both, and the angles at D right (by Const.), the sides FA and FB shall be equal (by Prop. 4, B. 1). It can be similarly demonstrated that the right lines

F

FA and FC are equal, therefore the three right lines, FA, FB, and FC are equal (by Ax. 1), and therefore the circle described from F, with the interval FA, shall pass through B and C, and therefore is circumscribed about the given triangle BAC.

SCHOL. This Problem is the same as to describe a circle through three given points not placed in directum.