THE ELEMENTS OF EUCLID. 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 upon the sides of the figure in which it is inscribed each upon each. 2. In like manner, a figure is said to be circumscribed about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is circumscribed, each through each. 3. A Rectilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are upon the circumference of the circle. 4. A Rectilineal figure is said to be circumscribed about a circle, when each side of the circumscribed figure touches the circumference of the circle. 5. In like manner, a circle is said to be inscribed in a rectilinear figure, when the circumference of the circle touches each side of the figure. 6. A circle is said to be circumscribed about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is circumscribed. 7. A straight line is said to be placed in a circle, when its extremities are in the circumference of the circle. оо B SCHOLIUM. A regular polygon is one which has all its sides or angles equal; in the first case it is said to be equilateral, and in the second, equiangular. Polygons further receive particular names, according to the number of sides which they possess, thus: PROBLEM.-In a given circle (ABC) to inscribe a straight line, equal to a given straight line (D), which is not greater than the diameter of the circle. SOLUTION. Draw a diameter BC of the circle; and if this be equal to the given line D, the problem is solved; but if it is not, take in it the segment CE equal to D (a), and from C as a center, with the radius CE, describe the circle AEF, and join CA. DEMONSTRATION. Because C is the center of the circle AEF, CA is equal to CE (b); but D is equal to CE (c), therefore D is equal to CA (d). SCHOLIUM. It should be observed that in the enunciation of the above proposition, the word D (a) I. 3. (d) Ax. 1. "given" is used in a different sense as applied to the circle and to the straight line, the former being given both in position and magnitude, while the latter is given only in magnitude. PROPOSITION II. PROBLEM. In a given circle (ABC) to inscribe a triangle equiangular to a given triangle (DEF). E SOLUTION. Draw the straight line GAH touching the circumference of the circle in the point A (a), and at the point ▲ in the straight line AH, and on the same side of it with the circle form the angle HAC equal to the angle E (b), and at the same point in the straight line AG, and on the same side of it, form the angle GAB equal to the angle F (b); and since AC and AB are drawn from A between the tangent and the circumference, they must cut the circumference (c); let them do so respectively in the points C and B; then join B and C. (a) III. 17. f) Ax. 1. DEMONSTRATION. Because HAG touches the circle ABC, and AC is drawn from the point of contact, the angle HAC is equal to the angle B in the alternate segment of the circle (d); but the angle HAC is equal to the angle E (e); therefore the angle B is equal to the angle E (f); and in the same manner it may be shown that the angle is equal to the angle F; therefore the remaining angle D is equal to the angle BAC (g), and therefore the triangle ABC, inscribed in the given circle, is equiangular to the given triangle DÉF. SCHOLIUM. In the solution of this problem, Euclid has omitted to state that the lines AC and AB must be drawn on the same side of the tangent as the circle, and he has assumed that these lines will cut the circumference, without showing the reason of their doing so. PROPOSITION III. PROBLEM.-About a given circle (ABC) to circumscribe a triangle equiangular to a given triangle (DEF). N III. 1. (b) I. 23. straight line, form the angle CKB equal to the anyle DFH (b); through the points A, B, and C draw the straight lines ML, MN, and NL, touching the circle ABC (c), then shall they meet in the points M, N, and L, and form the triangle required. DEMONSTRATION. Join A and B, then because KAM and KBM are right angles (d), the angles BAM and ABM are less than two right angles, and therefore the lines AM and BM must meet one another, if produced far enough (e), let them meet in M, and in a similar manner it may be shown that AL and CL must meet in some point L, and that BN and CN must meet in some point N. Because the four angles of the quadrilateral figure AKBM are together equal to four right angles (f), and the angles KAM and KBM are right angles (d), the other two M and AKB are together equal to two right angles; but the angles DEG and DEF are together equal to two right angles (9), therefore the angles AKB and M are together equal to the angles DEG and DEF; but AKB and DEG are equal (h), and therefore M and DEF are equal (2). In the same manner it may be shown that the angle N is equal to DFE; therefore the remaining angle L is equal to the remaining angle D (k); and therefore the triangle LMN circumscribed about the circle ABC is equiangular to the given triangle. SCHOLIUM. The demonstration of this proposition has been somewhat altered from that of Euclid, who omits to prove that the lines MN, LM, and LN must necessarily meet when produced. |