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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:

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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 (α).

SCHOLIUM. It should be observed that in the enunciation of the above proposition, the word

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"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).

SOLUTION. Draw the straight line GAH touching the circumference of the circle in the point A (a), and at the point A 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

D

B

III. 17.

b) I. 23.
c) III. 16.
d) III. 32.
(e) Solution.
(f) Ax. 1.
(g) I. 32 B, cor. 3.

respectively in the points C and B; then join B and C.

H

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 C 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).

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H

straight line, form the
angle CKB equal to the
anyle DFH (b); through
the points A, B, and C A
draw the straight lines
ML, MN, and NL, touch-
ing 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,

M

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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 (g), 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.

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perpendicular to any side BC (d), and from Das a center, and with the distance DF describe a circle EFG which shall be inscribed in the given triangle.

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B

A

F

(d) I. 12.
(e) Solution.

III. 18.

DEMONSTRATION. From D draw DE and DG perpendicular to AB and AC. Then the angle ABC being bisected by DB (e), the angles EBD and FBD are equal, and the angles DEB and DFB being both right angles (f) are also equal, therefore the triangles EBD and FBD have two angles of the one respectively, equal to two angles of the other, and the side BD common to both, and therefore their other sides ED and FD are equal (g). In the same manner it may be shown that GD is equal to FD; therefore the three lines ED, FD, and GD are equal (h), and therefore the circle described from the center D, with the radius DF, passes through the points E and G, and because the angles at F, E, and G are right angles, the lines BC, AB, and AC are tangents to the circle (i); therefore the circle FEG is inscribed in the given triangle (k).

SCHOLIUM. The above proposition is only a particular case of the more general problem," To describe a circle touching three given straight lines."

1°. If

the three given lines are parallel to each
other; or 2o. If they intersect at the same
point the problem is impossible; 3°. If
two of the lines, AB and CD, are parallel,
and the third, EF, intersect them, it is pos-
sible to describe two equal circles, each ful-
filling the conditions of the
problem, one on either side of
the line EF; 4°. If the three
given lines intersect so as to
form a triangle, four circles
may be described, touching
them, one inscribed as above,
and the other three touching
each of the sides of the tri-

angle externally, and the
other sides produced.

COROLLARY 1. The straight lines bisecting the three angles of a triangle meet in the center of the inscribed circle.

COROLLARY 2. A triangle is equal in area to the rectangle under the radius of the inscribed circle, and half the sum of the three sides or perimeter of the triangle.

C

(g) I. 26.
(h) Ax. 1.
(i) III. 16.
(k) III. Def.

B

For the area of the whole triangle ABC is equal to the areas of the three triangles AEB, BEC, and AEC, and the area of each of these triangles is respectively equal to that of the rectangle, under the radius and half the sides AB, BC, and AC.

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PROPOSITION V.

PROBLEM. To circumscribe a circle about a given triangle (ABC).

SOLUTION. The three angular points, A, B, and C, of the triangle, are not in the same straight line, therefore a circle may be described passing through them in the manner demonstrated in the theorem attached to III. 1.

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above-mentioned.

2. If the center F fall within the triangle all its angles are acute, for each of them is in a segment greater than a semicircle. If the center be in any side of the triangle, the angle opposite that side is a right angle, because it is in a semicircle. And if the center fall without the triangle, the angle opposite to the side which is nearest to the center is an obtuse angle, because it is in a segment greater than a semicircle.

3. The two following propositions are here introduced, in order to simplify the demonstration of several of the subsequent problems.

PROPOSITION V. A.

THEOREM.-If a rectilineal figure

(ABCDE) be equilateral and equiangular, [1] it may have one circle circumscribed about it, [2] and another inscribed in it; [3] and the same point is the center of both circles.

CONSTRUCTION. Bisect the angles BCD and CDE (a), by the straight lines CF and DF, then because the angles FCD and FDC

B

H

(a) I. 9.

2

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