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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 described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, 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 described 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 rectilineal figure, when the circumference of the circle touches each side of the figure.

6. A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described. 7. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle.

ODO

8. A regular polygon has all its sides equal, and also all its

angles.

9. A polygon of five sides is called a pentagon; of six, a hexagon; of seven, a heptagon; of eight, an octagon; of nine, a nonagon; of ten, a decagon; of eleven, an undecagon; of twelve, a dodecagon; and of fifteen, a quindecagon.

10. The centre of a regular polygon is a point equally distant from its sides, or from its angular points.

11. The apothem of a regular polygon is a perpendicular from its centre upon any of its sides.

12. The perimeter or periphery of any figure is its circumference or whole boundary.

PROPOSITION 1.-PROBLEM.

In a given circle, to place a straight line, equal to a given straight line, which is not greater than the diameter of the circle.

(Reference-Prop. I. 3.)

Given.-Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle.

Sought. It is required to place in the circle ABC, a straight line equal to D.

Construction.—1. Draw BC, a diameter of the circle ABC. 2. Then, if BC is equal to D, the

thing required is done; for in the circle ABC, a straight line is placed equal to D.

3. But, if it is not, BC is greater than D. (hyp.)

4. Make CE equal to D (I. 3), and from the centre C, at the distance CE, describe the circle A EF, and join CA. Then CA shall be equal to D.

D

F

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Proof.-1. Because C is the centre of the circle AEF, CA is equal to CE;

2. But CE is equal to D; (const.)

3. Therefore CA is equal to D. (ax. 1.)

Conclusion. Therefore, in the circle ABC, a straight line CA is placed, equal to the given straight line D, which is not greater than the diameter of the circle. Q. E. F.

PROPOSITION 2.-PROBLEM.

In a given circle, to inscribe a triangle equiangular to a given

triangle.

(References-Prop. I. 23, 32; III. 17, 32.)

Given.-Let ABC be the given circle, and DEF the given triangle.

Sought. It is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.

Construction.-1. Draw the straight line GAH touching the circle in the point A. (III. 17.)

2. At the point A, in the straight line AH, make the angle HAC equal to the angle DEF. (I. 23.)

3. At the point A, in the straight line AG, make the angle GAB equal to the angle DFE; and join BC.

Then ABC shall be the tri

angle required.

D

B

Proof.-1. Because GAH touches the circle ABC, and AC is drawn from the point of contact A; (const.)

2. Therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle. (III. 32.)

3. But the angle HAC is equal to the angle DEF; (const.) 4. Therefore the angle ABC is equal to the angle DEF. (ax. 1.)

5. For the same reason, the angle ACB is equal to the angle DFE.

6. Therefore the remaining angle BAC is equal to the remaining angle EDF. (I. 32, and ax. 3.)

Conclusion. Therefore the triangle ABC is equiangular to the triangle DEF, and is inscribed in the circle ABC. Q. E. F.

PROPOSITION 3.-PROBLEM.

About a given circle, to describe a triangle equiangular to a given triangle.

(References-Prop. I. 13, 23, 32; III. 1, 17, 18.)

Given.-Let ABC be the given circle, and DEF the given triangle.

Sought. It is required to describe a triangle about the circle ABC, equiangular to the triangle DEF.

Construction.-1. Produce EF both ways to the points

G, H.

2. Find the centre K of the circle ABC (III. 1), and from it draw any straight line KB.

3. At the point K, in the straight line KB, make the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DFH. (I. 23.)

A

B

4. And through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching the circle ABC. (III. 17.) Then LMN shall be the triangle required.

Proof.-1. Because LM, MN, NL, touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC;

2. Therefore the angles at the points A, B, C, are right angles. (III. 18.)

3. 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 because two of them, KAM, KBM, are right angles;

4. Therefore the other two, AKB, AMB, are together equal to two right angles. (ax. 3.)

5. But the angles DEG, DEF, are together equal to two right angles; (I. 13.)

6. Therefore the angles AKB, AMB, are equal to the angles DEG, DEF (ax. 1), of which AKB is equal to DEG; (const.) 7. Therefore the remaining angle AMB is equal to the remaining angle DEF. (ax. 3.)

8. In the same manner the angle LNM may be shown to be equal to the angle DFE;

9. Therefore the remaining angle MLN is equal to the remaining angle EDF. (I. 32, and ax. 3.)

Conclusion. Therefore, the triangle LMN is equiangular to the triangle DEF, and it is described about the circle ABC. Q. E. F.

PROPOSITION 4.-PROBLEM.

To inscribe a circle in a given triangle.

(References-Prop. I. 9, 12, 26; III. 16, cor.) Given.-Let ABC be the given triangle.

Sought. It is required to inscribe a circle in the triangle ABC.

Construction.-1. Bisect the angles ABC, ACB, by the straight lines BD, CD, meeting one another in the point D. (I. 9.)

2. From D draw DE, DF, DG, perpendiculars to AB, BC, CA. (I. 12.)

3. And from the centre D, at the distance DE, DF, or DG, describe the circle EFG.

Then the circle EFG shall be inscribed in the triangle ABC.

Proof.-1. Because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD (const.), and that the right angle BED is equal to the right angle BFD; (ax. 11.)

2. Therefore the two triangles EBD, FBD, have two angles of the one equal to two angles of the other, each to each;

3. And the side BD, which is opposite to one of the equal angles in each, is common to both;

4. Therefore their other sides are equal; (I. 26.)

5. Therefore DE is equal to DF.

6. For the same reason DG is equal to DF;

7. Therefore DE is equal to DG. (ax. 1.)

8. 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 one of them, will pass through the extremities of the other two.

9. And 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 the circle; (III. 16, cor.)

10. Therefore the straight lines AB, BC, CA, do each of them touch the circle.

Conclusion. Therefore, the circle EFG is inscribed in the triangle ABC. Q. E. F.

PROPOSITION 5.-PROBLEM.

To describe a circle about a given triangle.

(References-Prop. I. 4, 10, 11; III. 31.)

Given.-Let ABC be the given triangle.

Sought. It is required to describe a circle about ABC.

H

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