Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

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 on the sides of the figure in which it is inscribed, each on 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 on 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.

[ocr errors][merged small]

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.

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.

Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle: it is required to place in the circle ABC, a straight line equal to D.

Draw BC, a diameter of the circle ABC.

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.

[blocks in formation]

But, if it is not, BC is greater
than D.
[Hypothesis.

Make CE equal to D,

[I. 3.

and from the centre C, at the distance CE, describe the circle AEF, and join CA.

Then, because C is the centre of the circle AEF, CA is equal to CE; [I. Definition 15.

but CE is equal to D;

[Construction. [Axiom 1.

therefore CA is equal to D.

Wherefore, 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.

Let ABC be the given circle, and DEF the given triangle: it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.

Draw the straight line GAH touching the circle at the point A; [III. 17. at the point A, in the straight line AH, make the angle HAC equal to the angle DEF; [I. 23. and, at the point A, in the straight line AG, make the angle GAB equal to the angle DFE;

and join BC. ABC shall be the triangle required.

Because GAH touches the circle ABC, and AC is drawn from the point of contact A, [Construction. therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle. [III. 32. But the angle HAC is equal to the angle DEF. [Constr. Therefore the angle ABC is equal to the angle DEF. [Ax. 1. For the same reason the angle ACB is equal to the angle DFE.

E

G

H

Therefore the remaining angle BAC is equal to the remaining angle EDF. [I. 32, Axioms 11 and 3. Wherefore the triangle ABC is equiangular to the triangle DEF, and it 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.

Let ABC be the given circle, and DEF the given triangle it is required to describe a triangle about the circle ABC, equiangular to the triangle DEF.

L

Produce EF both ways to the points G, H; take K the centre of the circle ABC; [III. 1.

from K draw any radius KB;

K

G E

D

N

F H

M

B

at the point K, in the straight line KB, make the angle BKÁ equal to the angle DEG, and the angle BKC equal to the angle DFH ; [I. 23. and through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching the circle ABC. [III. 17. LMN shall be the triangle required.

Because LM, MN, NL touch the circle ABC at the points A, B, C, [Construction. to which from the centre are drawn KA, KB, KC, therefore the angles atthepointsA,B,Carerightangles.[III.18. And because the four angles of the quadrilateral figure AMBK are together equal to four right angles,

for it can be divided into two triangles,

and that two of them KAM, KBM are right angles, therefore the other two AKB, AMB are together equal to two right angles. [Axiom 3. But the angles DEG, DEF are together equal to two right angles. [I. 13. Therefore the angles AKB, AMB are equal to the angles DEG, DEF;

of which the angle AKB is equal to the angle DEG ; [Constr. therefore the remaining angle AMB is equal to the remaining angle DEF. [Axiom 3. In the same manner the angle LNM may be shewn to be equal to the angle DFE.

Therefore the remaining angle MLN is equal to the remaining angle EDF [I. 32, Axioms 11 and 3.

[ocr errors][ocr errors]

Wherefore 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.

Let ABC be the given triangle: it is required to inscribe a circle in the triangle ABC.

Bisect the angles ABC, ACB, by the straight lines BD, CD, meeting one another at the point D; [I. 9. and from D draw DE, DF, DG perpendiculars to AB, BC, CA. [I. 12.

Then, because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD, [Construction. and that the right angle BED is equal to the right angle BFD; [Axiom 11. therefore the two triangles EBD, FBD have two angles of the one equal to two angles of the other, each to each; and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore their other sides are equal; therefore DE is equal to DF.

[I. 26.

E

G

B

For the same reason DG is equal to DF. Therefore DE is equal to DG. [Axiom 1. 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;

and it will touch the straight lines AB, BC, CA, 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. Therefore the straight lines AB, BC, CA do each of them touch the circle, and therefore the circle is inscribed in the triangle ABC.

Wherefore a circle has been inscribed in the given triangle. Q.E. F.

« ΠροηγούμενηΣυνέχεια »