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BOOK IV.

DEFINITIONS.

I.

A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angular points of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each.

II.

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.

III.

A rectilineal figure is said to be inscribed in a circle, when all the angular points of the inscribed figure are upon the circumference of the circle.

IV.

A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle.

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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.

VI.

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.

VII.

A straight line is said to be placed in a circle, when the extremities

of it are in the circumference of the circle.

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PROPOSITION I. 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.

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Draw BC the 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 BC is placed equal to D.
But, if it is not, BC is greater than D; (hyp.),
make CE equal to D, (1. 3.)

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

Then CA shall be equal to D.

Because C is the center of the circle AEF,
therefore CA is equal to CE: (1. def. 15.)
but CE is equal to D; (constr.)

therefore D is equal to CA. (ax. 1.)

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 II. PROBLEM.

In a given oirole 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

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D

H

E F

Draw the straight line GAH touching the circle in the point Д, (ш. 17.)
and at the point A, in the straight line AH,
make the angle HAC equal to the angle DEF; (1. 23.)
and 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 triangle required.
Because HAG touches the circle ABC,

and 40 is drawn from the point of contact,

therefore the angle HAC is equal to the angle ABC in the alternate, segment of the circle: (I. 32.)

but HAC is equal to the angle DEF; (constr.)

therefore also the angle ABC is equal to DEF: (ax. 1.)

for the same reason, the angle ACB is equal to the angle DFE: therefore the remaining angle BAC is equal to the remaining angle EDF: (1. 32. and ax. 3.)

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

PROPOSITION HI. 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 equian-
gular to the triangle DEF

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Produce EF both ways to the points G, H;
find the center K of the circle ABC, (II. 1.)
and from it draw any straight line KB;
at the point K in the straight line KB,
make the angle BKA equal to the angle DEG, (1. 23.)
and the angle BKC equal to the angle DFH;

and through the points A, B, C, draw the straight lines LAM, MBN,
NCL, touching the circle ABC. (ш. 17.)

Then LMN shall be the triangle required.

Because LM, MN, NL touch the circle ABC in the points A, B, Cr

to which from the center are drawn KA, KB, KC,

therefore the angles at the points A, B, C are right angles: (II. 18.) 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 that two of them KAM, KBM are right angles, therefore the other two AKB, AMB are equal to two right angles: (ax. 3.)

but the angles DEG, DEF are likewise equal to two right angles; (1. 13.)

therefore the angles AKB, AMB are equal to the angles DEG, DEF; (ax. 1.)

of which 4KB is equal to DEG; (constr.) wherefore the remaining angle AMB is equal to the remaining angle DEF. (ax. 3.)

In like manner, the angle LNM may be demonstrated to be equal to DFE;

and therefore the remaining angle MLN is equal to the remaining angle EDF: (1. 32 and ax. 3.)

therefore the triangle LMN is equiangular to the triangle DEF:

and it is described about the circle ABC.

Q. E. F.

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

from which draw DE, DF, DG perpendiculars to AB, BC, CA. (1. 12.) And because the angle EBD is equal to the angle FBD,

for the angle ABC is bisected by BD,

and that the right angle BED is equal to the right angle BFD; (ax. 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; (1. 26.)

wherefore DE is equal to DF:

for the same reason, DG is equal to DF:
therefore DE is equal to DG: (ax. 1.)

therefore the three straight lines DE, DF, DG are equal to one another;

and the circle described from the center D, at the distance of any of them, will pass through the extremities of the other two, and 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.)

therefore the straight lines AB, BC, CA do each of them touch the circle,

and therefore the circle EFG is inscribed in the triangle ABC. Q.E. F.

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B

Bisect AB, AC in the points D, E, (1. 10.)

and from these points draw DF, EF at right angles to AB, AC; (1.11.)

DF, EF produced meet one another:

for, if they do not meet, they are parallel,

wherefore AB, AC, which are at right angles to them, are parallel; which is absurd :

let them meet in F, and join FA;

also, if the point F be not in BC, join BF, CF.

Then, because AD is equal to DB, and DF common, and at right angles to AB,

therefore the base AF is equal to the base FB.

(1. 4.)

In like manner, it may be shewn that CF is equal to FA;
and therefore BF is equal to FC; (ax. 1.)

and FA, FB, FC are equal to one another :

wherefore the circle described from the center F, at the distance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC. Q. E. F.

COR.-And it is manifest, that when the center of the circle falls within the triangle, each of its angles is less than a right angle, (III. 31.) each of them being in a segment greater than a semicircle; but when the center is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, (III. 31.) is a right angle; and if the center falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, (II. 31.) is greater than a right angle: therefore, conversely, if the given triangle be acuteangled, the center of the circle falls within it; if it be a right-angled triangle, the center is in the side opposite to the right angle; and if it be an obtuse-angled triangle, the center falls without the triangle, beyond the side opposite to the obtuse angle.

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Draw the diameters AC, BD, at right angles to one another, (. ́1. and I. 11.)

and join AB, BC, CD, DA.

The figure ABCD shall be the square required.

Because BE is equal to ED, for E is the center, and that EA is common, and at right angles to BD;

the base BA is equal to the base AD: (1. 4.) and, for the same reason, BC, CD are each of them equal to BA, or AD;

therefore the quadrilateral figure ABCD is equilateral.

It is also rectangular ;

for the straight line BD being the diameter of the circle ABCD,

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