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tre : Take the centre F, and through E the intersection of the straight lines AC, DB draw the diameter GEFH: And because the rectangle AE, EC is equal, as has been shewn to the rectangle GE, EH; and, for the same reason, the rectangle BE, ED is equal to the same rectangle GE, EH; therefore the rectangle AE, EC is equal to the rectangle BE, ED Wherefore, if two straight lines, &c. Q. E. D.

Proposition XXXVI. Theorem. If from any point without a circle two straight lines be drawn, ono of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.

Let D be any point without the circle ABC, and DCA, DB two straight lives drawn from it, of which DCA cuts the circle, and DB touches the same : The rectangle AD, DC is equal to the square of DB.

Either DCA passes through the centre, or it does not ; first, let it pass through the centre E, and join EB : therefore the angle EBD is a right (18. 3.) angle : And because the straight line AC is bisected in E, and produced to the point D the rectangle AD, DC, together with the square o EC, is equal (6. 2.) to the square of ED, and CE i equal to EB : Therefore the rectangle AD, DC, to gether with the square of EB, is equal to the square of ED: But the square of ED is equal (47. 1.) to

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E the squares of EB, BD, because EBD is a right angle : Therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: Take away the coinmon square of EB : therefore the remaining rectangle AD, DC is equal to the square of the tangent DB.

But if DCA does not pass through the centre of the circle ABC: take (1.3.) the centre E, and draw EF perpendicular (12. 1.) to AC, and join EB, EC, ED : And because the straight line EF, which

passe. through the centre, cuts the straight line AC which does not pass through the centre, at right angles, it shall likewise bisect (3. 3.) it; therefore AF is equal to FC · And because the straight line AC is bisected in F, and produced to D, the rectangle AD, DC, to gether with the squarc of FC is equal (6. 2.) to the square of FD: To each of these equals add the square of FE; therefore the rectangle AD, DC, logether with the squares of CF, FE, is equal

squares of DF, FE : But the square of ED is equal (47. 1.) to the squares of DF, FE, because EFD is a right angle : and the square of EC is equal to the squares of CF, FE; therefore the rectangle AD, DC, together with the square of EC, is equal to the square of ED, and CE is equal

to the

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to EB ; therefore the rectargle AD, DC, together with the square of EB is equal to the square of ED : But the squares of EB, BD are equal to the square (47. 1.) of ED, because EBD is a right angle ; therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: Take away the common square of EB ; therefore the remaining rectangle AD, DC is equal to the square of DB. Wherefore, if from any point, &c. Q. E. D.

Cor. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts. of them without the circle, are equal to one another, viz. the rectangle BA, AE to the rectangle CA, AF: For each of them is equal to the square of the straight Ine AD which touches the circle.

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Proposition XXXVII. Theorem. If from a point without a circle there be drawn two straight lines one of which cuts the circle, and the other meets it ; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle be equal to the square of the line which meets it, the line which meets shall touch the circle.

Let any point D be taken without the circle ABC, and from it let two straight liues DCA and DB be drawn, of which DCA cuts the cir. cle, and DB meets it ; if the rectangle AD, DC be equal to the square of DB ; DB touches the circle.

Draw (17.3.) the straight line De, touching the circle ABC, find its centre F, and join FE, FB, FD; then FED is a right (18.3) angle : And because DE touches the circle ABC, and DCĂ cuts it, the rectangle AD, DC is equal (36. 3.) to the square of DE : But the rectangle AD, DC is, by hypothesis, equal to the square of DB : Therefore the square of DE is equal to the square of DB; and the straight line DE equal to the straight line DB. And FE is equal to FB, wherefore DE, EF are equal to DB, BF; and the base FD is common to the two triangles DEF, DBF ; therefore the angle DEF, is equal (8. 1.) to the angle DBF; but DĚF is a right angle, therefore also DBF is a right angle : And FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from the extremity of it touches (16. 3.) the circle : There fore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D.

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BOOK IV.

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DEFINITIONS.
I.

scribed about a circle, when cach A IECTILINEAL figure is said to side of the circumscribed figure be inscribed in another rectilineal fie touches the circumference of the gure, when all the angles of the in circle. acribed figure are upon the sides of the figure in which it is inscribed, In like manner, a circle cach upon each.

is said to be inscribII.

ed in a rectilineal fi. In like manner, a figure

gure, when the cir. is said to be described

cumference of the about another figure,

circle touches each when all the sides of

side of the figure. the circumscribed fi

VI. gure pass through the angular A circle is said to be described about points of the figure about which a rectilineal figure, it is described, each through each. when the circumIII.

ference of the circle A rectilineal figure is

passes through all said to be inscribed

the angular points in a circle, when all

of the figure about the angles of the in

which it is described. scribed figure are

VII. upon the circumfer

A straight line is said to be placed ence of the circle.

in a circle, when the extremities IV.

of it are in the circumference of A rectilineal figure is said to be de. the circle.

PROP. I. PROB.

In a given circle to place a straight line, equal to a given straight

line not greater than the diameter of the circle.

Let ABC be the given circle, and BC is greater D the given straight line, not greater than D; make than the diameter of the circle. CE equal (3.1.)

Draw BC the diameter of the circle to D, aud from ABC; then, if BC is equal to D, the the centre C, thing required is done ; for in the at the distance circle ABC a straight' line BC is CE, describe placed equal to D: But, if it is not, the circle AEF, and join CA: There

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ore, because C is the centre of the equal to the given straight line D, circle AEF, CA is equal to CE, but which is not greater than the diaD is equal to CE; therefore D is meter of the circle. Which was to equal to CA: Wherefore, in the be done. circle ABC, a straight line is placed

PROP. II. PROB.

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

triangle.

D

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

Draw (17. 3.) the straight line GAH touching the circle in the point A, and at the point A, in the the alternate segment of the circle': straight line AH, make (23. 1.) the But HAC is equal to the angle DEF: angle HAC equal to the angle DEF; therefore also the angle ABC is equal and at the point A, in the straight to DEF: For the same reason, the line AG, make the angle GAB equal angle ACB is equal to the angle to the angle DFE, and join BC: DFE ; therefore the remaining angle Therefore, because HAG touches the BAC is equal (32. 1.) to the remain. circle ABC, and AC is drawn from ing angle EDF: Wherefore the triangle the point of contact, the angle HAC ABC is equiangular to the triangle is equal (32. 3.) to the angle ABC in DEF, and it is inscribed in the circle

ABC. Which was to be done.

PROP. III. PROB.

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

triangle. Let ABC be the given circle, and (17. 3.) the circle ABC; Therefore, DEF the given triangle ; it is require because LM, MN, NL touch the ed to describe a triangle about the circle ABC in the points A, B, C, to circle ABC equiangular to the tri- which from the centre are drawn KA, angle DEF.

KB, KC, the angles at the points A, Produce EF both ways to the B, C, are right (18. 2.) angles : And points G, H, and find the centre K because the four angles of the quaof the circle ABC, and from it draw drilateral figure AMBK are equal to any straight line KB; at the point K, four right angles, for it can be dividin the straight line KB, make (23. 1.) ed into two triangles; and that two the angle BKA equal to the angle of them KAM, KBM are right angles, DEG, and the angle BKC equal to the other two AKB, AMB are equal the angle DFH; and through the to 'two right angles : But the angles points A, B, C, draw the straight DEG, DEF are likewise equal (13. 1.) lines LAM, MBN, NCL, touching to two right angles; therefore the

A

maining angle AMB is equal to the remaining angle DEF : In like manner, the angle LNM may be demonstrated to be equal to DFE ; and therefore the remaining angle MLN is equal (32. 1.) to the remaining angle EDÈ: Wherefore the triangle

LMN is equiangular to the triangle angles AKB, AMB are equal to the DEF: And it is described about the angles DEG, DEF of which AKB is circle ABC. Which was to be done. eq nal to DEG; wherefore the re

M

B

N

PROP. IV. PROB.

A

To inscribe a circle in a given triangle. Let the given triangle be ABC, it equal angles in each, is common to is required to inscribe a circle in both; therefore their other sides shall ABC.

be equal; (26. 1.) wherefore DE is Bisect (9. 1.) the angles ABC, BCA equal to DF: For the same reason by the straight lines BD, CD meet. DG is equal to DF; therefore the ing one another in the point D, from three straight lines DE, DF, DG are which draw (12. 2.) DE, DF, DG equal to one another, and the circle perpendiculars

described from the centre D, at the to AB, BC,

distance of any of them, shall pass CA: And be

through the extremities of the other cause the an

two, and touch the straight lines AB, gle EBD is e

BC, CA, because the angles at the qual to the an

points E, F, G, are right angles, and gle FBD, for

the straight line which is drawn froin the angle ABC

the extremity of a diarneter at right is bisected by BD, and that the right angles to it, touches (16. 3.) the angle BED is equal to the right angle circle; therefore the straight lines • BÉD, the two triangles EBD, FBD AB, BC, CA do each of them touch have two angles of the one equal to the circle, and the circle EFG is intwo angles of the other, and the side scribed in the triangle ABC. Which BD, which is opposite to one of the was to be done.

D

PROP. V. PROB.

To describe a circle about a given triangle, Let the given triangle be ABC; it Bisect (10. 1.) AB, AC in the is required to describe a circle about points D, E, and from these points ABC.

draw DF, EF at right angles (il. 1.)

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