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Let ABC be a triangle, and from the angles at B, C draw Book. III. BD, CE perpendiculars to AC, AB, and let them meet in F, and join AF meeting BC in G: AC is perpendicular to BC.

E

Join DE: and because ADF, AEF are right angles, the circumference of the circle, of which AF is the diameter, fhall pass through the points E, D; thereA fore the angle DEF is equal to the angle DAF, because they are in the fame fegment DAEF. Again, because BDC, BEC are right angles, the circumference of which BC is the diameter, fhall pass through D, E 2; therefore the angle DEC is equal to the angle DBC, because they are in the fame fegment DEBC: but it has been proved, that the angle DEC is equal to the angle DAF; therefore the angle CBF is equal to DAF: and the angle BFG is equal to DFA; therefore the remaining angle BGF is

d

b

B

G

D

a 2. Cor. A. 3.

b 21. 3.

C 15. I.

equal to the remaining angle ADF; that is, to a right angle; d 32. 1. and AFG is therefore perpendicular to BC. Wherefore, &c. Q. E. D.

THE

BOOK IV.

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A

DEFINITIONS.

I.

RECTILINEAL figure is faid to be infcribed in another

rectilineal figure, when all the angles of the infcribed figure are upon the fides of the figure in which

it is inscribed, each upon each.

II.

In like manner, a figure is faid to be defcribed
about another figure, when all the fides of
the circumfcribed figure pafs through the an-

gular points of the figure about which it is described, each
through each.

III.

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VI.

A circle is faid to be described about a rectilineal figure, when the circumference of the circle paffes through all the angular points of the figure about which it is described.

VII.

Aftraight line is faid to be placed in a circle, when the extremities of it are in the circumference of the circle.

IN

PROP. I. PROB.

N a given circle to place a straight line, equal to a given ftraight line, not greater than the diameter of the circle.

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

A

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 ftraight line BC is placed equal to D: But, if it is not, BC is greater than D; make CE equal a to D, and from the centre C, at the distance CE, defcribe the circle AEF, and join CA: Therefore, because C is the centre of the circle AEF, CA is equal to CE; but D is equal to CE;

D

F

BOOK IV.

E

a 3. I.

B

therefore D is equal to CA: Wherefore, in the circle ABC, a ftraight line is placed equal to the given straight line D, which is not greater than the diameter of the circle. Which was to be done.

PROP. II. PROB.

IN

Na given circle, to infcribe a triangle equiangular to a given triangle.

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

Draw the straight line GAH touching the circle in the a 17. 1. point A, and at the point A, in the ftraight line AH, make

the angle HAC equal to the angle DEF; and at the point A, b 23. 1. in the ftraight line AG, make the angle GAB equal to the angle DFE, and join BC: Therefore, because HAG touches the circle ABC, and AC is drawn from the point of contact,

the

c

Book IV. the angle HAC is equal to the angle ABC in the alternate segment of the circle: But HAC is equal to the angle DEF;

€ 32. 3. therefore also the angle ABC

is equal to DEF: For the fame reafon, the angle ACB is equal to the angle DFE; therefore the remaining angle BAC is d 32. 1. equal to the remaining angle EDF: Wherefore the triangle B ABC is equiangular to the triangle DEF, and it is infcribed

See N.

G

in the circle ABC. Which was to be done.

A

PROP. III. PROB.

a given

H

D

E

F

BOUT 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 defcribe a triangle about the circle ABC equiangular to the triangle DEF.

Produce EF both ways to the points G, H, and find the centre K of the circle ABC, and from it draw any straight line a 23. 1. KB; at the point K, in the straight line KB, make a the angle

:

b

C

BKA equal to the angle DEG, and the angle BKC equal to the angle DFH; and through the points A, B, C, draw the straight b 17. 3. lines LAM, MBN, NČL, touching the circle ABC: Therefore, because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, c 18. 3. the angles at the points A, B, C, are right angles: And because the angles AKB, BKC are equal to DEG, DFH, they d 16. 1. are greater than DEG, DEF, that is, greater than two right e 13. 1. angles; therefore, if AC be joined, it fhall fall between K and f 2. Cor. L and the angles ACL, CAL fhall therefore be less than two 15. 1. right angles; wherefore AL, CL fhall meet &, if produced: For g12.AX.1. the fame reason, they fhall, each of them, meet MN: let them meet in the points L, M, N: 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, the other M

two

A

L

K

GE

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e

two AKB, AMB are equal to two right angles: But the angles Book. IV. DEG, DEF are likewife equal to two right angles; therefore the angles AKB, AMB are equal to the angles DEG, DEF, e 13. 1. of which AKB is equal to DEG; wherefore the remaining angle AMB is equal to the remaining angle DEF: In like manner, the angle LNM may be demonftrated to be equal to DFE; and therefore the remaining angle MLN is equal to the re- h 32. I. maining angle EDF: Wherefore the triangle LMN is equiangular to the triangle DEF: And it is defcribed about the circle ABC. Which was to be done.

PROP. IV. PROB.

To inferibe a circle in a given triangle.

h

Let the given triangle be ABC; it is required to infcribe a circle in ABC.

a

b

a 9. I.

Bifect the angles ABC, BCA by the ftraight lines BD, CD meeting one another in the point D, from which draw DE, b 12. x. DF, DG perpendiculars to AB, BC, CA: And becaufe the angle EBD is equal to the angle FBD, for the angle ABC is bifected by BD, and that the right angle BED is equal to the right angle BFD, the two triangles EBD, FBD have two angles of the one equal to two angles of the other, and the fide BD, which is oppofite to one of the equal angles in each, is common to

B

D

F

both; therefore their other fides fhall be equal; wherefore c 26. 1. DE is equal to DF: For the fame reafon, DG is equal to DF; therefore the three ftraight lines DE, DF, DG are equal to one another, and the circle defcribed from the centre D, at the distance of any of them, fhall pass through the extremities of the other two; and because the angles at the points E, F, G, are right angles, and the ftraight line which is drawn from the extremity of a diameter at right angles to it, touches the circle; d 16. 3. therefore the straight lines AB, BC, CA do each of them touch the circle, and the circle EFG is inferibed in the triangle ABC. Which was to be done.

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