Book IV.

a. 17.3.

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.

a

Draw the straight line GAH touching the circle in the point A, b. 23. 1. and at the point A, in the straight line AH, make b the angle HAC equal to the angle DEF; and at the point A, in the straight line AG, make the angle.

G

A

H

F

B

C

ternate segment of the circle. but HAC is equal to the angle DEF, therefore alfo the angle ABC is equal to DEF. for the fame reafon the angle ACB is equal to the angle DFE; therefore the remaining d. 32. 1. angle BAC is equal to the remaining angle EDF. wherefore the triangle ABC is equiangular to the triangle DEF, and it is infcribed in the circle ABC. Which was to be done.

A

PROP. III. PROB.

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 center K of the circle ABC, and from it draw any straight line KB; at the a. 23. 1. 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; and thro'

a

the points A, B, C draw the straight lines LAM, MBN, NCL touchb. 17. 3. ing b the circle ABC. therefore becaufe LM, MN, NL touch the circle ABC in the points A, B, C to which from the center are drawn c. 18. 3. KA, KB, KC, the angles at the points A, B, C are right angles. and because the four angles of the quadrilateral figure AMBK are

equal

équal to four right angles, for it can be divided into two triangles; Book IV. and that two of them KAM, KBM are right angles, the other two

AKB, AMB are equal to two right angles. but the an

L

equal to DEG; wherefore the remaining angle AMB is equal to the
remaining angle DEF. in like manner the angle LNM may be dè-
monftrated to be equal to DFE; and therefore the remaining angle
MLN is equal to the remaining angle EDF.
angle LMN is equiangular to the triangle DEF.
about the circle ABC. Which was to be done.

e

wherefore the tri- c. 3. źl and it is described

O infcribe a circle in a given triangle.

то

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

See Ni

Bifect the angles ABC, ECA by the ftraight lines BD, CD a. 9. i. meeting one another in the point D, from which draw b DE, DF, b. iz. t. DG perpendiculars to AB, BC,CA.

and because 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 com

D

B

F

G

mon to both; therefore their other fides shall be equal; where- c. 10. 2

Book IV. fore DE is equal to DF. for the fame reason, DG is equal to DF;

therefore the three ftraight lines DE, DF, DG are equal to one

another, and the circle defcribed

from the center D, at the distance

A

of any of them, fhall pass thro' the
extremities of the other two, and
touch the straight lines AB, BC,
CA, becaufe the angles at the
points E, F, G are right angles,
and the straight line which is
drawn from the extremity of a di-
ameter at right angles to it, touches

d. 16. 3d the circle. therefore the straight

See N.

lines AB, BC, CA do each of them touch the circle, and the circle EFG is infcribed in the triangle ABC. Which was to be done.

PROP. V. PROB.

To defcribe a circle about a given triangle.

Let the given triangle be ABC; it is required to describe a circle about ABC.

a

Bisect AB, AC in the points D, E, and from these points draw b. 11. 1. DF, EF at right angles to AB, AC; DF, EF produced meet one

C. 4.1.

another. for if they do not meet they are parallel, wherefore AB, AC which are at right angles to them are parallel; which is abfurd. let them meet in F, and join FA; alfo, 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, the base AF is equal to the base FB. in like manner it may be fhewn that CF is equal to FA; and therefore BF is equal to FC; and FA, FB, FC are therefore equal to one

another.

another. wherefore the circle defcribed from the center F, at the Book IV. distance of one of them, fhall pafs thro' the extremities of the other two; and be described about the triangle ABC. Which was to be done.

COR. And it is manifeft that when the center of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a fegment greater than a femicircle. but when the center is in one of the fides of the triangle, the angle oppofite to this fide, being in a femicircle, is a right angle. and if the center falls without the triangle, the angle oppofite to the fide beyond which it is, being in a fegment lefs than a femicircle, is greater than a right angle. Wherefore, if the given triangle be acute angled, the center of the circle falls within it; if it be a right angled triangle, the cen-. ter is in the fide opposite to the right angle; and if it be an obtufe. angled triangle, the center falls Without the triangle, beyond the fide oppofite to the obtufe angle.

PROP. VI. PROB.

To inscribe a square in a given circle.

Let ABCD be the given circle; it is required to infcribe a fquare in ABCD.

a

A

Draw the diameters AC, BD at right angles to one another; and join AB, BC, CD, DA. because BE is equal to ED, for E is the cen ter, and that EA is common, and at right. angles to BD; the bafe BA is equal to the bafe AD. and for the fame reafon, BC, CD are each of them equal to BA or AD; therefore the quadrilateral B figure ABCD is equilateral. It is alfo rectangular; for the ftraight line BD being the diameter of the circle ABCD,

BAD is a femicircle; wherefore the

E

D

C

angle BAD is a right angle. for the fame reafon each of the angles b. 1. Al ABC, BCD, CDA is a right angle. therefore the quadrilateral figure ABCD is rectangular. and it has been fhewn to be equilateral, therefore it is a fquare; and it is infcribed in the circle ABCD: Which was to be done.

Book IV.

PROP. VII. PROB.

'O defcribe a fquare about a given circle.

Let ABCD be the given circle; it is required to defcribe a fquare about it.

Draw two diameters AC, BD of the circle ABCD, at right angles a. 17. 3. to one another, and thro' the points A, B, C, D draw a FG, GH, HK, KF touching the circle. and becaufe FG touches the circle ABCD, and EA is drawn from the center E to the point of contact b. 18. 3. A, the angles at A are right bangles. for the fame reason, the angles at the points B, C, D are right angles. and because the angle AEB is a right angle, as likewife is EBG, GH c. 28. 1. is parallel to AC. for the fame reason, AC is parallel to FK. and in like manner GF, HK may each of them be demonftrated to be parallel to BED. therefore B the figures GK, GC, AK, FB, BK are parallelograms, and GF is therefore e

J

A

F

G

E

D

H

C

K

d. 34. 1. qual to HK, and GH to FK. and becaufe AC is equal to BD, and that AC is equal to each of the two GH, FK; and BD to each of the two CF, HK; GH, FK are each of them equal to GF or HK. therefore the quadrilateral figure FGHK is equilateral. It is alfo rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB is likewite a right angle. in the fame manner it may be fhewn that the angles at H, K, F are right angles. therefore the quadrilateral figure FGHK is rectangular. and it was demonstrated to be equilateral; therefore it is a fquare; and it is described about the circle ABCD. Which was to be done.

2. 10. I.

PROP. VIII. PROB.

TO infcribe a circle in a given square.

Let ABCD be the given fquare; it is required to infcribe a circle in ABCD.

Bifect each of the fides AB, AD, in the points F, E, and thro' b. 31. г. E draw b EH parallel to AB or DC, and thro' F draw FK parallel to

AD