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tilineal figure, when the circumference
mities of it are in the circumference of the circle.
IN a given circle to place a straight line, equal to a gi
ven straight line 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.
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
Α. is greater than D; make CE equal to D, and from the cen
a 3. 1, ter C, at the distance CE, de
E scribe the circle AEF, and join
B CA: Therefore, because Č is
F the center of the circle AEF, CA is equal to CE; but Dis D equal to CE; therefore D is equal to CA: Wherefore in the circle ABC a straight 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.
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 a the straight line GAH touching the circle in the point A, and at the point A, in the ftraight line AH, make b the angle HAC equal to the angle DEF; and at the point A, in the straight line
C 32. 3.
d 32. I.
BOUT a given circle to describe a triangle equian, gular to a given triangle.
a 13. I.
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.
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 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 through the points A, B, C, draw the Araight lines LAM, MBN, NCL touching the circle ABC: Therefore, because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the center are drawn KA, KB, KC, the angles at the points A, B, C, are right angles : And because the four angles of the quadrilateral fi
b 19. 3.
c. 18. 3.
gure AMBK are equal to four right angles, for it can be divided in- Book IV. to two triangles; and that two of them KAM,KBM are right angles, the other two AKB, AMB are
L equal to two right angles: But the
D angles DEG, DEF are likewise
d 13. s. equal d to two A K K right angles ; therefore the an
GE FH gles AKB, AMB are equal to the
M angles DEG,
B DEF, 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 demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal o to the c 32, 1. remaining angle EDF: Wherefore the triangle LMN is equian
gular to the triangle DEF: And it is described about the circle - ABC. Which was to be done.
PRO P. IV.
To infcribe a circle in a given triangle.
Let the given triangle be ABC ; it is required to inscribe a circle in ABC.
Bisect * the angles ABC, BCA by the straight lines BD, CD a. 9. 1. meeting one another in the point D, from which draw DE, DF, 12. 1. DG perpendiculars to AB,
BC, CA: And because the angle EBD
A 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
G have two angles of the one e. D qual to two angles of the o
E ther, and the fide BD, which is a oppofite to one of the equal angles in each, is common to boch; therefore their other B fides Thall be equal“; where,
c 26. fore
Book IV. fore DE is equal to DF: For the fame reason, DG is equal to
DF; 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, shall 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 d the circle : Therefore the straigh. lines AB, BC, CA do each of them touch the circle, and the circle EFG is inscribed in the triangle ABC. Which was to be done.
PRO P. V. PROB.
To describe a circle about a given triangle.
Let the given triangle be ABC; it is required to describe a
circle about ABC. a 10. I. Bisecta AB, AC in the points D, E, and from these points bil. S. draw DF, EF at right angles b to AB, AC; DF, EF produced
meet one another: For, if they do not meet, they are parallel, wherefore AB, AC, which are at right angles to ihcm, are pa. rallel; which is abfurd : 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, the base AF is equal to the bafe FB: In like mannır, it may be shewn that CF is equal to FA; and therefore BF is equal to FC: and FA, FB, FC are equal to one an.
other; wherefore the circle described from the center F, at the Book IV. diftance of one of them, shall pass through 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 segment greater than a semicircle ; but, when the center is in one of the sides of the triangle, the angle opposite to this fide, being in a semicircle, 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 fegment less than a semicircle, 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 center is in the Gide opposite to the right angle; and, if it be an obtufe angled triangle, the center falls without the triangle, beyond the fide oppolite to the obtufe angle.
Let ABCD be the given circle ; it is required to inscribe square in ABCD.
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 center, and that EA is common, and at right angles to
A BD; the base BA is equal to the
a 4. in base AD; and, for the same reason, BC, CD are each of them equal to
E BA or AD; therefore the quadri- B lateral figure ABCD is equilateral. It is also rectangular; for the straight line BD, being the diameter of the circle ABCD, BAD is a semicircic; wherefore the angle BAD is a right
с bangle ; for the same reason each of the angles ABC, BCD, 1 31. 3. CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular, and it has been shewn to be equilateral ; therefore it is a square ; and it is infcribed in the circle ABCD. Which was to be done.