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1. A Right lined figure is said to be inscribed in a right lined Book IV,
every one of the angles of the inscribed figure touches every one of the sides of the figure wherein it is inscribed.
gure, when every one of the sides of the circumscribed figure
the sides of the circumscribed figure touches the circumfe-
ference of the circle touches all the fides of the figure in
VI. A circle is described about a right lined figure when the circumference of the circle touches all the angles of the figure.
VII. MA right line is applied in a circle when its extremes are in the
circumference of the circle.
PRO P. I. PRO B.
To apply a right line in a circle, equal to a given right line, not
It is required to apply a right line in the circle ABC, equal to a given right line D, not greater than the diameter of the circle.
Draw the diameter BC; if equal to D, what was required is done; if not, the diameter BC is greater than D; put CE equal to Da; about the center C, with the distance CE, describe the circle AEF; then CA is equal to CE ; but CE is equal to D; therefore CA is equal to D. Wherefore, there is drawn, &c.
IN a given circle, to inscribe a triangle equiangular to a given
It is required to inscribe a triangle, in a given circle ABC,
equiangular to a given triangle DEF: Draw the right line a 37. 3. GAH, touching the circle in the point A*; with the right line
AH, at the point A, make the angle HAC equal to the angle b 23, 1. DEFb, and the angle GAB equal to DFE; join BC.
Then the angle HAC is equal to the angle ABC; but the ¢ 32. 31 angle HAC is equal to DEF; therefore, ABC is equal to DEF:
And BAG is equal to ACB C; but BAG is equal to DFE;
therefore ACB, is equal to DFE; therefore the remaining third & Cor. 34. angles BAC, EDF d are equal; therefore the triangle ABC, is
inscribed in the circle ABC, and equiangular to the triangle DEF, which was required. Wherefore, &c.
P R O P. III. PRO B.
BOUT a given circle, to describe a triangle equiangular
to a triangle given. It is required to describe a triangle about the given circle ABC equiangular to the given triangle DEF ; produce the fide EF both ways to G, H; find the center of the circle K, and draw KB any how; at the point K, with the right line KB, make the angles BKA, BKC, equal to the angles DEG, DFH , ' 23. si each to each ; at the points A, B, C, draw the right lines LAM, MBN, LCN, tangents to the circle, in the points A, B, Ch.
Then the angles that LM, MN, LN, make with the right lines KA, KB, KC, are right angles; therefore the angles © 18. 3. AKB, AMB, are equal to two right anglesd, and equal to d 32. 1. DEF, DEGʻ; but BKA was made equal to DEG; therefore e 13. 1. the remainder DEF is equal to AMBf. For the same reason f Ax. 8. 31 DFE is equal to LNM, and the remaining angle M N equal . to EDF; wherefore the triangles LMN, DEF, are equiangular. Which was required.
b 16. 3:
It is required to inscribe a circle in the given triangle ABC. Bifect the angles ABC, ACB, by the right lines BD, DC, a 9. si meeting each other in the point D; from which let fall DF, DE, DG, perpendiculars, upon the right lines AB, BC, ACÓ. Then, because the two angles DFB, DBF, in the triangle DBF, are equal to the two angles DEB, PBE, in the triangle DBE, and iħe fide DB common to both; the remaining fides BE, ED, are equal to BF, FD, each to each. For the fame c 26. 1. reason DF is equal to DG; therefore D is the center, and with any of the distances the circle EFGd may be infcribed in the gi-d 9. 3. ven triangle ABC: Which was required.