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Book IV. gle; it is required to inscribe in the circle ABC a triangle
the angle HAC equal to the angle DEF; and at the point A,
of contact, the angle • 34. 3. HAC is equal e to the
angle ABC in the alternate segment of the circle : But HAC is equal to the angle DEF:therefore also the angle ABC is equal to DEF: For the same reason, the angle
ACB is equal to the angle DFE; therefore the remaining • 32. 1. angle BAC is equald to the remaining angle EDF: Where
fore the triangle ABC is equiangular to the triangle 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 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
centre K of the circle ABC, and from it draw any straight • 23. 1. line KB; at the point K in the straight line KB, make a
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 straight lines LAM, MBN, NCL, • 17. 3. touching b 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, the angles at the C18. 3. points A, B, C, are right angles: 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 Boox IV: that two of them KAM, KBM are right angles, the other two AKB, AMB are equal to two right angles : But the angles DEG, DEF are likewise equal
d 13. 1. to two right angles; therefore
F H the angles AKB, AMB are equal
the angles "
DEG, 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 to the remaining angle EDF :. 32. 1. Wherefore the triangle LMN is equiangular to the triangle DEF: And it is described about the circle ABC. Which was to be done.
PROP. IV. PROB.
Bisecta the angles ABC, BCA by the straight lines BD, 9. 1. CD meeting one another in the point D, from which drawb o 12. 1. DE, DF, DG perpendiculars to A AB, BC, CA: And because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD, and that the right angle BED is equal to the right angle BFD, the two triangles E EBD, FBD have two angles of the one equal to two angles of the other, and the side BD, K which is opposite to one of the B equal angles in each, is common to both ; therefore their
Boo'k IV. other sides shall be equalc; wherefore DE is equal to DF: W ten
For the same 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 centre 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 d 16. 3. a diameter at right angles to it, touches d the circle: There
fore the straight 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,
PROP. V. PROB.
See N. To describe a circle about a given triangle.
Let the given triangle be ABC; it is required to describe
a circle about ABC. *10. 1. Bisecta AB, AC, in the points D, E, and from these points b11. 1. draw DF, EF at right angles b to AB, AC; DF, EF, pro
duced meet one another : For, if they do not meet, they are parallel, wherefore AB, AC, which are at right angles to them, are parallel ; which is absurd : 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, • 4. 1. and at right angles to AB, the base AF is equalc to the
base FB. In like manner, it may be shown that CF is equal to FA; and therefore BF is equal to FC; and FA, FB, FC are equal to one another ; wherefore the circle described from the centre F, at the distance of one of them, Book IV. 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 manifest, that when the centre 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 centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, is a right angle; and, if the centre falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle: Wherefore, if the given triangle be acute angled, the centre of the circle falls within it; if it be a right angled triangle, the centre is in the side opposite to the right angle; and, if it be an obtuse angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle.
PROP. VI. PROB.
* 4. 1.
To inscribe a square in a given circle."
Let ABCD be the given circle; it is required to inscribe a 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 ÉD, for E' is the centre, and that EA is common, and at right angles to BD; the base BA is equal a to the base AD; and, for the same reason, BC, CD are each of them equal to BA, or AD; therefore the quadrilateral figure ABCD is equilateral. It is also rectangular; for the straight line BD, being the diameter of the circle ABCD, BAD is a semicircle; wherefore the angle BAD is a right b an- 1 31. 3. gle; for the same reason each of the angles ABC, BCD, CDA, is a right angle; therefore the quadrilateral figure ABCD is rectangular, and it has been shown to be equilateral; therefore it is a square; and it is inscribed in the circle ABCD. Which was to be done.
PROP. VII. PROB.
Let ABCD be the given circle; it is required to describe a square about it.
Draw two diameters AC, BD of the circle ABCD, at
right angles to oue another, and through the points A, B, a 17. 3. C, D, draw a FG, GH, HK, KF touching the circle ; and
because FG touches the circle ABCD, and EA is drawn
from the centre E to the point of contact A, the angles at b 18. 3. A are right b angles; for the same reason, the angles at the
points B, C, D are right angles; and
angle, as likewise is EBG, GH is
son AC is parallel to FK, and in like
GC, AK, FB, BK are parallelo-
СК to HK, and GH to FK; and because AC is equal to BD, and that AC is equal to each of the two GH, FK; and BD to each of the two GF, HK: GH, FK are each of them equal to GF or HK; therefore the quadrilateral figure FGHK is equilateral. It is also rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB d is likewise a right angle: In the same manner it may be shown 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 square; and it is described about the circle ABCD. Which was to be done,
PROP. VIII. PROB.
Let ABCD be the given square; it is required to inscribe : . a circle in ABCD.
* 10. 1. Bisecta each of the sides AB, AD, in the points F, E, and | 31. 1. through E drawb EH parallel to AB or DC, and through