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FE is equal to FB, wherefore DE, EF are equal to DB, BF; Book III. and the base FD is common to the
D two triangles DEF, DBF; therefore the angle DEF is equal d to the angle
d 8.1. DBF; but DEF is a right angle, therefore also DBF is a right angle: and FB, if produced, is a diameter,
E and the straight line which is drawn. B at right angles to a diameter, from the extremity of it, touches e the circle:
e 16. 3. therefore DB touches the circle ABC.
ELEMENTS OF EUCLID.
I. Book IV. A RECTILINEAL figure is said to be inscribed in another
rectilineal figure, when all the angles of the inscribed figure See N
are upon the sides of the figure in which it is
about another figure, when all the sides of
in a circle, when all the angles of the in-
each side of the circumscribed figure
ed in a rectilineal figure, when the cir-
Book IV. A circle is said to be described about a rec
tilineal figure, when the circumference of
ties of it are in the circumference of the circle.
PROP. I. PROB.
IN a given circle to place a straight line, equal to a given straight line not greater than the diameter of the circle.
Lét 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
А equal to D; but, if it is not, BC is greater than D; make CE equal a to D, and from the cen
a 3. 1.
E tre C, at the distance CE, de
C scribe the circle AEF, and join CA: therefore, because C is
F the centre of the circle AEF, CA is equal to CE; but D is
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.
PROP. II. PROB.
IN a given circle to inscribe a triangle equiangular. to a given triangle.
Book IV. 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. a 17. 3. Draw a the straight line GAH touching the circle in the point b 23. 1. A, 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 an
G gle GAB equal to the
H angle DFE, and join D BC: therefore because HAG touches the circle ABC, and AC is drawn from the point
F of contact, the angle
с c 32. 3.' HAC is equalc to the
angle ABC in the alter-
is equal to the angle DFE; therefore the remaining angle BAC d 32. 1. is equald to the remaining angle EDF: wherefore 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 line KB; a 23. 1. 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 ► 17. 3. lines LAM, MBN, NCL touching the circle ABC: there
fore because LM, MN, NL touch the circle ABC in the
points A, B, C, to which from the centre are drawn KA, KB, < 18.3. KC, the angles at the points A, B, C are rightc angles : and
because the four angles of the quadrilateral figure AMBK are equal to four right angles, for it can be divided into two tri- Book IV. angles: and that two of them KAM, KBM are right angles, the other two AKB, L AMB are equal to two right angles :
D but the angles DEG, DEF are likewise equal to A
d 13. 1. two right angles;
K therefore the an
GE FH gles AKB, AMB are equal to the angles DEG, DEF
N. of which AKB is MB 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:e 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.
TO inscribe a circle in a given triangle.
Let the given triangle be ABC; it is required to inscribe a circle in ABC.
Bisect a the angles ABC, BCA by the straight lines BD, CD a 9.1. meeting one another in the point D, from which drawb DE, b 12. 1. DF, DG perpendiculars to AB, А 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
B F fore their other sides shall be