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Book II. to the square of GH: But the squares of HE, EG are equalc
to the square of GH: Therefore the rectangle BE, EF, to€ 47, 1.
gether with the square of EG, is equal to the squares of HE, EG: Take
of EG, which is common to both; and the remaining rectangle BE, EF is equal to the square of EH: But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the square of EH; but BD is equal to the rectilineal figare A, therefore the rectilineal figure A is equal to the square of EH. Wherefore a square has been made equal to the given rectilineal figure A, viz. described
EH. Which was to be done.
E U C L I D.
Equal circles are those of which the diameters are equal, Book IIS. or from the centres of which the straight lines to the circumferences are equal.
This is not a definition, but a theorem, the truth of which is evident; for, if the circles be applied to one another
, so that their centres coincide, the circles must likewise coincide, since the straight lines from the centres are equal.
touch a circle, when it
another, which meet but
greater perpendicular falls, is said to be further from the centre.
contained by a straight line and
contained by two straight lines
upon the circumference intercept-
tained by two straight lines drawn
cle are those in which
PROP. I. PROB.
See N. To find the centre of a given circle.
Let ABC be the given circle ; it is required to find its
centre. a 10.1.
Draw within it any straight line AB, and bisecta it in D; 11. 1. from the point D drawb DC at right angles to AB, and proEG
duce it to E, and bisect CE in F: The point F is the centre of the circle ABC.
For, if it be not, let, if possible, G be the centre, and join Book III. GA, GD, GB: Then, because DA is equal to DB, and DG e common to the two triangles ADG, BDG, the two sides AD, DG, are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are drawn from the centre G*: Therefore the angle ADG is equal to the angle
° 8. 1. GĎB : But when a straight line standing upon another straight line A
B makes the adjacent angles equal to
E one another, each of the angles is a right angled : Therefore the angle GDB is a right angle:' 10 Def. I. But FDB is likewise a right angle : wherefore the angle FDB is equal to the angle GDB, the greater to the less, which is impossible : Therefore G is not the centre of the circle ABC. In the same manner it can be shown, that no other point but F is the centre ; that is, F is the centre of the circle ABC: Which was to be found.
Cor. From this it is manifest, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line wbich bisects the other.
PROP. II. THEOR. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.
Let ABC be a circle, and A, B any two points in the circumference; the straight line drawn from A to B shallfall within the circle.
For, if it do not, let it fall, if possible, without, as AEB; find a D the
* 1. 3. centre of the circle ABC; and join AD, DB, and produce DF, any straight line meeting the circumference AB to E: Then because DA is
F equal to DB, the angle DAB is equalh
5. i. to the angle DBA; and because AE,
A E B * N. B. Whenever the expression straight lines from the centre" or “ drawn from the centre" occurs, it is to be understood that they are drawn to the circumference.
Book III. a side of the triangle DAE, is produced to B, the angle
DEB is greater than the angle DAE;. but DAE is equal 616. 1.
to the angle DBE; therefore the angle DEB is greater than
the angle DBE: But to the greater angle the greater side * 19. 1. is opposited; DB is therefore greater than DE: But DB
is equal to DF; wherefore DF is greater than DE, the less
be demonstrated that it does not fall upon the circumference; it falls therefore within it, Wherefore, if any two points, &c. Q.E.D.
PROP. III. THEOR.
* 1. 3.
If a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre, it shall cut it at right angles; and if it cuts it at right angles, it shall bisect it.
Let' ABC be a circle ; and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre, in the point F: It cuts it also at right angles.
Takea Ę the centre of the circle, and join EA, EB. Then, because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other, and the base
angle BFE: But when a straight
fore each of the angles AFE, BFE, is
В line CD, drawn through the centre,
But let CD cut AB at right angles; CD also bisects it, that is, AF is equal to FB.
The same construction being made, because EA, EB,
from the centre are equal to one another, the angle EAF d 5. 1. is equald to the angle EBF: and the right angle "AFÉ is
equal to the right angle BFE: Therefore, in the two tri