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E L E M E N T S
G E O M E T R Y.
Β Ο Ο Κ ΙΙΙ. .
A. *HE radius of a circle is the straight line drawn from Book III. the centre to the circumference.
1. A straight line is said to
touch a circle, when it
another, which meet, but
stant from the centre of a circle,
greater perpendicular falls, is said to
tained by a straight line, and the
contained by two straight lines
upon the arch intercepted between
Let ABC be the given circle ; it is required to find its centre.
a 10. 1.
· Draw within it any straight line AB, and bisect a it in D; Book III. from the point D drawb DC at right angles to AB, and produce it to E, and bisect CE in F: The point F is the centre d u. i. of the circle ABC.
For, if it be not, let, if pollible, G be the centre, and join GA, GD, GB: Then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two fides 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 radii of the same circle : therefore the angle ADG is equal c
c 8. 1. to the angle GDB : But when a straight line standing upon another straight line makes the adjacent angles equal to one another, each
В of the angles is a right angle d.
d 7. def. s. Therefore the angle GDB is a right
I angle : 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 which bifects the other,
F any two points be taken in the circumference of
a circle, the straight line which joins them Thall fall within the circle.
Let ABC be a circle, and A, B any two points in the ciscumference; the straight line drawn
Take any point in AB as E; find
B В a s. I. angle DAB is equal to the angle
the triangle DAE, is produced to B, the angle DEB is b 16. I, greater b than the angle DAE ; but DAE is equal to the
angle DBE ; therefore the angle DEB is greater than the
angle DBE: Now to the greater angle the greater fide is opc 19. 1. polited; DB is therefore greater than DE: but BD is equal
to DF; wherefore DF is greater than DE, and the point E is therefore within the circle. The same may be demonstrated of any other point between A and B, therefore AB is within the circle. Wherefore, if any two points, , &c. Q. E. D.'
F a straight line drawn through the centre of a
circle bitect a straight line in the circle, which does not pass through the centre, it will cut that line at right angles ; and, if it cut it at right angles, it will bifect 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.
Take a E the centre of the circle, and join EA, EB. Then, because AF is equal to FB, and FE common to the two
a 1. 3.