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Book II to the square of GH: But the squares of HE, EG are equal to the square of GH: Therefore the rectangle BE, EF, together with the square of EG, is equal to the squares of HE, EG; Take away the square 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 figure A; therefore the rectilineal figure A is equal to the square of EH. Wherefore asquare has been made equal to the given rectilineal figure A, viz. the square described upon EH. Which was to be done.
EQUAL circles are those of which the diameters are equal, Book III. or from the centres of which the straight lines to the circumferences are equal.
'This is not 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 like'wise coincide, since the straight lines from the centres ' are equal.'
A straight line is said to touch a circle, when it meets the circle, and being produced does not
Circles are said to touch one another, which meet but do not cut one another.
Straight lines are said to be equally
And the straight line on which the
A segment of a circle is the figure
"The angle of a segment is that which is contained by the straight line and the circumference."
An angle in a segment is the angle
And an angle is said to insist orstand
upon the circumference intercept
ed between the straight lines that contain the angle.
See N. To find the centre of a given circle.
a 10. 1.
Let ABC be the given circle; it is required to find its
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 produce it to E, and bisect CE in F: The paint 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 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 GDB: But when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a
c 8. 1.
right angled: Therefore the angle GDB is a right angle: 410 Def. 1. 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 bisects the other.
PROP. II. THEOR.
Ir 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 shall fall within the circle.
For, if it do not, let it fall, if possible, without, as AEB; find a D the 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 equal to DB, the angle DAB is equal to the angle DBA; and because AE,
*N. B. Whenever the expression." straight lines from the centre "drawn from the centre" occurs, it is to be understood that they are drawn to the circumference.
a 1. 3.
b 5. 1.
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 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 than the greater, which is impossible: Therefore the straight line drawn from A to B does not fall without the circle. In the same manner, it may 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.
Ir 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.
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 triangles AFE, BFE, there are two sides in the one equal to two sides in the other, and the base EA is equal to the base EB; there8. 1. fore the angle AFE is equal to the angle BFE: But when a straight line standing upon another makes the adjacent angles equal to one another, 10 Def. 1. each of them is a right angle: Therefore each of the angles AFE, BFE, is a right angle; wherefore the straight line CD, drawn through the centre, bisecting another AB that does not
pass through the centre, cuts the same at right angles.
The same construction being made, because EA, EB, from the centre are equal to one another, the angle EAF 5. 1. is equal to the angle EBF: and the right angle AFE is equal to the right angle BFE: Therefore, in the two tri