« PreviousContinue »
the same manner it can be shown that no other point out of the line CE is the centre; and since CE is bisected in F, any other point in CE divides it into unequal parts, and cannot be the centre; therefore no point but 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.
of a circle, the straight line which joins them
DF, and produce it to E: then +15Def.
because DA is equal t to DB, the angle DAB is
equal * to the angle DBA : and because AE, a *5.1.
side of the triangle DAE, is produced to B, the +16. 1. exterior angle DEB is greatert than the angle
but DAE was proved equal to the angle , DBE; therefore the angle DEB is greater than
the angle DBE; but to the greater angle the * 19. 1. greater side is * opposite, therefore DB is greater +15Def. 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,
strated that it does not fall upon the circumference; therefore it falls within it. Wherefore, if any two points, &c. Q. E. D.
PROP. III. THEOR.
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 shall cut AB at right angles.
Take * E the centre of the circle, and join EA, *1.3.. EB. Then, because AF is equalf to FB, and † Hype FE common to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other, each to each; and the base EA is equal to the base EB; therefore the angle AFE is equal * to the angle BFE: E but when a straight line standing upon another straight line makes
B the adjacent angles equal to one another, 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. But let CD cut AB at right angles;
CD shall also bisect it, that is, AF shall be equal to FB.
The same construction being made, because EA, EB, from the centre are equal f to one an- 1.
†15 Def. other, the angle EAF is equal * to the angle *5. 1; EBF; and the right angle AFE is equal t to ļ0 Def the right angle BFE: therefore, in the two tri
angles, EAF, EBF, there are two angles in the one equal to two angles in the other,each to each; and the side EF, which is opposite to one of the
equal angles in each, is common to both; there26. 1. fore the other sides are* equal; therefore AF
is equal to FB. Wherefore, if a straight line, &c. Q. E. D.
PROP. IV. THEOR.
If in a circle two straight lines cut one another,
which do not both pass through the centre, they do not bisect euch other.
Let ABCD be a circle, and AC, BD two straight lines in it which cut one another in the point E, and do not both pass through the centre: AC, BD, shall not bisect one another.
For, if it is possible, let AE be equal to EC, and BE to ED. If one of the lines pass through the centre, it is plain that it carnot be bisected by the other which does not pass through the centre: but if neither of them pass through
through the centre, bisects anHyp. other+ AC which does not pass through the 3.3. centre, it cuts it at right* angles : wherefore
FEA is a right angle: again, because the straight Hyp. line FE bisectst the straight line BD, which
does not pass through the centre, it cuts it at right * angles: wherefore FEB is a right
angle: but FEA was shown to be a right angle; | Ax. therefore FEA is equal + to the angle FEB, the
less to the greater, which is impossible : Therefore AC, BD do not bisect one another. Wherefore, if in a circle, &c. Q. E. D.
PROP. V. THEOR.
the same centre.
For, if it be possible, let E be their centre : join EC, and draw any straight line EFG meeting them in F and G: and because E is the centre of the circle ABC, EC is equal † to EF:
A again, because E is the centre of D G 1. the circle CDG, EC is equal t
E to EG: but EC was shown to be equal to EF; therefore EF is equalt to EG, the less to
+1 the greater, which is impossible.
Therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q.E.D.
†15 Def, 1.
PROP. VI. THEOR.
shall not have the same centre.
For, if they have, let it be F: join FC and draw any straight line FEB, meeting them in E and B: and because F is the centre of the circle ABC, FC is equal+ A
B +15 Def. to FB; also, because Fis the centre of the circle CDE, FC
D is equal t to FE : but FC was shown to be equal to FB; therefore FE is equal to FB, the less to the greater, which is †1 Ax.
impossible: therefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, &c. Q. E. D.
PROP. VII. THEOR,
which is not the centre, of all the straight lines
Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre: let the centre be E: of all the straight lines FB, FC, FG, &c. that can be drawn from F to the circumference, FA shall be the greatest, and FD, the other part of the diameter AD, shall be the least: and of the others, FB shall be greater than FC, and FC greater than FG.
Join BE, CE, GE: and because two sides of a triangle are greater* than the third, therefore
BE, EF are greater than BF: but AE is 715 Def. equalf to BE ; therefore AE, EF, that is, AF is greater than
K +9Ax.i. the angle BEF is greater t than GDH
the angle CEF; therefore the * 24. 1. base BF is greater * than the base CF: for the
same reason CF is greater than GF: again, be* 20.1. cause GF, FE are greater* than EG, and EG
* 20. 1.