PROPOSITION I. PROBLEM.

To find the center of a given circle.

Let ABC be the given circle: it is required to find its center.

Draw within it any straight line AB to meet the circumference in A, B; and bisect AB in D; (1. 10.) from the point D draw DC at right angles to AB, (I. 11.) meeting the circumference in C, produce CD to E to meet the circumference again in E, and bisect CE in F.

Then the point Fshall be the center of the circle ABC. For, if it be not, if possible, let G be the center, and join GA, GD, GB. Then, because DA is equal to DB, (constr.)

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, (I. def. 15.) because they are drawn from the center G:

therefore the angle ADG is equal to the angle GDB: (1. 8.) but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle; (1. def. 10.)

therefore the angle GDB is a right angle:

but FDB is likewise a right angle; (constr.)

wherefore the angle FDB is equal to the angle GDB, (ax. 1.) the greater angle equal to the less, which is impossible; therefore G is not the center of the circle ABC.

In the same manner it can be shewn that no other point out of the line CE is the center;

and since CE is bisected in F,

any other point in CE divides CE into unequal parts, and cannot be the center.

Therefore no point but F is the center of the circle ABC.

Which was to be found.

COR. From this it is manifest, that if in a circle a straight line bisects another at right angles, the center of the circle is in the line which bisects the other.

PROPOSITION II. THEOREM.

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. Then the straight line drawn from A to B shall fall within the circle.

For if AB do not fall within the circle,

let it fall, if possible, without the circle as AEB; find D the center of the circle ABC, (III. 1.) and join DA, DB; in the circumference AB take any point F, join DF, and produce it to meet AB in E. Then, because DA is equal to DB, (1. def. 15.) therefore the angle DBA is equal to the angle DAB; (1. 5.) and because AE, a side of the triangle DAE, is produced to B, the exterior angle DEB is greater than the interior and opposite angle DAE; (I. 16.)

but DAE was proved to be equal to the angle DBE;
therefore the angle DEB is greater than the angle DBE;
but to the greater angle the greater side is opposite, (1. 19.)
therefore DB is greater than DE:

but DB is equal to DF; (1. def. 15.)
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;

therefore it falls within it.

Wherefore, if any two points, &c. Q. E.D.

If a straight line drawn through the center of a circle bisect a straight line in it which does not pass through the center, it shall cut it at right angles: and conversely, if it cut it at right angles, it shall bisect it.

Let ABC be a circle; and let CD, a straight line drawn through the center, bisect any straight line AB, which does not pass through the center, in the point F.

Then CD shall cut AB at right angles.

Take E the center of the circle, (III. 1.) and join EA, EB.
Then, because AF is equal to FB, (hyp.)

and 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; (1. def. 15.) therefore the angle AFÈ is equal to the angle BFE; (1. 8.) but when a straight line standing upon another straight line makes the adjacent angles equal to one another,

each of them is a right angle; (1. def. 10.)

therefore each of the angles AFE, BFE, is a right angle: wherefore the straight line CD, drawn through the center, bisecting another AB that does not pass through the center, cuts the same at right angles.

Conversely, let CD cut AB at right angles.

Then CD shall also bisect AB, that is, AF shall be equal to FB. The same construction being made,

because, EB, EA, from the center are equal to one another, (1. def. 15.)

therefore the angle EAF is equal to the angle EBF; (1.5) and the right angle AFE is equal to the right angle BFE; (1. def. 10.) therefore, in the two triangles, 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;

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therefore the other sides are equal; (1. 26.)
therefore AF is equal to FB.
Wherefore, if a straight line, &c. Q.E.D.

If in a circle two straight lines cut one another, which do not both pass through the center, they do not bisect each 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 center. Then AC, BD, shall not bisect one another.

For, if it be possible, let AE be equal to EC, and BE to ED.
If one of the lines pass through the center,

it is plain that it cannot be bisected by the other which does not
pass through the center:

but if neither of them pass through the center,

find F the center of the circle, (III. 1.) and join EF.

Then because FE, a straight line drawn through the center, bisects another AC which does not pass through the center, (hyp.) therefore FE cuts AC at right angles: (III. 3.) wherefore FEA is a right angle.

Again, because the straight line FE bisects the straight line BD, which does not pass through the center, (hyp.)

therefore FE cuts BD at right angles: (III. 3.)
wherefore FEB is a right angle:

but FEA was shewn to be a right angle;

therefore the angle FEA is equal to the angle FEB. (ax. 1.) the less equal to the greater, which is impossible: therefore AC, BD do not bisect one another. Wherefore, if in a circle, &c. Q. E.D.

PROPOSITION V. THEOREM.

If two circles cut one another, they shall not have the same center. Let the two circles ABC, CDG, cut one another in the points B, C. They shall not have the same center.

If possible, let E be the center of the two circles; join EC, and draw any straight line EFG meeting the circumferences in Fand G. And because E is the center of the circle ABC, therefore EF is equal to EC: (1. def. 15.) again, because E is the center of the circle CDG, therefore EG is equal to EC: (1. def. 15.) but EF was shewn to be equal to EC; therefore EF is equal to EG, (ax. 1.)

the less line equal to the greater, which is impossible. Therefore E is not the center of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E.D.

PROPOSITION VI. THEOREM.

If one circle touch another internally, they shall not have the same center. Let the circle CDE touch the circle ABC internally in the point C. They shall not have the same center.

If possible, let F be the center of the two circles: join FC, and draw any straight line FEB, meeting the circumferences in E and B. And because F is the center of the circle ABC, FB is equal to FC; (1. def. 15.)

also, because F is the center of the circle CDE,
FE is equal to FC: (1. def. 15.)
but FB was shewn to be equal to FC;
therefore FE is equal to FB, (ax. 1.)

the less line equal to the greater, which is impossible:
therefore Fis not the center of the circles ABC, CDE.

Therefore, if two circles, &c.

Q. E.D.

PROPOSITION VII. THEOREM.

If any point be taken in the diameter of a circle which is not the center, of all the straight lines which can be drawn from it to the circumference, the greatest is that in which the center is, and the other part of that diameter is the least; and, of the rest, that which is nearer to the line which passes through the center is always greater than one more remote: and from the same point there can be drawn only two equal straight lines to the circumference one upon each side of the diameter.

Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the center:

let the center be E.

Then, of all the straight lines FB, FC, FG &c. that can be drawn from F to the circumference,

FA, that in which the center is, shall be the greatest,

and FD, the other part of the diameter AD, shall be the least: and of the rest, FB, the nearer to FA, shall be greater than FC the more remote, and FC greater than FG.

Because two sides of a triangle are greater than the third side, (1. 20.) therefore BE, EF are greater than BF: but AE is equal to BE; (1. def. 15.) therefore AE, EF, that is, AF is greater than BF. Again, because BE is equal to CE,

and FE common to the triangles BEF, CEF,

the two sides BE, EF are equal to the two CE, EF, each to each; but the angle BEF is greater than the angle CEF; (ax. 9.) therefore the base BF is greater than the base CF. (1. 24.) For the same reason CF'is greater than GF. Again, because GF, FE are greater than EG, (1. 20.) and EG is equal to ED;

therefore GF, FE are greater than ED:

take away the common part FE,

and the remainder GF is greater than the remainder FD. (ax. 5.)