greater than the angle DBE: But to the greater angle the greater side is opposite; therefore DB is greater than DE: But DB is equal to DF; therefore DF is greater than DE, the less than the greater-which is absurd; therefore the straight line drawn from A to B does not fall without the circle: And, in like 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.

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.

E

F

C

B

Take E the centre of the circle, and join EA, EB: Then, because AF is equal to BF, and FE common to the two triangles AFE, BFE, the two sides AF, FE, of the one are equal to the two, BF, FE, of the other, each to each; and the base AE is equal to the base BE -therefore the angle AFE is equal to the angle BFE: 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; therefore each of the angles AFE, BFE is a right angle; and therefore the straight line CD, drawn through the centre, bisecting another AB that does not pass through the centre, cuts the same at right angles.

D

Next, 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, drawn from the centre, are equal to one another, the angle EAF is equal to the angle EBF: And the right angle AFE is equal to the right angle BFE; 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-therefore the other sides are equal, and 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 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 centre: AC, BD, shall not bisect each other.

A

B

F

E

C

If one of the lines pass through the centre, it is plain that it cannot be bisected by the other which does not pass through the centre: But if neither of them pass through the centre, let, if possible, AE be equal to EC, and BE to ED; and take F the centre of the circle, and join FE: Then, because FE, a straight line through the centre, bisects another AC which does not pass through the centre, it cuts it at right angles (3. 3); therefore FEA is a right angle: Again, because the straight line FE bisects the straight line BD, which does not pass through the

centre, it cuts it at right angles; therefore also FEB is a right angle: but FEA was shown to be a right angle; therefore the angle FEA is equal to the angle FEB, the less to the greater-which is absurd: Therefore AC, BD do not bisect one another.

If two circles cut one another, they shall not have the same

centre.

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

A

C

G

E

B

For, if possible, let E be their centre; join EC, and draw any straight line EFG cutting the circles in F and G: Then, because E is the centre of the circle ABC, EC is equal to EF: Again, because E is the centre of the circle CDG, EC is equal to EG: But EC was shown to be equal to EF; therefore EF is equal to EG, the less to the greater-which is absurd: Therefore E is not the centre of the circles ABC, CDG.

If two circles touch one another internally, they shall not have the same centre.

Let the two circles ABC, CDE, touch one another internally in the point C: they shall not have the same

centre.

B

F E

D

For, if possible, let F be their centre: join FC, and draw any straight line FEB, cutting the circles in E and B: Then, because F is the centre of the circle ABC, FC is equal to FB: Again, because F is the centre of the circle CDE, FC is equal to FE: But FC was shewn to be equal to FB; therefore FE is equal to FB, the less to the greater-which is absurd: therefore F is not the centre of the circles ABC, CDE.

Wherefore, If two circles &c. Q. E.D.

A

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

Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre, and let the centre be E: of all the straight lines FB, FC, FG, &c., that can be drawn from F to the circumference, FA, which passes through E, 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 than FG.

Join BE, CE, GE: Then, because any two sides of a triangle are greater than the third, therefore BE, EF are to

D H

gether greater than BF: but BE is equal to AE; therefore AE, EF, that is AF, is greater than BF: Again, because BE is c equal to CE, and EF common to the triangles BEF, CEF, the two sides BE, EF are equal to the two CE, EF: but the angle BEF is greater than the angle CEF; therefore (1. 24) the base BF is greater than the base CF : And, in like manner, it may be shewn that CF is greater than GF: Again, because EF, FG are greater than EG, and that EG is equal to ED, therefore EF, FG are greater than ED; take away the common part EF, and the remainder FG is greater than the remainder FD: Therefore AF is the greatest, and FD the least, of all the straight lines from F to the circumference, and FB is greater than FC, and FC than FG.

Also there can be drawn two equal straight lines from the point F to the circumference, one upon each side of the shortest line FD: For, at the point E, in the straight line EF, make the angle FEH equal to the angle FEG, and join FH: Then, because GE is equal to HE, and EF common to the two triangles GEF, HEF, the two sides GE, EF are equal to the two HE, EF, each to each, and the angle GEF is equal to the angle HEF— therefore the base FG is equal to the base FH.

But, besides FH, no other straight line can be drawn from F to the circumference, equal to FG: For, if there can, let it be FK: Then, because FK is equal to FG, and FG to FH, therefore also FK is equal to FH, that is, a line nearer to that which passes through the centre, is equal to one which is more remote-which is impossible.