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other point but F is the centre; that is, F is the centre of the circle ABC.

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.

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

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Take any point in AB as E; find D the centre of the circle ABC; join AD, DB, and DE, and let DE meet the circumference in F. Then, because DA is equal to DB, the angle DAB is equal to the angle DBA (I. 5); and because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater than the angle DAE (I. 16); but DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE; DB is therefore greater than DE (I. 19). But DB 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.

Schol. When the two points are at the extremities of a diameter, or diametrically opposite, the truth of the proposition is evident. It is obvious, from the nature of the circle, when they are at a less distance. When the points are exceedingly near, however, it is not then so evident that the chord lies wholly within the circle, and no part of it on the arc; and therefore, in this case, it is necessary to demonstrate the proposition. Some parts of the figure, however, in this extreme case, would be so small as to be indistinct, and therefore it is necessary to assume the points at a sufficient distance; and as the reasoning in this case applies to every case, the proposition will therefore be established also in the extreme case. A similar observation applies to se

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, and of bisecting a chord--if a line lfil the third. I 28 9863 90

hords cut one another which do not centre, they do not bisect each other. rele, and AC, BD, two chords in it, in the point E, and do not both pass C, BD, do not bisect one another. let AE be equal to EC, and BE to s pass through the centre, it is plain d by the other which

E

F

ets another AC which does not pass hall cut it at right angles (III. 3); t angle. Again, because the straight ight line BD, which does not pass hall cut it at right angles; wherefore and FEA was shown to be a right is equal to the angle FEB, the less impossible; therefore AC, BD, do 8 of lampe

THEORga ente

e another, they shall not have the

ABC, CDG, cut one another in the ve not the same centre.

e, let E be their centre; join EC, and e EFG meeting them

ause E is the centre of

is equal to EF. Again, Af tre of the circle CDG, but CE was shown to herefore EF is equal to

E

greater, which is impossible; therefore of the circles ABC, CDG.

veral propositions, which appear to be axiomatic, except in the extreme case, as I. 20.

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If a straight line drawn through the centre of a circle bisect a chord 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 chord AB, which does not pass through the centre, in the point F; it cuts it also at right angles.

Take E the centre of the circle (III. 1), 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; therefore the angle AFE is equal to the angle BFE (I. 8); therefore (I. Def. 10) 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 also bisects it, that is, AF is equal to FB.

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The same construction being made, because EA, EB, from the centre are equal to one another, the angle EAF is equal to the angle EBF (I. 5); and the right angle AFE is equal to the right angle BFE; therefore, in the two triangles EAF, EBF, there are two angles in one equal to two angles in the other, 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 (I. 26); AF therefore is equal to FB. COR.-If a line bisect a chord of a circle and be perpen

dicular to it, it will pass through the centre.

For a line from the centre bisecting the chord is perpendicular to it, and therefore coincides with the former perpendicular; therefore it must also pass through the centre.

Schol. It appears, therefore, that of the three conditions of passing through the centre of a circle, of being

perpendicular to a chord, and of bisecting a chord-if a line fulfil two, it will also fulfil the third.

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If in a circle two chords 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 chords in it, which cut one another in the point E, and do not both pass through the centre; AC, BD, do not bisect one another.

A

B E

F

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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 cannot be bisected by the other which does not pass through the centre. But if neither of them pass through the centre, take F the centre of the circle, and join EF (III. 1); and because FE, a straight line through the centre, bisects another AC which does not pass through the centre, it shall cut it 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 centre, it shall cut it at right angles; wherefore FEB is a right angle; and FEA was shown to be a right angle; therefore FEA is equal to the angle FEB, the less to the greater, which is impossible; therefore AC, BD, do not bisect one another.

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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 have not 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, CE is equal to EF. Again, A because E is the centre of the circle CDG, CE is equal to EG; but CE was shown to be equal to EF; therefore EF is equal to

D

EG, the less to the greater, which is impossible; therefore E is not the centre of the circles ABC, CDG.

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