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PROPOSITION XIV. THEOREM. Equal chords in a circle are equally distant from the centre; and conversely, those which are equally distant from the centre, are equal to one another.

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Let the chords AB, CD in the O ABDC be equal.
Then must AB and CD be equally distant from the centre 0.
Draw OP and OQ 1 to AB and CD; and join AO, CO.
Then P and Q are the middle pts. of AB and CD: III. 3.

and :: AB=CD, .. AP=CQ.
Then :: AP=CQ, and AO=CO,
in the right-angled As AOP, COQ,
.:: OP=OQ;

I. E. Cor. p. 43. and .. AB and CD are equally distant from 0. Def. 8. Next, let AB and CD be equally distant from 0.

Then must AB=CD.
For :: OP=OQ, and AO=CO,
in the right-angled As AOP, COQ,
.. AP=CQ,

I. E. Cor. and .. AB=CD.

Q. E. D.

Ex. In a circle, whose diameter is 10 inches, a chord is drawn, which is 8 inches long. If another chord be drawn, a a distance of 3 inches from the centre, shew whether it is equal or not to the former.

PROPOSITION XV. THEOREM.

The diameter is the greatest chord in a circle, and of all others that which is nearer to the centre is always greater than one more remote ; and the greater is nearer to the centre than the less.

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Let AB be a diameter of the O ABD whose centre is O, and let CD be any other chord, not a diameter, in the o, nearer to the centre than the chord EF. Then must AB be greater than CD, and CD greater than EF. Draw OP, OQ I to CD and EF; and join OC, OD, OE.

Then :: A0=CO, and OB=OD, I. Def. 13.

.: AB=sum of CO and OD, and .. AB is greater than CD.

I. 20. Again, :: CD is nearer to the centre than EF, .: OP is less than 0Q.

Def. 8. Now :- sq. on OC=sq. on OE, .: sum of sqq. on OP, PC=sum of sqq. on OQ, QE. I. 47.

But sq. on OP is less than sq. on OQ ;
.: sq. on PC is greater than sq. on QE;

.. PC is greater than QE ;
and .. CD is greater than EF.

Next, let CD be greater than EF.
Then must CD be nearer to the centre than EF.

For :: CD is greater than EF,

:: PC is greater than QE.

Now the sum of sqq. on OP, PC=sum of sqq. on OQ, QE.

But sq. on PC is greater than sq. on QE;
.. sq. on OP is less than sq. on OQ ;

.. OP is less than OQ; .
and .:. CD is nearer to the centre than EF.

Q. E. D. Ex. 1. Draw a chord of given length in a given circle, which shall be bisected by a given chord.

Ex. 2. If two isosceles triangles be of equal altitude, and the sides of one be equal to the sides of the other, shew that their bases must be equal.

Ex. 3. Any two chords of a circle, which cut a diameter in the same point and at equal angles, are equal to one another.

DEF. IX. A straight line is said to be a TANGENT to, or to touch, a circle, when it meets and, being produced, does not cut the circle.

From this definition it follows that the tangent meets the circle in one point only, for if it met the circle in two points it would cut the circle, since the line joining two points in the circumference is, being produced, a secant. (III. 2.)

DEF. X. If from any point in a circle a line be drawn at right angles to the tangent at that point, the line is called a Normal to the circle at that point.

DEF. XI. A rectilinear figure is said to be described about a circle, when each side of the figure touches the circle.

And the circle is said to be inscribed in the figure.

PROPOSITION XVI. THEOREM.

The straight line drawn at right angles to the diameter of a circle, from the extremity of it, is a tangent to the circle.

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Let ABC be a 0, of which the centre is 0, and the diameter AOB.

Through B draw DE at right angles to AOB. I. 11.

Then must DE be a tangent to the o.
Take any point P in DE, and join OP.

Then, :: ZOBP is a right angle,
.:LOPB is less than a right angle, I. 17.
and .. OP is greater than OB.

I. 19. Hence P is a point without the o ABC. Post.

In the same way it may be shewn that every point in DE, or DE produced in either direction, except the point B, lies without the O; .: DE is a tangent to the o.

Def. 9.

Q. E. D.

PROPOSITION XVII. PROBLEM. To draw a straight line from a given point, either WITHOUT or on the circumference, which shall touch a given circle.

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I. A.

I. A.

Let A be the given pt., without the o BCD.

Take O the centre of o BCD, and join 0A.
Bisect 0 A in E, and with centre E and radius EO describe
O ABOD, cutting the given o in B and D.
Join AB, AD. These are tangents to the o BCD.

Join BO, BE.
Then :: 0E=BE, ..ZOBE= 2 BOE ;
... AEB=twice 2 OBE ;

I. 32.
and ::: AE=BE, ..LABE=_BAE;
..LOEB=twice 2 ABE;

I. 32. .. sum of 2s AEB, OEB=twice sum of 2 S OBE, ABE, that is, two right angles=twice 2 OBA;

:: ZOBA is a right angle,

and .. AB is a tangent to the o BCD. III. 16. Similarly it may be shewn that AD is a tangent to o BCD. Next, let the given pt. be on the Oce of the o, as B.

Then, if BA be drawn I to the radius OB,
BA is a tangent to the o at B. III. 16.

Q. E. D. Ex. 1. Shew that the two tangents, drawn from a point without the circumference to a circle, are equal.

Ex. 2. If a quadrilateral ABCD be described about a circle, shew that the sum of AB and CD is equal to the sum of AD and BC.

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