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Of the Circle.

THEOREM I.

Every chord is less than a diameter.

Let AD be any chord. Draw the radii CA, CD to its extremities. We shall then have, AD less than AC+ CD (Book I. Th. x*).

But

AC+CD is equal to the diameter

AB: hence, the chord AD is less than the diameter.

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THEOREM II.

If from the centre of a circle a line be drawn to the middle of a chord,

I. It will be perpendicular to the chord; II. And it will bisect the arc of the chord. Let C be the centre of a circle, and AB any chord. Draw CD through D, the middle point of the chord, and produce it to E: then will CD be perpendicular to the chord, and the arc AE equal to EB.

First. Draw the two radii CA, CB Then the two triangles ACD DOS,

B

E

have the three sides of the one equal to toe three sides of the

*Note. When reference is made from one theorem to another, in the same Book, the number of the theorem referred to is a me given but when the theorein referred to is found in a preceding Pk the name of e Book is also given.

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Of the Circle.

other, each to each: viz. AC equal to CB, being radii, AD equal to DB, by hypothesis, and CD common: hence, the corresponding angles are equal (Book I. Th. viii): that is, the angle CDA equal to CDB, and the angle ACD equal to the angle DCB.

But, since the angle CDA is equal

B

E

to the angle CDB, the radius CE is perpendicular to the chord AB (Bk. I. Def. 20).

Secondly. Since the angle ACE is equal to BCE, the arc AE will be equal to the arc EB, for equal angles must have equal measures (Bk. I. Def. 29).

Hence, the radius drawn through the middle point of a chord, is perpendicular to the chord, and bisects the arc of the chord.

Cor. Hence, a line which bisects a chord at right angles, bisects the arc of the chord, and passes through the centre of the circle. Also, a line drawn through the centre of the circle and perpendicular to the chord, bisects it.

THEOREM III.

If more than two equal lines can be drawn from any point within a circle to the circumference, that point will be the centre.

Let D be any point within the circle ABC. Then, if the three lines DA, DB, and DC, drawn from the point D to the circumference, are equal, the point D will be the centre.

For, draw the chords AB, BC, bisect them at the points E and F, and join DE and DF

B

Of the Circle.

Then, since the two triangles DAE and DEB have the side AE equal to EB, AD equal to DB, and DE common, they will be equal in all respects; and consequently, the angle DEA is equal to the angle DEB (Bk. I. Th. viii); and therefore, DE is perpendicular to AB (Bk. I. Def. 20) But, if DE bisects AB at right angles, it will pass through the centre of the circle (Th. ii. Cor).

In like manner, it may be shown that DF passes through the centre of the circle, and since the centre is found in the two lines ED, DF, it will be found at their common intersection D.

THEOREM IV.

Any chords which are equally distant from the centre of a circle,

are equal.

Let AB and ED be two chords equally distant from the centre C: then will the two chords AB, ED be equal to each other.

Draw CF perpendicular to AB, and CG perpendicular to ED, and since these perpendiculars measure the distances from the centre, they will be equal. Also draw CB and CE.

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Then, the two right angled triangles CFB and CEG hav ing the hypothenuse CB equal to the hypothenuse CE, and the side CF equal to CG, will have the third side BF equal to EG (Bk. I Th. xix). But, BF is the half of BA, and EG he half of DE (Th. ii. Cor); bence BA is equal to DE (Ax 6).

Of the Circle.

THEOREM V.

A line which is perpendicular to a radius at its extremity, is tangent to the circle.

Let the line ABD be perpendicular to the radius CB at the extremity B: then will it be tangent to the circle at the point B.

For, from any other point of the line, as D, draw DFC to the centre, cutting the circumference in F.

Then, because the angle B, of the

A

triangle CDB, is a right angle, the angle at D is acute (Bk. I. Th. xvii. Cor. 3), and consequently less than the angle B. But the greater side of every triangle is opposite to the greater angle (Bk. I. Th. xi); therefore, the side CD is greater than CB, or its equal CF. Hence, the point D is without the circle, and the same may be shown for every other point of the line AD. Consequently, the line ABD has but one point in common with the circumference of the circle, and therefore is tangent to it at the point B (Def. 13)

Cor. Hence, if a line is tangent to a circle, and a radius be drawn through the point of contact, the radius will be perpen dicular to the tangent.

THEOREM VI.

If the distance between the centres of two circles is equal the sum of their radii, the two circles will touch each other externally.

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