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Let the two circles ABC, ADF touch one another internally at the point A; then will A and the centres of those circles be all in the same straight line.

Let G be the centre of the circle ABC, through which draw the diameter AGC; then, if the centre of the other circle can be

G

out of this line AC, let it be supposed at some other point H, through which draw the line GH, cutting the two circles in B and D join AH.

:

Now, in the triangle AGH, the sum of the two sides GH, HA is greater than the third side AG (B. I, Prop. XVI), or greater than its equal, the radius GB. From each of these take away the common part GH, and the remainder HA will be greater than the remainder HB. But the point H being supposed the centre of the inner circle, its two radii HA, HD are equal to each other; conse quently HD will also be greater than HB.

Again, ADF being the inner circle, HD is necessarily less than HB; so that HD is both greater and less than HB, which is absurd. To remove this absurdity, we must abandon the supposition which led to it, which was that the centre H might be out of the line AGC. Consequently the centre H cannot be out of the line AGC; that is, the the line GH, joining the centres, passes through A the point of contact.

PROPOSITION III.

THEOREM. If two circles touch one another externally, the centres of the circles and the point of contact will be all in the same straight line.

Let the two circles ABC, ADF touch one another externally at the point A; then will the point of contact A, and the centres of the circles, be all in the same straight line.

Let G be the centre of the circle ABC, through which draw the diameter AGC, and produce it to meet the other circle at the point F. Then, if the centre of the other circle ADF can be out of the line CF, let it, if possible, be supposed to be at some other point as H; and draw the lines AH, GBDH, cutting the two circles in B and D.

A B

G

H

Then, in the triangle AGH, the sum of the two sides AG, AH is greater than the third side GH (B. I, Prop. XVI). But since G and H are the centres of the two circles, the two radii HA, HD are equal; so also are the two radii GA, GB; hence the sum of AG, AH is equal to the sum of GB, HD. We have just shown that the sum of AG, AH is greater than the distance GH; consequently the sum of GB, HD is greater than GH, which is absurd. Consequently, as in the last proposition, the centre H cannot be out of the line FC.

PROPOSITION IV.

THEOREM. Chords in a circle, which are equally distant from the centre, are equal to each other. Conversely, if chords in the same circle are equal to each other, they will be equally distant from the centre.

Let AB, CD be any two chords equally distant from the centre H; then will these two chords be equal to each other.

Draw the two radii HA, HC, and the two perpendiculars HF, HG, which are the equal distances of the

B

G

H

chords from the centre H. Then the two right-angled triangles HAF, HCG have the sides HA and HC equal (being radii), the side HF equal to HG, and the angle HFA equal to the angle HGC, each being a right angle; therefore those two triangles are equal (B. II, Prop. vIII, Cor. 2), and consequently AF is equal to CG. But AB is the double of AF, and CD the double of CG (B. III, Prop. 1); therefore AB is equal to CD (Ax. VI).

Conversely, if the chord AB is equal to the chord CD, then will their distances from the centre, HF, HG, be equal to each other.

For, since AB is equal to CD, we must have AF the half of AB equal to CG the half of CD. We also have the radii HA, HC equal, and the angle HFA equal to the angle HGC, each being a right angle; therefore the triangle HAF is equal to the triangle HCG (B. II, Prop. vIII, Cor. 2), and consequently HF is equal to HG.

PROPOSITION V.

THEOREM. A line perpendicular to a radius, at its extremity, is tangent to the circumference.

Let the line ADB be perpendicular to the radius CD at its extremity; then will AB touch the circumference at the point D only.

From any other point, as F, in the line AB, draw FGC to the centre, cutting the circle at the point G. Then because the angle

A

F

B

CDF of the triangle CDF is a right angle, the angle DFC is acute (B. I, Prop. xxiv, Cor. 1), and consequently less than the angle CDF. But the longer side is always opposite the greater angle (B. I, Prop. xv); therefore the side CF is longer than the side CD, or its equal CG. Hence the point F is without the circle; and the same may be shown of any other point of the line AB, except the point D. Consequently the line AB touches the circumference at only one point, and is therefore tangent to it.

PROPOSITION VI.

THEOREM. When a line is tangent to the circumference of a circle, a radius drawn to the point of contact is perpendicular to the tangent.

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line, as CF, drawn from the centre C to meet the line AB at any point different from D, must have its extremity F without the circumference. Hence the radius CD is the shortest line that can be drawn from the centre to meet the tangent AB, and therefore CD is perpendicular to AB (B. I, Prop. xxvi).

Cor. Hence, conversely, a line drawn perpendicular to a tangent at the point of contact, passes through the centre of the circle.

PROPOSITION VII.

THEOREM. The angle formed by a tangent and chord is

measured by half the arc of that chord.

Let AB be a tangent, and CD A

a chord drawn from the point of contact C; then the angle BCD will be measured by half the arc CGD, and the angle ACD will be measured by half the arc CLD.

Draw the radius FC to the

B

G

point of contact, and the radius FG perpendicular to the chord CD, meeting it at the point K. Then the radius FG, being perpendicular to the chord CD, bisects the arc CGD (B. III, Prop. 1); therefore CG is half the arc CGD.

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