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D.

But the greater side of every triangle is opposite to the greater angle (I. 17.); whence ce, or its equal ce, will be greater than cd.

The point d, therefore, falls within the circle ; and the same may be shewn of any other point in AB; consequently the whole line AB must fall within the circle.

Q. E, D,

PROP. III. THEOREM.

If a right line, which passes through the centre of a circle, bifect a chord, it will be perpendicular to it; and if it be perpendicular to the chord, it will bisect it.

B

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Let ABC be a circle, and ce a right line which passes through the centre D, and bisects the chord AB in £; then will ce be perpendicular to AB.

For join the points AD, DB:

Then, because Ad is equal to DB (II. Def. 13.), AE TO EB (by Hyp.), and ed common to each of the triangles ADE, BDE, the angle DEA will be equal to the angle DEB (I. 7.)

But one line is said to be perpendicular to another, when it makes the angles on both sides of it equal to each other (I. Def.8.); confequently ce is perpendicular to the chord AB.

Again,

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Again, let the right line de be drawn from the centre D, perpendicular to the chord AB; then will AB be bilected in the point E.

For join the points AD, DB, as before :

Then, since the angle DAB is equal to the angle DBA (1. 5.), and the angle AED to the angle Dęb, (being each of them right angles) the angle ADE will also be equal to the angle EDB (I. 28. Cor. 1.)

And, because the triangles DEA, DEB are mutually equiangular, and have the fide de common, the side AE will also be equal to the side EB (I. 21.); whence AB is bisected in the point E, as was to be shewn.

COROLL. If a right line be drawn from the vertex of an isosceles triangle, to the middle of the base, it will be perpendicular to it; and if it be perpendicular to the base, it will bisect both it and the vertical angle.

PRO P. IV. THEORE M.

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

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Let ABDC be a circle, and o a point within it; then if any three right lines OA, OB, oc, drawn from the point o to the circumference, be equal to each other, that point will be the centre.

For draw the lines AB, AC, and bisect them in the points F, G (I. 10.); and through the centre o, draw FD, Ge, cutting the circumference in D and E.

Then, since af is equal to FB (by Conft.), AO tO OB (by Hyp.), and of common to each of the triangles AOF, BOF, the angle AFO will be equal to the angle bro (I. 7.)

And because the right line of falls upon the right line AB, and makes the adjacent angles equal to each other, DF will be perpendicular to AB (I. Def. 8.)

But when a right line bisects any chord at right angles, it passes through the centre of the circle (III. 1. Cor.); whence the centre must be somewhere in the line FD.

And, in the fame manner, it may be thewn, that the centre must be somewhere in the line ge.

But the lines FD, GE have no other point but o which is common to them both; therefore o is the centre of the circle ABD,

as was to be shewn.

PRO P. V. THEOR E M.

Circles of equal radii are equal to each other; and if the circles are equal, the radii will be equal.

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Let ABC, der be two circles, of which the radii GA, GB are equal to the radii HF, ME; then will the circle ABC be equal to the circle Def.

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For conceive the circle DEF to be applied to the circle ABC, so that the centre h may coincide with the centre

Then, since the radii HF, HE are equal to the radii GA, GB (by Hyp.), the points F, E will fall in the circumference of the circle ABC (I. Def. 13.); and the same may be shewn of any other point D.

But since any number of points, taken in the circumference of the circle DEF, fall in the circumference of the circle ABC, the two circumferences must coincide, and consequently the circles are equal to each other.

Again, let the circle ABC be equal to the circle def; then will the radii GA, GB be equal to the radii HF, HE.

For if they be not equal, they must be either greater or less : let them be greater; and apply the circles te each other as before.

Then, since the radii GA, GB are greater than the radii HF, HE, the points F, E will fall within the circle ABC; and the same may be shewn of any other point D.

But, since any number of points, taken in the circumference of the circle DEF, fall within the circle ABC, the whole circle DEF muft, also, fall within the circle ABC.

The circle Der is, therefore, less than the circle ABC, and equal to it at the same time (by Hyp.), which is absurd: whence the radii GA, GB are not greater than the radii HF, HE.

And in the same manner it may be sewn that they cannot be less; confequently they are equal to each other.

Q. E. D. COROLL. Equal circles, or such as have equal radii, or diameters, have equal circumferences.

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PRO P. VI. THEOREM.

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If two circles touch each other internally, the centres of the circles and the point of contact will be all in the same right line.

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Let the two circles beg, BDF touch each other inter-
nally at the point B; then will the centres of those cir.
cles and the point o be in the same right line.

For let a be the centre of the circle Beg, and draw
The diameter GB.

And if the centre of the circle BDF be not in GB, let,
if possible, some point c, out of that line, be the centre;
and join A, C, C,B; and produce ac to cut the circles
in D and E.

Then, since ACB is a triangle, the sides AC, CB, taken together, are greater than the side AB (I. 18.), or its equal AE.

And if, from these equals, the part AC, which is common, be taken away, the remainder CB will be greater than the remainder CE.

But, fince c is the centre of the circle BDF (by Hyp.), CB is equal to CD (I. Def. 13.); whence cd will also be greater than ce, which is imposible.

The point c, therefore, cannot be the centre of the
circle bdf; and the same may be thewn of any other
point out of the line AB.

Q.E.D.
G

PROP.

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