For conceive the circle DEF to be applied to the circle ABC, fo that the centre н may coincide with the centre G.

Then, fince 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 fame may be fhewn of any other point D.

But fince any number of points, taken in the circumference of the circle DEF, fall in the circumference of the circle ABC, the two circumferences muft coincide, and confequently 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 lefs: let them be greater; and apply the circles to each other as before.

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

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

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

And in the fame manner it may be fhewn that they cannot be lefs; confequently they are equal to each other.

Q.E.D.

COROLL. Equal circles, or fuch as have equal radii,

or diameters, have equal circumferences.

PROP. VI. THEOREM.

1

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

B

Let the two circles BEG, BDF touch each other internally at the point B; then will the centres of those circles and the point в be in the fame 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 poffible, fome 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, fince ACB is a triangle, the sides AC, CB, taken together, are greater than the fide AB (I. 18.), or its equal AE.

And if, from thefe 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 impoffible.

The point c, therefore, cannot be the centre of the circle BDF; and the fame may be fhewn of any other point out of the line AB.

Ꮐ

Q. E. D.

PROP.

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

ӨӨ

Let the two circles BEG, BDF touch each other externally at the point B; then will the centres of those circles and the point B, be in the fame right line.

For, let A be the centre of the circle BEG, and draw the diameter GB, which produce till it cuts the circle BDF in F.

And, if the centre of the circle BDF be not in the line AF, let, if poffible, fome point c, out of that line, be the centre; and join C, A, C, B.

Then, fince A is the centre of the circle BEG, AE is equal to AB (1. Def. 13.)

And because c is the centre of the circle BDF (by Hyp.) CD is equal to CB (I. Def. 13.)

But AB, BC, together, are greater than AC (I. 18.); therefore AE, CD, together, are alfo greater than AC; which is abfurd.

The point c, therefore, cannot be the centre of the circle BDF; and the fame may be fhewn of any other point out of the line AF.

Q. E.D.

PRO P. VIII. THEOREM.

Any two chords in a circle, which are equally diftant from the centre, are equal to each other; and if they be equal to each other, they will be equally diftant from the

centre.

Let ABED be a circle, whofe centre is o; then will any two chords AB, DE, which are equally distant from o, be equal to each other.

For join the points AO, OD, and let fall the perpendiculars OC, OF (I. 12.)

Then, fince a right line, drawn from the centre of a circle, at right angles to any chord, bifects it (III. 3.), AC will be equal to CB, and DF to FE.

And, because the angles ACO, DFO are right angles, the fquares of AC, CO will be equal to the fquare of AO (II. 14.), and the fquares of DF, FO to the fquare of OD.

But the fquare of AO is equal to the fquare of OD (II. 2.); confequently the fquares of Ac, co will be equal to the fquares of DF, FO.

And fince oc is equal to or (III. Def. 8.), the fquare of oc will be equal to the fquare of oF (II.2.); whence

the remaining square of AC will also be equal to the remaining fquare of DF; or AC equal to Dr (II. 3.), and AB to DE (I. Ax. 6.)

Again, let the chord AB be equal to the chord DE; then will oc, or, or their diftances from the centre, be equal to each other.

For the fquares of AC, CO are equal to the square of OA (II. 14.), and the squares of DF, FO to the fquare of OD.

But the fquare of OA is equal to the fquare of OD (II. 2.); therefore the fquares of AC, CO are equal to the fquares of DF, FO.

And fince AC is the half of AB (III. 3.), and DF is the half of DE (III. 3.), the square of AC is equal to the fquare of DF (II. 2.)

The remaining square of co is, therefore, equal to the remaining square of Fo; and confequently co is equal to FO (II. 3.), as was to be fhewn.

COROLL. If two right angled triangles, having equal hypotenuses, have two other fides alfo equal, the remaining fides will likewise be equal, and the triangles will be equal in all respects.