PROP. P.. PROBLEM.

To find the centre of a given circle.

FG

Let ABC be the given circle; it is required to find its

centre.

Draw any chord AB, and bifect it in D (I. 10.); and through the point D draw CE at right angles to AB (L. 11.), and bisect it in F: then will the point F be the centre of the circle.

For if it be not, fome other point must be the centre, either in the line EC, or out of it.

But it cannot be any other point in the line Ec, for if it were, two lines drawn from the centre of the circle to its circumference would be unequal, which is abfurd.

Neither can it be any point out of that line; for if it can, let & be that point; and join GA, GD and GB.

Then, because GA is equal to GB (1. Def. 13.), AD to DB (by Conft.), and GD common to each of the triangles AGD, BGD, the angle ADG will be equal to the angle BDG (I. 7.)

But when one line falls upon another, and makes the adjacent angles equal, thofe angles are, each of them, right angles (1. Def. 8 and 9.)

The

The angle ADG, therefore, is equal to the angle ADC (I. 8.), the whole to the part, which is abfurd; confequently no point but F can be the centre of the circle.

Q.E.D. COROLL. If any chord of a circle be bisected, a right line drawn through that point, perpendicular to the chord, will pass through the centre of the circle.

PROP. II. THEOREM.

If any two points be taken in the circumference of a circle, the chord, or right line which joins them, will fall wholly within the circle.

Let ABE be a circle, and A, B any two points in the circumference; then will the right line AB, which joins those points, fall wholly within the circle.

For find c, the centre of the circle ABE (III. 1.), and join C, A, C, B; and through any point D, in AB, draw the right line CE, cutting the circumference in E.

Then, because CA is equal to CB (I. Def. 13.), the angle CAB will be equal to the angle CBA (I. 5.)

And, fince the outward angle CDB of the triangle ACD, is greater than the inward oppofite angle CAB (I. 16.), it will also be greater than the angle CBA.

But the greater fide of every triangle is opposite to the greater angle (I. 17.); whence CB, or its equal CE, will be greater than CD.

The point D, therefore, falls within the circle; and the fame may be fhewn of any other point in AB; consequently the whole line AB must fall within the circle.

Q.E.D.

PRO P. III. THEOREM.

If a right line, which paffes through the centre of a circle, bifect a chord, it will be

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perpendicular to it; and if it be perpendicular to the chord, it will bifect it.

E

Let ABC be a circle, and CE a right line which passes through the centre D, and bifects the chord AB in E; 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 (1.7.)

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

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

For join the points AD, DB, as before :

Then, fince the angle DAB is equal to the angle DBA (1. 5.), and the angle ABD to the angle DEB, (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 fide aɛ will also be equal to the fide EB (1.21.); whence AB is bifected in the point E, as was to be shewn.

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

PRO P. IV. THEOREM.

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

G

Let ABDC be a circle, and o 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 bifect them in the points F, G (I. 10.); and through the centre o, draw FD, CE, cutting the circumference in D and E.

Then, fince AF is equal to FB (by Conft.), AO to OB (by Hyp.), and or common to each of the triangles AOF, BOF, the angle AFO will be equal to the angle Bro (1.7.)

And because the right line or 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 bifects any chord, at right angles, it paffes 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 fhewn, 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 fhewn.

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

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

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