and comparing the two triangles ACD,. BCD, which have CA equal to CB (Def. XXIII,) and CD common, as also AD equal to DB by hypothesis, we find that the three sides of the one are equal to the three sides of the other; therefore the angle CDA is equal to the angle CDB, (B. I, Prop. vi,) and each of these angles is a right-angle, and CD is perpendicular to AB, (Def. X.)

Again, if CD be perpendicular to AB, then will the chord AB be bisected at the point D, or have AD equal to DB; and the arc AFB will be bisected at the point F, or have AF equal to FB.

For, drawing CA, CB as before, we have the triangle CAB isosceles, and consequently the angle CAD is equal to CBD, (B. I, Prop. v ;) we also have the angles CDA and CDB equal, cach being a right-angle, (Def. X ;) therefore the third angles of these triangles are equal, (B. I, Prop. xxiv, Cor. 1,) and, having the side CD common, they must also have the side AD equal to the side DB. (B. I, Prop. IV.)

Also, since the angle ACF is equal to the angle BCF, the arc AF which measures the former is equal to the arc BF which measures the latter.

Cor. Hence, a line bisecting any chord at right-angles, passes through the centre of the circle.

(68.) By means of this proposition, we may find the centre of a given circle, or of a given arc of a circle.

Let ABC be the arc of a circle: it is

required to find its centre.

Draw any two chords AB, BC; bisect them with the perpendiculars DF and GH; then will the point K, in which they intersect, be the centre sought.

A

H

For, by the corollary of the preceding proposition, each of the perpendiculars DF and GH passes through the centre, and therefore the centre is at the point K.

(69.) It is also obvious that the above operation is equivalent to finding the centre of a circle whose circumference will pass through three given points A, B, C. Hence, by this means, a circumference of a circle may always be made to pass through any three points, not situated in the same straight line.

(70.) The same method is applicable, in case it is required to describe a circle about a triangle.

they are perpendicular be parallel; that is, CA, CB would be parallel, which is not the case: consequently the perpendiculars DG, FG must meet. Join AG, GB, GC; and comparing the two right-angled triangles DGC, DGA, we have DG common, and the side DC equal to DA, each being half the side AC; therefore, GA is equal to GC, (B. II, Prop. viii, Cor. 2.) By comparing the two triangles HGA, HGB, we may, for similar reasons,

show that GA is also equal to GB. Therefore, the point G is equidistant from A, B and C.

Hence, if, with G as a centre, a circumference be described passing through A, it will also pass through B and C, and consequently circumscribe the triangle ABC.

Cor. 1. The three perpendiculars which bisect the three sides of a triangle, meet in the same point.

Cor. 2. This problem is equivalent to describing a circumference of a circle through any three points not situated in the same straight line.

PROPOSITION III.

THEOREM. Chords in a circle, which are equally distant from the centre, are equal to each other.

versely, 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

chords from the centre H. Then the two right-angled triangles HAF, HCG have the sides HA and HC equal, they 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. vii, 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. ;) 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 IV.

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 circum

ference at the point G. Then because CD is perpendicular to AB it is shorter than the oblique line CF, (B. I, Prop. xxvi, Cor.) 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 V.

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.

Let the line AB be tangent to the circumference of a circle at the point D; then will the radius CD be perpendicular to the tangent AB,

For, the line AB being wholly without the circumference except at the point D, it follows

A

B

that any 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. 1. Hence, conversely, a line drawn perpendicular to a tangent at the point of contact, passes through the centre of the circle.

Cor. 2. If two circumferences touch each other either externally or internally; their centres and point of contact will be in the same straight line. For, if we draw a tangent at the point of contact, and from the same point of contact draw a line perpendicular to this tangent, it will pass through the centre of each circle, (Cor. 1, of this Proposition.)