parallel, the line AB which is perpendicular to DE, is also perpendicular to FG, and the angle K is a right angle (B. I., P. 20, c. 1). But BK, the prolongation of AB, is a different line from BF, because the three points A, B, C, are not in the same straight line; hence, there would be two perpendiculars, BF, BK, let fall from the same point B, on the same straight line, which is impossible (B. I., P. 14); hence, DE, FG, are not parallel, and consequently, will intersect in some point 0.

Moreover, since the point O lies in the perpendicular DE, it is equally distant from the two points, A and B (B. I., P. 16); and since the same point O lies in the perpendicu lar FG, it is also equally distant from the two points B and C: hence, the three distances OA, OB, OC, are equal; therefore, the circumference described from the centre 0, with the radius OB, will pass through the three given points, A, B, C.

We have now shown that one circumference can always be nade to pass through three given points, not in the same straight line: we say farther, that but one can be described through them.

For, if there were a second circumference passing through the three given points A, B, C, its centre could not be out of the line DE, for any point out of this line is unequally distant from A and B (B. I., P. 16); neither could it be out of the line FG, for a like reason; therefore, it would be in both the lines DE, FG. But two straight lines cannot cut each other in more than one point; hence, there is but one circumference which can pass through three given points.

Cor. Two circumferences cannot meet in more than two points; for, if they have three common points, there will be two circumferences passing through the same three points; which has been shown, by the proposition, to be impossible.

PROPOSITION VIII. THEOREM.

Two equal chords are equally distant from the centre; and of two unequal chords, the less is at the greater distance from the centre.

Suppose the chord AB to be equal to the chord DE. From the centre of the circle, draw CF, and CG respectively perpendicular to the chords: then will CF be equal to CG.

E

M

G

and the side

D

AB (P. 6), is

Draw the radii CA, CD; then in the right-angled triangles CAF, DCG, the hypothenuses CA, CD, are equal (D. 2); AF, the half of equal to the side DG, the half of DE: hence, the triangles are equal, and CF is equal to CG (B. I., P. 17); consequently, the two equal chords AB, DE, are equally distant from the

centre.

N

K

H

Secondly. Let the chord AH be greater than DE: then will DE be furthest from the centre C. Since the chord AH is greater than DE the arc AKH is greater than DME (P. 5). Cut off from the former, a part ANB, equal to DME; draw the chord AB, and draw CF perpendicular to this chord, and CI perpendicular to AH. It is evident that CF is greater than CO (B. I., A. 8), and CO than CI (B. I., P. 15); therefore, CF is still greater than CI. But CF is equal to CG, because the chords AB, DE, are equal: hence, CG is greater than CI; therefore, of two unequal chords, the less is the farther from the centre of the circle.

PROPOSITION IX. THEOREM.

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

Let the line BD be perpendicular to the radius CA at its extremity A; then will it be tangent to the circumfer

ence.

A

E

D

Ο

For, every oblique line CE, B is longer than the perpendicular CA (B. I., P. 15); hence, the point E is without the circle; therefore, the line BD has no point but A in common with the circumference; consequently, the line BD is a tangent (D. 9).

Cor. 1. Conversely, if a straight line be tangent to a circle, it will be perpendicular to the radius drawn to the point of contact.

Let BAD be a tangent, and CA a radius drawn through the point of contact A: then will BD be perpendicular to CA. For, through the centre C, suppose any other line, as COE, to be drawn. Then, since BD is a tangent, the point E will lie without the circle, and consequently CE will be greater than the radius CO or CA; therefore, the radius CA, measures the shortest distance from the centre C, to the tangent BD: hence, it is perpendicular to the tangent (B. I., P. 15, c. 1).

Cor. 2. At a given point of the circumference only one tangent can be drawn to the circle. For, let A be the given point, BD a tangent, and CA the radius drawn through the point of contact A. Now, if another tangent could be drawn, it would also be perpendicular to CA at the point A, by the last corollary: that is, we should have two lines perpendicular to CA, at the same point; which is impossible (B. I., P. 14, c).

once the middle of the arc MHP, and of the arc NHQ (P. 6); consequently, we shall have the arc MH=HP, and the arc NH=HQ; and therefore

MH-NH-HP-HQ;

in other words, MN=PQ.

Second. When, of the two parallels AB, DE, one is a secant, and the other a tangent, draw the radius CH to the point of contact H; it will be perpendicular to the tangent DE (P. 9, c. 1), and also to its parallel MP (B. I., P. 20, c. 1). But since CH is perpendicular to the chord MP, the point H must

be the middle of the arc MHP (P. 6); therefore, the arcs MH, HP, included between the parallels AB, DE, are equal.

t

Third. If the two parallels DE, IL, are tangents, the one at H, the other at K, draw the parallel secant AB; and, from what has just been shown, we shall have MH=HP, MK=KP:

and hence, the whole are HMK=HPK. It is further evident that each of these arcs is a semi-circumference.

Cor. Conversely: If the arc HM is equal to the arc HP, it is plain that the chord MP will be parallel to the tangent DE

PROPOSITION XI. THEOREM.

If two circumferences have one point common, out of the straight line which joins their centres, they will also have a second point in common; and the two points will be situated in a line perpendicular to the line joining the centres, and at equal distances from it.

Let the two circumferences described about the centres and D intersect each other at the point A; draw AF

perpendicular to CD, and prolong it till BF is equal to AF; then will the circumferences also intersect each other at B.

For, since AF is equal to FB, CF common and the ́angles at F right angles, the hypothenuses CB and CA are equal (B. I., P. 5): hence, the circumference described about the centre C, with the radius CA, will pass through B. In the same manner it may be shown, that the circumference described about the centre D, with the radius DA, will also pass through B.

Cor. If two circumferences intersect each other, they will intersect in two points, and the line which joins the centres will be perpendicular to the common chord at the middle point.

PROPOSITION XII. THEOREM.

If the circumferences of two circles intersect each other, the distance between their centres will be less than the sum of their radii, and greater than the difference.

Let two circumferences be described about the centres C and D, with the radii CA and DA: then, if these circumferences intersect each other, the triangle CAD can always be formed. Now, in this triangle,

CAD,

also,

CD<CA+AD (B. I., P. 7),
CD>DA-AC (B. I., P. 7, c.)

B