the chord AD will coincide with EG (A. 11), and is, therefore, equal to it; which was to be proved.

2o. Let the chords AD and EG be equal: then will the arcs AMD and ENG be equal.

Draw the radii CD and OG. The triangles ACD and EOG have all the sides of the one equal to the corresponding sides of the other; they are, therefore, equal in all respects: hence, the angle ACD is equal to EOG. If, now, the sector ACD be placed upon the sector EOG, so that the angle ACD shall coincide with the angle EOG, the sectors will coincide throughout; and, consequently, the arcs AMD and ENG will coincide: hence, they are equal; which was to be proved.

PROPOSITION V. THEOREM.

In equal circles, a greater arc is subtended by a greater chord; and conversely, a greater chord subtends a greater

For, place the circle EGK upon AHL, so that the centre O shall fall upon the centre C, and the point E upon A; then, because the arc EGP is greater than AMD, the point P will fall at some point H, beyond D, and the chord EP will take the position AH.

Draw the radii CA, CD, radii CA, CD, and CH. Now, the sides AC, CH, of the triangle ACH, are equal to the sides AC, CD, of the triangle ACD, and the angle ACH is

greater than ACD: hence, the side AH, or its equal EP, is greater than the side AD (B. I., P. IX.); which was to be proved.

to AMD, the chord AH would be equal to the chord AD (P. IV.); which contradicts the hypothesis. And, if the arc ADH were less than AMD, the chord AH would be less than AD; which also contradicts the hypothesis. Then, since the arc ADH, subtended by the greater chord, can neither be equal to, nor less than AMD, it must be greater than AMD; which was to be proved.

PROPOSITION VI. THEOREM.

The radius which is perpendicular to a chord, bisects that chord, and also the arc subtended by it.

Let CG be the radius which is perpendicular to the chord AB: then this radius bisects the chord AB, and also the arc AGB.

For, draw the radii CA and CB. Then, the right-angled triangles CDA and CDB have the hypothenuse CA equal to CB, and the side CD com

C

G

D'

mon; the triangles are, therefore, equal in all respects: hence, AD is equal to DB. Again, because CG is perpen

dicular to AB, at its middle point, the chords GA and GB are equal (B. I., P. XVI.); and consequently, the arcs GA and GB are also equal (P. IV.): hence, CG bisects the chord AB, and also the arc AGB; which was to be proved.

Cor. A straight line, perpendicular to a chord, at its middle point, passes through the centre of the circle.

Scholium. The centre C, the middle point D of the chord AB, and the middle point G of the subtended arc, are points of the radius perpendicular to the chord. But two points determine the position of a straight line (A. 11): hence, any straight line which passes through two of these points, passes through the third, and is perpendicular to the chord.

Through any three points, not in the same straight line, one circumference may be made to pass, and but one.

Let A, B, and C, be any three points, not in a straight line: then may one circumference be

through them, and but one.

Join the points by the lines AB, BC, and bisect these lines by perpendiculars DE and FG: then will these perpendiculars meet in some point O. For, if they do not meet, they are parallel. Draw DF. The sum of the angles EDF and GFD

made to pass

is less than the sum of the angles EDB and GFB, i. e.,

less than two right angles: therefore, DE and FG are not parallel, and will meet at some point, as O (B. I., P. XXI.) Now, O is on a perpendicular

D

O

to AB at its middle point; it is, therefore, equally distant from A and B (B. I., P. XVI.). For a like reason, O is equally distant from B and C. If, therefore, a circumference be described from O as a centre, with a radius equal to the distance from O to A, it will pass through A, B, and C. Again, O is the only point which is equally distant from A, B, and C: for, DE contains all of the points which are equally distant from A and B; and FG all of the points which are equally distant from B and C; and consequently, their point of intersection O, is the only point that is equally distant from A, B, and C: hence, one circumference may be made to pass through these points, and but one; which was to be proved.

Cor.

Two circumferences can not intersect in more than two points; for, if they could intersect in three points, there would be two circumferences passing through the same three points; which is impossible.

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

1o. In the equal circles ACH and KLG, let the chords AC and KL be equal; then are they equally distant from the centres.

S

For, let the circle KLG be placed upon ACH, so that the centre R shall fall upon the centre O, and the point K upon the point A: then will the chord KL coincide with AC (P. IV.); and consequently, they are equally distant from the centre; which was to be proved.

B

M

D

K

R

H

G

2o. Let AB be less than KL: then is it at a greater distance from the centre.

For, place the circle KLG upon ACH, so that R shall fall upon O, and K upon A. Then, because the chord KL is greater than AB, the arc KSL is greater than AMB; and consequently, the point L will fall at a point C, beyond B, and the chord KL will take the direction AC.

Draw OD and OE, respectively perpendicular to AC and AB; then OE is greater than OF (A. 8), and OF than OD (B. I., P. XV.): hence, OE is greater than OD. But, OE and OD are the distances of the two chords from the centre (B. I., P. XV., C. 1): hence, the less chord is at the greater distance from the centre; which was to be proved.

Scholium. All the propositions relating to chords and arcs of equal circles, are also true for chords and arcs of one and the same circle. For, any circle may be regarded as made up of two equal circles, so placed that they coincide in all their parts.