PROPOSITION III. THEOREM. A straight line cannot meet a circumference in more than two points. Let AEBF be a circumference, and AB a straight line: then AB cannot meet the circumference in more than two points.. F A B For, suppose that they could meet in three points. We should then have three equal straight lines drawn from the same point to the same straight line; which is impossible (B. I., P. XV., C. 2) hence, AB cannot meet the circumference in more than two points; which was to be proved. : PROPOSITION IV. THEOREM. In equal circles, equal arcs are subtended by equal chords and conversely, equal chords subtend equal arcs. Draw the diameters AB and EF If the semi-circle ADB be applied to the semi-circle EGF, it will coincide with it, and the semi-circumference ADB will coincide with the semi-circumference EGF. But the part AMD is equal to the part ENG, by hypothesis: hence, the point D will fall on G; therefore, 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 their parts: 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 will be 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 0 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 П, beyond D, and the chord EP will take the position AH. Draw the 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. AMD, the chord. AH For, if ADH were equal to would be equal to the chord AD (P. IV.); which is contrary to the hypothesis. And, if the arc ADH were less than AMD, the chord AH would be less than AD; which is also contrary to 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 to the chord AB: perpendicular to the then will this radius bisect the chord AB, and also the arc AGB. For, draw the radii CA and CB. Then, the right-angled triangles CDA and CDB will have the hypothenuse CA equal to CB, and the side CD common; the triangles are, therefore, parts hence, AD is equal to DB. Α equal in all their Again, because CG is perpendicular 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 mid dle 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, will pass through the third, and be perpendicular to the chord. PROPOSITION VII. THEOREM. 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 made to pass. 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 0. For, if they do not meet, they are parallel; and if they are parallel, H C F D K the line ABK, which is perpendicular to DE, is also perpendicular to KG (B. I., P. XX., C. 1); consequently, there are two lines BK and BF, drawn through the same. point B, and perpendicular to the same line KG; which is impossible hence, DE and FG meet in some point 0. Now, is on a perpendicu lar to AB at its middle point, If, therefore, a circumference be de F D K scribed from O as a centre, with a radius equal to OA, 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 eircumference may be made to pass through these points, and but one; which was to be proved. Cor. Two circumferences cannot 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. PROPOSITION VIII. THEOREM. 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 will they be equally distant from the centres. |