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. For, if ADH were equal to AMD, the chord AH 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 perpendicular to the chord AB: 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. A G B 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 F 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, O is on a perpendicu lar 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 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, 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 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. 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 will be equally distant from the centre ; which was to be proved. B M D A K R H 2o. Let AB be less than KL: then will it be at a greater distance from the centre. For, place the circle KLG upon ACH, so that R shall fall upon 0, and K upon A. Then, because the chord KL is greater than AB, the arc KSL is greater than AMB; and consequently, the point point C, beyond B, and the chord KL will take the direction AC. Draw OD and OE, respectively perpendicular to AC and AB; then will OE be 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. PROPOSITION IX. THEOREM. If a straight line is perpendicular to a radius at its outer extremity, it will be tangent to the circle at that point; conversely, if a straight line is tangent to a circle at any point, it will be perpendicular to the radius drawn to that point. 1o. Let BD be perpendicular to the radius CA, at A then will it be tangent to the circle at A. For, take any other point of BD, as E, and draw CE: B E D then will CE be greater than CA (B. I., P. XV.); and consequently, the point E will lie without the circle hence, BD touches the circumference at the point A; it is, therefore, tangent to it at that point (D. 11); which was to be proved. 2o. Let BD be tangent to the circle at A: then will it be perpendicular to CA. For, let E be any point of the tangent, except the point of contact, and draw CE Then, because BD is a tangent, E lies without the circle; and consequently, CE is greater than CA: hence, CA is shorter than any other line that can be drawn from C to BD; it is, therefore, perpendicular to BD (B. I., P. XV., C. 1); which was to be proved. Cor. At a given point of a circumference, only one tangent can be drawn. For, if two tangents could be drawn, they would both be perpendicular to the same radius at the same point; which is impossible (B. I., P. XIV.). |