14. A Polygon is circumscribed about the circle when all of its sides are tangents to the circle. The circle is at the same time inscribed in the polygon. 15. From the conception of a circle and its parts we perceive the following truths to be self-evident. AXIOMS. 1. The radii of a circle, or of equal circles, are equal. 2. Every diameter of a circle is double the radius, or is equal to the sum of two radii. 3. Every point within a circle is at a distance from the centre which is less than the radius of the circle. 4. Every point without the circle is at a distance from the centre which is greater than the radius of the circle. POSTULATE. 1. A circle may be described from any point as a centre and with any length as a radius. NOTE. In referring to one of the above axioms, the number of the Book will be given; in referring to an axiom of Book I., the number of the Book will be understood. Thus, "III. Ax. 2," means Book III. Ax. 2. Notice also that "I. 12," means Book I. Th. 12; "II. 14" means Book II. Th. 14; and “III. 15, 2," means Book III. Th. 15, Cor. 2. PROPOSITION I.-THEOREM. The diameters of a circle, or of equal circles, are equal. PROPOSITION II.—THEOREM. The diameter of a circle is greater than any other chord. Given. Let AD be a diameter of the circle ABD, and AB be any other chord. To Prove. Then we are 'to prove that AD is greater than AB. Proof. From the centre C draw the radius CB. Then in the triangle ACB, C Every diameter bisects the circle and its circumference. Given.-Let AB be a diameter of the circle ACBD, whose centre is 0. To Prove. Then we are to prove that AB bisects the circle ACBD and its circumference. Proof. Revolve the figure ADB around the line AB as an axis until it coincides with the plane ACB. Then the arc ADB will coincide C B with the arc ACB, since every point of each is equally distant from the centre 0. The two figures will then coincide throughout, and are therefore equal in all respects. Hence ABC or ABD is one-half of the circle; and ACB or ADB is one-half of the circumference. Therefore, etc. SCHOLIUM.-A segment equal to one-half of the circle, as the segment ABC, is called a semi-circle. PROPOSITION IV.-THEOREM. A straight line cannot intersect the circumference of a circle in more than two points. Given.-Let a straight line MN intersect the circumference of a circle whose centre is O in the two points A and B. To Prove. Then we are to prove that MN can intersect the circumference in no other points. M Proof. If it were possible, suppose MN to intersect the N Ᏼ circumference also in some other point, as P. Draw the radii OA, OB, and OP. Then the radii OA, OB, and OP are all equal. Now, if MN could intersect the circumference in three points, we should have three equal straight lines, OA, OB, and OP, drawn from the same point to the same straight line, which is impossible (B. I., Th. IX. C. 2). Hence it is impossible for MN to intersect the circumference in the point P. Therefore, etc, PROPOSITION V.-THEOREM. In the same circle, or in equal circles, equal angles at the centre intercept equal arcs on the circumference; and, CONVERSELY, equal arcs subtend equal angles at the centre. Given. In the two equal circles, ABC and EFC', let the angle COD be equal to the ABC to its equal EFC', so that the equal angles coincide. Then, since the radii are equal, the point C will coincide with C', and the point D with D'; hence, the arc CD will coincide with the arc C'D', since every point of each is equally distant from the common centre 0. CONVERSELY.-Let the arc CD equal the arc C'D'. angle COD angle C'O'D'. = Proof. Apply the circle ABC to the equal circle EFC', so that the arc CD coincides with the equal arc C'D'. Then the radius OC will coincide with the radius O'C', the radius OD with O'D', and the angle COD will coincide with the angle C'O'D'. Therefore, etc. SCHOLIUM. If the angles are in the same circle the demonstration is similar. |