the supposition is false; :that ac does not fall within the O, 2. that similarly ac does not fall on the O, and :. that it falls without the o. Part 2nd. That between AE and o no right line can be drawn which does not cut O. Suppose that ca is between them without cutting the O; and prove, on that supposition, 1. that, in A DAG, DA > DG, 2, that DH > DG, i.e., less > greater, which shows the supposition to be false. Part 1st. To draw a tangent to O BCD, from point A without it. Steps of the Demonstration. (base DF = base AB, 1. Prove that (in AS FED, AEB){ A FED = 1 AEB, Z EDF = LEBA, 2. that 2 EBA is a right Z, 3. that all touches the O BCD. Part 2nd, in which given point d is on the o of the o, is proved by Cor. to xvi. 3. PROPOSITION XVIII. ( Argument ad absurdum). Theorem. If a straight line touches a circle, the straight line drawn from the centre to the point of contact shall be perpendicular to the line touching the circle. F D B В Steps of the Demonstration. Suppose that fc is not I DE, but that some other line drawn from F, as FG, is I DE. Then prove, on that supposition, 1. that < GCF <rt. FGC, 2. that FC > FG, 3. that FB > FG, i. e., less > greater; which shows the supposition to be false, and :. FG is not I DE, 4. that, similarly, no line but rc, drawn from F, is I DE. PROPOSITION XIX. Theorem. If a straight line touches a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line. B D с E prove, on that Suppose that the centre of the O is not in ac, but in some other direction, as F; and supposition, 1. that FCE is a rt. Zo 2. that FCE = ACE, i. e., less = greater, which shows that the supposition is false, and :: F is not the centre, 3. that the centre is in AC. PROPOSITION XX. Theorem. The angle at the centre of a circle is double of the angle at the circumference, upon the same base, that is, upon the same part of the circumference. B Steps of the Demonstration to Case 1st, In which the centre E is within 2 BAC. 1. Prove that EAB = _ EBA, 2. that 2 _ EAB = ex. _ BEF of A BAE, 3. that similarly 2 _ EAC = _ FEC, 4. that :. whole _ BEC = 2 whole Z BAC. Steps of the Demonstration to Case 2nd, In which centre E is without _ BDC. B 1. Prove that 2 EDC = _ ECD, 2. that 2 Z EDC – Z GEC, 3. that similarly 2 % GDB = L GEB, 4. that :: remaining Z BEC = 2 remaining 2 BDC. PROPOSITION XXI. E Theorem. The angles in the same segment of a circle are equal to one another. B Steps of the Demonstration to Case 1st, In which the segment is > 1 O. 1. Prove that _ BFD = 2 2 BAD, 2. that similarly / BFD = 2 Z BED, 3. that :. _ BAD = Z BED. E Steps of the Demonstration to Case 2nd, In which the segment is < d 0. B F 1. Prove that / BAC Z BEC, PROPOSITION XXII. Theorem. The opposite angles of any quadrilateral figure inscribed in a circle are together equal to two right angles. B Steps of the Demonstration. 1. Prove that whole / ADC = US BAC + ACB, 2. that ZS ADC + CBA = the 3 Zs of the A CBA; which = 2 rt. 28. 3. that, similarly, Z8 BAD + DCB = 2 rt. _8. |