the supposition is false; . that ac does not fall within the O, and :: that it falls without the O. Part 2nd. That between AE and can be drawn which does not cut O. no right line 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. PROPOSITION XVII. Problem. To drawn a straight line from a given point, either a without or in the circumference, which shall touch a given circle. 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 = A AEB, IZ EDF = Z EBA, 2. that EBA is a right 2, 3. that ab touches the O BCD. Part 2nd, in which given point d is on the of the o, is proved by Cor. to xvi. 3. PROPOSITION XVIII. 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. 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. Suppose that the centre of the O is not in ac, but in some other direction, as F; and prove, on that supposition, 1. that FCE is a rt. 2, 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. Steps of the Demonstration to Case 1st, In which the centre E is within _ BAC. 1. Prove that / EAB = _ EBA, that 2 % EAB = ex. / BEF of A BAE, 3. that similarly 2 / EAC = _ FEC, that :. whole BEC = 2 whole < BAC. |