B PROPOSITION X. Problem. To bisect a given finite straight line; that is, to divide it into two equal parts. COR. (Argument ad absurdum). Two right lines cannot have a common segment. B PROPOSITION XII. Problem. To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. PROPOSITION XIII. Theorem. The angles which one straight line makes with another on one side of it, are either two right angles, or are together equal to two right angles. Steps of the Demonstration. (In fig. 1, the DBA = ≤ ABC :. each is a rt. ≤). In fig. 2. prove, 1. that s CBE + EBD2S CBA + ABE + EBD, 2. that s DBA + ABC = ≤S DBE + EBA + ABC, 3. that LS DBA + ABC = the two rt. Ls CBE + EBD. PROPOSITION XIV. (Argument ad absurdum). Theorem. If, at a point in a straight line, two other straight lines on op posite sides of it make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. -D B Steps of the Demonstration. Suppose that BD is not in the same st. line with BC, but that BE is in the same st. line with it. Then prove, on that supposition, 1. that ZS ABE + ABC 2 rt. Zs, 2. that S ABE + ABCLS ABC + ABD, 3. that ABE = ABD, i. e. less greater, which shows the supposition to be false, 4. that similarly* no other line but BD is in the same right line with BC. PROPOSITION XV. Theorem. If two straight lines cut one another, the vertical, or opposite, angles shall be equal. B Steps of the Demonstration. 1. Prove that S AEC + AED = 2 right /s, that S AED + DEBS AEC + AED, 2. * In every case in which we have employed the word 'similarly," it has reference to the complete demonstration of the steps as given in Euclid. PROPOSITION XVI. Theorem. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. B B Steps of the Demonstration. Jbase AB = base FC, 1. Prove that (in As EAB, ECF) and BAC=≤ACF, 2. 3. that ACD > < BAC, that similarly ▲ BCG or ACD > ≤ abc. PROPOSITION XVII. Theorem. Any two angles of a triangle are together less than two right angles. Steps of the Demonstration. 1. Prove that ZS DCA + ACB > <S ACB + CBA, that LS ACB + CBA < 2 right Ls, 2. 3. that similarly each of the other pairs of Ls is 2 right Ls. PROPOSITION XVIII. Theorem. The greater side of every triangle is opposite to the greater angle. B PROPOSITION XIX. (Argument ad absurdum). Theorem. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it. B B Steps of the Demonstration. 1. Prove that AC cannot be AB, that AC cannot be < AB, 2. 4. 5. that. BA + AC > BC, that similarly each of the other pairs of sides is the remaining side. |