squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle. Steps of the Demonstration to Case 1st. 1. Prove that BC2 + BD2 = 2 BC X BD + DC2, 2. 3. that AB2+BC2 = 2 BC X i. e., that ac2 alone < CB2 + BA (AD2 + DC2), BD + AC2*, by 2 BC × BD. * Obs. The learner should particularly mark the substitution of AB2 for BD2 + AD2, and of ac2 for AD2 + DC2. Steps of the Demonstration to Case 2nd, (in which AD falls without the ▲.) 1. Prove that ACB > right <, 2. 3. that. AB AC2 + CB2 + 2 BC × CD, that AB2 + BC2 = ac2 + (2 bc2 +2 BC × cd), 4. Prove that AB2 + BC2 = AC2 × (2 DB × BC)*, i. e., that ac2 alone < AB2 + вc2, by 2 db × BC. * Obs. Mark particularly the substitution of 2 DB X BC, for 2 BC2 + 2 BC × CD. The last case, (in which the side ac of the ▲ is 1 BC) is proved by xlvii. I. PROPOSITION XIV. Problem. To describe a square which shall be equal to a given rectilineal figure. AA B G Steps of the Demonstration. 1. Prove that BE X EF + EG2 = GF2; and that this 2. 3. 4. = EH2 + EG2, that BE XEF, i. e., □m BD = EH3, that rectilineal fig. A EH. Suppose that F is not in the centre of the O, but that some other point, as G, is the centre. Then prove, on that supposition, shows the supposition to be false, 4. that F is the centre of the . D D E B PROPOSITION II. (Argument ad absurdum). Theorem. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. Steps of the Demonstration. Suppose that AB falls without the O, and prove, on 4. that similarly AB does not fall upon the ; and .. that it falls within it. PROPOSITION III. Theorem. If a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre, it shall cut it at right angles; and if it cuts it at right angles, it shall bisect it. Part 1st is proved (from viii. 1.) by showing that (in AS AFE, BFE) ≤ AFE = ≤ BFE, and.. that they are both right s. Part 2nd is proved (from xxvi. 1.), by showing that in the same AS AF FB. Suppose that AC, BD do bisect each other in E. (State that this is evidently impossible if one line pass through the centre, and the other not). If neither of them pass through the centre, prove, on the above supposition, 1. that 2. that FEA is a right, FEB is a right, and.. = FEA; i. e. greater less, which shows the above supposition to be false; and . that AC, BD do not bisect each other. |