IN Book H. PROP. A. THEOR. See N. IN any triangle, the difference between the squares of two of the sides is equal to twice the rectangle contained by the base, and the segment of it intercepted between the perpendicular drawn to it from the opposite angle, and the point in which it is bifected, Let ABC be a triangle, of which the fide AB is greater than a 10. 1. AC; and bisect a the base BC in D, and draw 6 AE perpendib 12. 1. cular to BC; the square of AB is greater than the square of AC, by twice the rectangle BC, DE. C 3. I. Produce BC to F, and make © EF equal to EB: and because BF is double of BE, and BC double of BD; the remainder CF rectangle BC, CF, together with the 3.5. 2. fquare of CE, is equal d to the square of BE: add to these equals the the rectangle BC, CF is double e of the rectangle BC, DE, bef 47. 1. cause CF is double of DE; and the square of AC is equal to the squares of CE, EA, because CEA is a right angle; and the square of AB is equal to the squares of BE, EA; therefore twice the re&tarigle BC, DE together with the square of AC, is equal to the square of AB; that is, twice the rectangle BC, DE is equal to the excess of the square of AB above the square of AC. Wherefore, &c. Q. E. D. g 4. 1. Cor. And if AF be joined, the triangle ABF is isosceles 5; hence it is manifest, that if any straight line AC be drawn froin the vertex to the base of an isosceles triangle ABF; the square of the fide AB is equal to the square of the line AC, together with the rectangle BC, CF of the segments of the base. D E e 1. 2. PROP. B. THEOR. "HÉ squares of two ficles of a triangle are toge ther double of the square' of half the base, and of the square of the straight line drawn from the vertex to bifect the base. Let a 10. I. b Let ABC be a triangle; and let the base BC be bise&ed a in Book II. If the angles at D be right angles, the square of BA is equal But, if the angles at D be not right angles, from A draw C 12. 1. d 9. 2. B PROP. C. THEOR. THE squares of the two diameters of a paral lelogram are together equal to the squares of its four fides. TH Let ABCD be a parallelogram, of which the diameters are Let AC, BD cut one another in E: and because AC meets the bare d B. 2. Book II. base BD; the squares of BA, AD' are double d of the squares mof BE, EA. For the same reason, the squares of BC, CD are double d of the squares of BE, EC, or of BE, EA, because EC is equal to EA: therefore the squares of BA, AD, BC, CD are quadruple of the squares of BE, EA: but the e 3 Cor.4.2. square of BD is quadruple of А D the square of BE, because BD is double of BE; and the square AC is. quadruple of the square of AE ; therefore the squares of BD, AC are quadruple of the squares of BE, EA: and it has B been proved, that the squares of the four sides are quadruple of BE, EA; therefore the squares of BD, AC are equal to the squares of the four fides AB, BC, CD, DA. Wherefore, &c. 0. E. D. Cor. Hence it is manifest, that the diameters of a parallelogram bisect one another. THE A. BOOK. III. I. Omitted. II. a circle, when it meets the III. another, which meet, but do IV. ftant from the centre of a circle, V. greater perpendicular falls, is said to VI. Book III. VI. tained by a straight line, and the part VII. Qmitted. VIII. tained by two straight lines drawn IX. arch intercepted between the straight X. tained by two straight lines drawn from XI. those in which the angles are PROP. I. PROB. To find the centre of a given circle. Let ABC be the given circle ; it is required to find its centre. Draw within it any straight line AB, and bisect a it in D; b) 31. 1. from the point D drawb DC at right angles to AB, and pro duce it to E, and bisect CE in F: The point F is the centre of the circle ABG. For, if it be not, let, if possible, G be the centre, and join GA, GD, GB: Then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two fides AD, DG are equal to the two BD, DG, each to each ; and bale GA is equal to the base GB, because they are drawn from the centre |