(2) Constr. B. 1. (1) Prop.6. square of EA is equal to the square of EF (1), or to B. 2. the square of EB which is equal to EF (2), and there(3) Prop 47 fore to the squares of EA and AB (3), take away the common square of EA and the rectangle under CF and FA is equal to the square of AB: but because AF and FG are equal, CG is the rectangle under CF and FA, therefore CG and AD are equal, and if the common rectangle CH be taken away, AG and HD are equal, but AG is the square of AH, for AH and (4) Const. AF are equal (4), and the angle A is a right angle; HD is the rectangle under AB and HB, for BD is equal to AB. Cor. The rectangle under the greater segment and the difference between the segments is equal to the square of the less segment, as is evident by taking away the rectangle under AH and HB from the equals AG and HD. Fig. 2 4. PROP. XII. THEOR. In any obtuse angled triangle (BAC) the square of the side (AB) subtending the obtuse angle, exceeds the sum of the squares of the sides (BC and CA)which contain the obtuse angle by double the rectangle under either of these sides (BC), and the external segment (CD) between the obtuse angle and the perpendicular let fall from the opposite angle. The square of BA is equal to the sum of the squares (1) Prop.47 of AD and DB (1), but the square of DB is equal to the squaresof DC and CB,together with double the rect(2) Prop 4. angle under DC and CB (2), therefore the square of B. 1. B. 2. (3) Prop.47 B. 1. AB is equal to the squares of AD, DC and CB, together with double the rectangle under DC and CB, but the square of AC is equal to the squares of AD and DC (3), and therefore the square of AB is equal to the squares of AC and CB, together with double the rectangle under BC and CD; therefore the square of AB exceeds the sum of the squares of AC and CB by double the rectangle under DC and CB. PROP. XIII. THEOR. In any triangle (ABC) the square of the side (AB) Fig. 25. subtending an acute angle is less than the sum of the squares of the sides (AC and CB), containing that angle by twice the rectangle under either of them (AC), and the segment between the acute angle and the perpendicular (BF), let fall from the opposite angle. The squares of AC and CF are equal to twice the rectangle under AC and CF, together with the square of AF (1), and, if the square of the perpen- (1) Prop. 7. dicular BF be added to both, the squares of AC, CF, B. 2. and BF, are equal to twice the rectangle under AC and CF, together with the squares of BF and AF, or with the square of AB, which is equal to them (2), (2) Prop.47 but the squares of BF and CF are equal to the square of BC (2), and therefore the squares of BC and AC are equal to twice the rectangle under AC and CF together with the square of AB, therefore the square of AB is less than the sum of the squares of AC and CB, by twice the rectangle under AC and CF. Schol. 1. If the angle CAB be a right angle, the points F and A coincide, and the rectangle under AC and CF is the square of AC; but it is evident that in this case the square of AB is less than the sum of the squares of AC and CB by twice the square of AC (1). Schol. 2. Hence, given in numbers the sides of any triangle we can find its area; for subtract the square of one of the sides AB, which is not the greatest, from the sum of the squares of the other sides AC and CB, and divide half the remainder, that is, the rectangle under AC and CF, by either of these sides AC, and subduct the square of the quote CF from the square of the remaining side CB, the square root, BF, of the remainder multiplied into half the former divisor AC gives the area of the triangle. B. 1. (1) Prop.47 B. 1. Cor. If from any angle A of a triangle BAP a right Fig. 26. line AC be drawn bisecting the opposite side the squares of the sides AB and AP, containing that angle, are H B. 2 double of the squares of the bisecting line AC, and of half the side subtending the angle. If the bisecting line be perpendicular to the side, it is manifest from prop. 47. B. 1. If not, draw from the angle A the line AF perpendicular to the opposite side; since one of the angles ACB and ACP is obtuse, let ACB be the obtuse angle, and the square of AB is equal to the sum of the squares of AC and CB, together with twice the (1)Prop.12 rectangle under BC and CF (1): the angle ACP is acute, and therefore the square of AP, together with twice the rectangle under PC and CF, or, as PC (2) hypoth. and BC are equal (2), with twice the rectangle under BC and CF, is equal to the sum of the squares of (3) Prop.13 AC and CP (3), therefore the squares of AB and AP, with twice the rectangle under BC and CF are equal to twice the square of AC with the squares of BC and CP, and twice the rectangle under BC and CF, take away from both twice the rectangle under BC and CF, and the sum of the squares of AB and AP is equal to twice the square of AC, with the squares of BC and CP, or with twice the square of BC, because CP and BC are equal, B. 2. Fig. 27. Schol. Hence it is evident, that the sum of the (1) Prop.29 squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. B. 1. (2) Prop.34 B. 1. (3) Prop 4, Because the angles DBC and BDA, ACB and CAD are equal (1), and the side BC is equal to AD (2), the .B. 1. diagonals mutually bisect each other (3), therefore (4) Cor. the sum of the squares of the sides is equal to four prec. (5) Cor. times the squares of the halves of the diagonals (4), or prop.4.B.2, to the sum of the squares of the diagonals (5), Fig. 28. PROP. XIV. PROB. To make a square equal to a given rectilineal figure (Z). Make a rectangle CI equal to the given rectilineal (1)Prop.45 figure (1), if the adjacent sides be equal, the problem is done. B. 1. (2) Prop. 3. B. 1. If not, produce either side IA, and make the produced part AL equal to the adjacent side AC (2), bisect IL in O, and from the centre O, with the radius OL, describe a semicircle LBI, and produce CA till it meet the periphery in B; the square of AB is equal to the given rectilineal figure. For draw OB, and because IL is bisected in O and cut unequally in A, the rectangle under IA and AL, together with the square of OA, is equal to the square (3) Prop.5. of OL (3), or of OB which is equal to OL, and there- B. 2. fore to the squares of OA and AB (4); take away (4) Prop.47 from both the square of OA, and the rectangle under B. 1. IA and AL is equal to the square of AB, but the rectangle under IA and AL is equal to IC, for AL and AC are equal (5), therefore the square of AB is (5) Constr. equal to the rectangle 1C, and therefore to the given rectilineal figure Z. THE ELEMENTS OF EUCLID. BOOK III. See N. Plate 3. Fig. 1. Fig. 2. Fig. 3. Fig. 9. See N. DEFINITIONS. 1. EQUAL circles are those whose diameters are equal. 2. A right line is said to touch a circle when it meets the circle, and being produced does not cut it. 3. Circles are said to touch one another, which meet, but do not cut one another. 4. Right lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. 5. And the right line on which the greater perpendicular falls, is said to be farther from the centre. 6. A segment of a circle is the figure contained by a right line, and the part of the circumference it cuts. off. 7. An angle in a segment is the angle contained by two right lines drawn from any point in the circumference of the segment to the extremities of the right line which is the base of the segment. 8. An angle is said to stand upon the part of the circumference, or the arch, intercepted between the right lines that contain the angle. 9. A sector of a circle is the figure contained by two radii and the arch between them. 10. Similar segments of circles are those which contain equal angles. |