square of B E is equal to the rectangle under the whole line A B, and the other part A E. Because c is the centre of the arc, c d is equal to c A, and Bc is equal to the half of A B; therefore c A is to B C, as the square root of 5 to 1; but B p is the remainder of c d after B C is taken away, therefore bd is equal to the square root of 5 minus 1, which is the part required; and again, because B is the centre of the arc D E, B E is equal to BD; and consequently A B is divided in the ratio which was required. LINES AND CIRCLES. In order to bring the circle properly within the province of elementary geometry, it is necessary to have the means of determining the length of the circumference and diameter in terms of each other, or in other words to find their ratio. We formerly stated the approximate value of this ratio; and we are not now to enter farther into the investigation of the circle, than merely to state one or two properties of some lines in and about a circle, which have their relations determined by that figure. 1. If two straight lines cross or cut each other in a circle, the rectangle or product of the two segments of the one is equal to that of the two segments of the other, without any regard whatever to the lengths of the lines, or their situation in the circle, provided that they cross each other, and the extremities of both are on the circumference. If both lines pass through the centre, the truth asserted is self-evident; because each of the segments of both is a radius of the same circle, and consequently they and their products or rectangles two and two, are equal. There are other three cases, one line passing through the centre, crossing the other at right angles, and consequently bisecting it; and the proof of this is also nearly self-evident. Let A B, passing through the centre, cut c D, which does not pass through the centre, at right angles, in the point E, then the rectangle A E X E B is equal to the rectangle C E X E D, that is, as c d is bisected the square of c E or ED. Join A C, C B, B D, D A; and the triangles A CB, A D B, are equal to each other, and right-angled at c and d, because each of them is a semicircle. C E and D Е, the segments of C D, are the perpendiculars drawn from the right angles of those triangles to the opposite side ; therefore cach of them is a mean proportionable between A E and E B, the segments of a B their common base; consequently, A E: EC = ED: EB; and, multiplying extremes and means, A E X EB = CE X E), that is, the rectangle of the segments of a B is equal to that of those Next, let a B, which passes through the centre, cut c D, not at right angles, and consequently not bisect it, the rectangle of the segments of the one is still equal that of the segments of the other. Let A B pass through the centre, and cut c p in the point E, but not at right angles, and consequently not into equal parts, Find the centre F by bisecting a B, and from draw G at right angles to c D, and F D. CD is equally divided in G, and unequally in E, and A B is equally divided in F and unequall in E; therefore, DG? = DE X EC + GE, Add the square of F G to both, and D G? + F G? = DE X EC + GE + G F2. But because F is the centre of the circle, D F is equal A F or F B; and because of the right angle at G, D F2 = D G? + G Fo; and because D F is equal to a F, and a F to F B, the square of D G + G F = A E XE B + FE?, but FE? = F G? + G Eo; therefore A E X E B + F G? + G E% = DG2 + f g?. But it has been already shown, that D G? + F G? = DE X EC + G F2 + G FR:. Therefore A E X E B + F G? + E G? = DE X EC + G F + gr. Take away the parts F G? + E G', which are common t both; and the remaining rectangles are equal, that is, A E X EB = D E X EC, and they are the segments of the lines. If neither of the lines passes through the centre of the circle, a line can be drawn through the point of section and the centre, meeting the circumference both ways; and as it can be shown (as in the last case) that the rectangle of the two segments of each of the lines, is equal to the rectangle of the segments of this one which passes through their point of section and the centre, it follows that their rectangles must be equal to each other. 2. As the rectangles of the segments of any two lines which intersect each other in a circle are equal, it follows conversely that lines which intersect each other so as to make the rectangles of the segments of each line equal, are so situated that a circle can always be described so as to touch all their four extremities; and the centre of this circle can in every case be found by bisecting each of the lines at right angles ; for, as both the bisecting lines must pass through the centre, that centre must be the point where the perpendiculars intersect each other. But it also follows, that when the rectangle formed by two lines is equal to that formed by other two, and the contained angle is equal in both, the sides about the equal angles must be reciprocally proportional, that is, in looking back to either of the last figures, A E : DE = EC: E B. Hence we have a general method of transforming any rectangle or parallelogram whatsoever into another of the same surface, but differing in the lengths of the sides and their ratios to each other; for we have only to place the sides of the given rectangle in a straight line, to draw from the point where they meet a line the length of one of the proposed sides of the equal but not similar figure; and by describing a circle through the three points, and producing the line which was made equal to one of the sides of the proposed figure until it meets the circumference, in order to get the remaining side of the rectangle sought. It may be worth while to illustrate this by a diagram. Let A and B represent two sides of a parallelogram, and o one of the angles which it contains, and let p be one side of another parallelogram which is to have the same area and the same angles, it is required to find the remaining side of this parallelogram. Make the line E F equal to A, and continue it to g till F G is equal to B. At the point F draw the line F h, equal to d, and making the angle E F 1 equal to c. Through the points E H and a describe a circle, and produce u F till it meet the circumference of this circle in the point 1, and F 1 is the side required. To complete the parallelograms, transfer f g to Fi by the arc G K, and transfer F i to Fm in f g produced, and the sides I F, F K, and F, F M, are the sides of the equal parallelograms, only they contain the supplements of the proposed angle c; complete the parallelograms, by drawing parallels meeting in L and in P, and 1 F K L and H F M P are the equal parallelograms, having IF : 1 F = FM: F K, which were required ; continue the parallels till they meet in o n as marked by the dotted lines, and the whole figure OL N P is a parallelogram ; and if on are joined, o n is a diameter of that parallelogram, and it passes through F, the point in the circle where Eg and hi intersect each other. The sides of the parallelograms E F K L and F M P, which are about their equal angles at F, are reciprocally proportional, and for this reason they are directly proportional in the reversette |