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which is evidently the diagonal of a square of which FE or EB is the side, cut off FG equal to FE, and join GE. Then it would be shown as before, that BG is less than BE; and therefore BE, the difference between the side and diagonal of the square AC, is contained twice in the side AB, with the remainder GB, which is itself the difference between the side FE or EB, and the diagonal FB, of another square. By repeating the process, we should find, in exactly the same manner, that BG would be contained twice in BE, with a remainder, which would be the difference between the side and diagonal of a square described on BG: and it is evident that a like process might be repeated continually, as no excess of a diagonal above a side would be contained in the side without remainder: and as this process has no termination, there is no line, however small, which will be contained without remainder in both AB and BD; they are therefore incommensurable.
1. A LINE which is such, that any point whatever in it fulfils certain conditions, is called the locus of that point.
2. A porism is a proposition affirming the possibility of finding such conditions as will render a certain problem indeterminate, or capable of innumerable solutions.
3. Isoperimetrical figures are such as have their perimeters, or bounding lines, equal.
Schol. 1. Several instances of loci have already occurred in the preceding books.
1. Thus it was proved in the 37th proposition of the first book, that all triangles on the same base, and between the same parallels, are equal in area : and hence, if only the base and area of a triangle be given, its vertex may be at any point in a straight line parallel to the base, and at a distance from it, which may be determined by applying (I. 45. cor.) to half the base a parallelogram equal to the given area ; and therefore the parallel is the locus of the vertex. Here the conditions fulfilled are, that straight lines drawn from any point in the parallel to the extremities of the given line, form with it a triangle having a given area.
2. It was proved in the 21st proposition of the third book, that all angles in the same segment of a circle are equal : and hence, if only the base and vertical angle of a triangle be given, the vertex may be at any point of the arc of a segment described on the base, in the manner pointed out in the 33d proposition of the third book : that arc, therefore, is the locus of the vertex.
3. It was shown in proposition G. of the sixth book, that straight lines drawn from any point whatever in the circumference of the circle ABGC to the points E, F, have the same ratio—that of EA to AF. Hence, therefore, if the base of a triangle, and the ratio of the sides be given, the locus of the vertex is the circumference of the circle described in the manner pointed out in the
corollary to that proposition ; unless the ratio be that of equality, in which case the locus is evidently a perpendicular bisecting the straight line joining the points.
4. It follows likewise, from the fourth corollary to the 47th proposition of the first book, that when the base of a triangle and the difference of the squares of the sides are given, if the point D be found * in the base BC, or its continuation, such that the difference of the squares of BD, CD is equal to the difference of the squares of the sides ; and if through D a perpendicular be drawn to BC, straight lines drawn from any point of the perpendicular to B, C, will have the difference of their squares equal to the given difference; and hence the perpendicular is the locus of the vertex, when the base and the difference of the squares of the sides are given.
5. It will appear, in a similar manner, from proposition A. of the second book, that if BC the base of a triangle, and the sum of the squares of the other sides AB, AC, be given, the locus of the vertex is the circumference of a circle described from D, the middle point of the base as centre, and with the radius DA. To find DA, take the diagonal of the square of BD as one leg of a right-angled triangle, and for its hypotenuse take the side of a square equal to the given sum of the squares of AB, AC: then the diagonal of the square described on half the remaining leg of that right-angled triangle will be the radius DA. The proof of this is easy, depending on the second and third corollaries to the 47th proposition of the first book, and on A. of the second book.
Schol. 2. Porisms may be regarded as having their origin in the solution of problems, which, in particular cases, on account of peculiar relations in their data, admit of innumerable solutions ; and the proposition announcing the property or relation which renders the problem indeterminate, is called a porism. This will be illustrated by the solution of the following easy problem.
Through a given point A, let it be required to draw a straight line bisecting a given parallelogram BCDE.
Suppose AFG to be the required line, and let it cut the sides BE, CD in F, G, and the diagonal CE in H. Then from the equal figures EBC, FBCG take FBCH, and the remaining triangles EHF, CHG are equal. Now, since (I. 29. and 15.) these triangles are equiangular, it is evident that they can be equal in area only when their sides are equal: wherefore H is the middle point of the diagonal. The construction, therefore, is effected by bisecting the diagonal EC in H, and drawing AFHG. For the triangles CHG, EHF are equiangular, and since CH, HE are equal, the triangles are equal. To each of them add the
The point D may be determined in the manner pointed out in the first ncte in page 59.
figure FBCH; then the figure FBCG is equal to the triangle EBC, that is, to half the parallelogram BD.
Now, since the diagonal CE is given in magnitude and position, its middle point H is given in position, and therefore H is always a point in the required line, wherever A is taken. Hence, so long as A and H are different points, the straight line AHG (I. post. 1.) is determined. This, however, is no longer so, if the given point A be the intersection of the diagonals, that is, the point H, as in that case only one point of the required line is known, and the problem becomes indeterminate, any straight line whatever, through H,'equally answering the conditions of the problem: and we are thus led by the solution of the problem to the conclusion, that in a parallelogram a point may be found, such that any straight line whatever, drawn through it, bisects the parallelogram ; and this is a porism.
The 32d proposition of the preceding book, when considered in a particular manner, affords another instance of a porism; as it appears that if a circle and a point D or E be given, another point E or D may be found, such that any circle whatever described through D and E will bisect the circumference of the given circle: and this may be regarded as the indeterminate case of the problem, in which it is required through two given points, to describe a circle bisecting the circumference of another given circle ;-a problem which is always determinate, except when the points are situated in the manner supposed in the proposition referred to.
PROP. I. PROB.*
Through two given points, to describe a circle touching a straight line given in position.
Through two given points A, B, let it be required to describe a circle, touching a given straight line CD.
This is one of the problems known under the name of tangencies. The general problem of the tangencies, as understood by the ancients, is as follows: Of three points, three straight lines, and three circles of given radii, any three being given in position; it is required to describe a circle passing through the points, and touching the straight lines and circles. This general problem comprehends ten subordinate ones, the data of which are as follows: (1.) three points ; (2.) two points and a straight line; (3.) two points and a circle; (4.) a point and two straight lines ; (5.) a point, a straight line, and a circle ; (6.) a point and two circles; (7.) three straight lines; (8.) two straight lines and a circle ; (9.) a straight line and two circles; and (10.) three circles. The first
Join AB. Then, if CD be parallel to AB, bisect AB by the perpendicular EF, and (III. l. schol.). througb A, F, B describe a circle: it is the circle required. For (III. 1. cor.) EF passes through the centre, and (I. 29.) is perpendicular to' CD; therefore (111. 16.) the circle touches CD; and it passes through A, B.
But if CD, BA be not parallel, let them meet in E: between AE, BE find a mean proportional, and make EF equal to it: through B, A, F describe a circle: it is the one required. For (const. and VI. 17.) the rectangle BE. ÈA is equal to the square of EF, and therefore (III. 37.) EF touches the circle; and the circle passes through the given points.
Schol. 1. In this latter case, since EF may
be taken on either side of E, there are two circles that answer the conditions of the problem ; and these are evidently unequal, unless BAE be perpendicular to CD.
Schol. 2. If one of the points A be in the given line, the preceding solution fails, as it does not give a third point in the required circumference. In that case two straight lines, one bisecting AB perpendicularly, and the other drawn through A perpendicular to CD, will intersect each other in the centre of the required circle. The proof is plain from the fourth proposition of the first book, and the sixteenth of the third book.
PROP. II. PROB. THROUGH two given points, to describe a circle touching a given circle.
Through the given points A, B, let it be required to describe a circle touching a given circle CDE.
and seventh of these are the fifth and fourth propositions of the fourth book of Euclid : the rest are given here.
If a circle be continually diminished, it may be regarded as becoming ultimately a point. By being continually enlarged, on the contrary, it may have its curvature so much diminished, that any portion of its circumference may be made to differ in as small a degree as we please from a straight line. Viewing the subject in this light, we may regard the first nine of the problems, now mentioned, as comprehended in the tenth. Thus, we shall have the first, by supposing the circles to become infinitely small; the seventh, by supposing them infinitely great; the fifth, by taking one of them infinitely small, one infinitely great, and one as a circle of finite magnitude: and so on with regard to the others. These views of the subject tend to illustrate it; but they do not assist in the solution of the problems.