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. Next, in the case of external section: The opposite tangents AD, BE being equal, the triangles AGD and BGE are evidently equal, and therefore DE passes through the centre. Hence the triangles BGE and FGC are also equal, and GC equal to GE. The modified construction is therefore to erect the perpendicular BE equal to the side of the given square, join GE, and where this cuts the circumference apply the tangent FC to meet AB produced : Then AC and CB are the required external segments of the given line AB. For it is evident that the rectangle AC, CB will be equal to the square of BE; which is also deduced from Prop. 26. cor. 2. Book III., since CF is now a tangent and AC.CB=CF or BE”. If AB be equal to BE, the construction will exactly correspond with what was before given.

In applying this problem to the construction of quadratic equations, it is necessary previously to ascertain the precise import of the ordinary signs used in Algebra, when extended to geometrical quantities. The signs--and-intimate, in general, nothing more than that the number, or the magnitude expressed by number, to which they are respectively prefixed, is to be added to, or taken away from, any other number, with which it comes to be combined. It would be more correct language, therefore, to call the quantities carrying such signs additive and subtractive, implying merely a casual and mutable relation; instead of the usual appellations of positive and negative, which seem to bestow a distinct and absolute charact..., and have hence led incautious reasoners into mystery and paradox. A similar degree of reserve is indispensable in Geometry. Following the European mode of writing from left to right, we might fancy it almost natural to draw a line in thesame direction: When we want to extend a line, we apply an additional line to the right; but when we seek to contract it, we retrace a deft.


cient line to the left. Thus, if NO be annexed to the right of MN, there results –H H o MO; or if NO’ be M O N O taken to the left of the extremity N, there will remain MO'. The position of NO or NO, to the right or left, will, therefore, in reference to a combination with any line MN, have the same effect as the signs of addition or subtraction produce in Algebra. Following out the same analogy, while lines drawn upwards may correspond to additive quantities, lines drawn downwards must express subtractive quantities. Quadratic equations are reducible to these four forms: 1. ** + air = + b c 2. r"—az = + be 3. ** + air = be 4. r" as: = — be. The two first may be constructed from the second case of Proposition seventeenth; and the two last will receive their construction from the first case of that problem. We shall resume the equations in their order:

1. **-Har = + be, then c= — #== V: +bc, these being two roots, the greater subtractive, and the less additive. Employing the construction of the second case of the problem, let AB = a, AD = b, and BE = — c, since it stretches below AB; if BC represent—r, then CA, in the reverse position, will be denoted by – a r. Wherefore BC X CA = (– a c) ra—a r—a", and consequently AD.BE =—bc=—ar—w", or, by inversion, r* + ar = + be. The roots are, consequently, the shorter segment BC which is additive, and the longer segment BC which is subtractive.

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likewise two roots, but the greater additive, and the less subtractive. Here AB, AD and BE being denoted by a, b, and —c, as before; if AC’ represent a, CB in a reverse position will be expressed by a-r. Consequently AC.CB = (a-r) + = ar—r", and therefore AD. BE= —bc = air—a", or ro–ar =+bc. The roots are hence the greater segment AC", which is additive, and the less segment AC, which is subtractive. In this case, the quadratic equation will always admit of a double solution, since the radical part of the root is both additive and subtractive, while the circle crossing AB must ne. cessarily cut it in two parts. The third and fourth forms of the equation are constructed by the application of the first case of the problem.

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likewise the same character, but additive. Let AB, AD, and BE be expressed as before by a, b and c; if AC represent r, CB will be denoted by a-v. Wherefore, AC.CB=(a—w).c=ar—a", and AD.BE=bc=ar—w". Consequently by transposition wo—ar=—bc. The roots of this equation are, therefore, expressed by AC and AC", both of them additive. *


When the rectangle under the perpendicular AD and BE, becomes equivalent to the square of half of AB, the circle touches AB, and the two points C and C’ merge in a single

point. At this limit, too, the radical part== w/– bc of the

value of a vanishes, and there results a single root, which is additive or subtractive according to the sign of the second term of the quadratic equation. If it were sought that the rectangle under AD, BE, or under the segments AC, CB, should exceed the square of the half of AB, the circle would not meet this straight line, while the radical would evidently become impossible, and thus betray the same incongruity of hypothesis. It may be observed, that the algebraical solution of these quadratic equations flows from the geometrical construction. For, suppose AB were bisected in O ; it is evident that AD.BE=AC.CB=AO”—OC*, or OC*—AO”, or OC*= AO*—AD.B.E, or AD.BE+ AO”, according as the intersection takes place within or without AB. Wherefore OC always

2 represents the radical part = # =le of the expression for

the values of a, which are formed by its combination with OA.

If the construction of Pappus be used, while the perpendiculars AD, BE, and the transverse line DE remain the same as before, the intersection of this with a circle described on AB determines the position of a perpendicular to it, dividing the diameter internally or externally into the required segments.

5. Proposition eighteenth. To this proposition might be added a corollary: That four times the area of a triangle is to the rectangle under any two sides, as the base to the radius of the circumscribing circle.

For the area of the triangle ABC is (Prop. 5. II.) equivalent to half the rectangle contained by the base AC and the perpendicular BD, and consequently four times this area is equivalent to twice the rectangle AC, BD. But (VI. 18.) the rectangle under the sides AB and BC is equivalent to the rectangle under the perpendicular BD and BE, the diameter of the circumscribing circle, or to A. twice the rectangle under BD and the radius of that circle. Whence four times the area of the triangle is to the rectangle under the sides AB and BC, as twice the rectangle under BD and AC to twice the rectangle under BD and the radius of the circumscribing circle, or as the base AC to that radius. Let a, b and c denote the three sides of a triangle, and S half their sum or the semiperimeter; then, combining Prop. 29. Book VI. with this corollary, the radius of the circumscribing

circle will be expressed * Avos. SES-F SE.s=== y Thus,

if the sides of the triangle be 13, 14, 15, the radius of the cir13. 14. 15. 2780 4v'( 336

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cumscribing circle = = 8*.

6. Proposition nineteenth. This well-known proposition is now rendered more general, by its extension to the case of the exterior angle of the triangle. The two cases combined afford an easy demonstration of the corollary to Proposition 7. Book VI:; for the straight lines bisecting the vertical and its adjacent angle form a right-angled triangle, of which the hypotenuse is the distance on the base between the points of internal and external section.

7. Proposition twenty-third. The latter part of the scholium was added to this proposition, with a view to explain the principle of the construction of the pantagraph, a very useful instrument contrived for copying, reducing, or even enlarging plans. It consists of a jointed rhombus DBFE, framed of wood or brass, and having the two sides BD and BF extended to double their length; the side DE and the branch DA are marked from D with successive divisions, DO being made to BO always in the ratio of DP to BC; small sliding

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