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Because CG cuts the chord EF at right angles, GE is equal to GF (III.4.); wherefore the squares of AE and AF are equivalent to twice the squares of AG and GE (II. 22.) But ACG being a right-angled triangle, the square of AG is equivalent to the squares of AC and CG (II. 10.), or twice the square of AG is equivalent to twice the squares of AC and CG. Wherefore the squares of AE and AF are equivalent to twice the three squares of AC, CG, and GE. Of these, the two squares of CG and GE are equivalent to the square of CE or CB, for the triangle CGE is right-angled. Consequently the squares of AE and AF are equivalent to twice the squares of AC and CB. But the straight line BD being cut equally at C and unequally at A, the squares of the unequal segments AB and AD are together equivalent to twice the squares of AC and CB (II. 18, cor.); whence the squares of AE and AF are together equivalent to the squares of AB and AD.

PROP. V. THEOR.

The rectangle under the segments of a chord is greater or less

than the rectangle under the segments into which a perpendicular

from the point of section divides a diameter, by the square of that perpendicular—according as it lies without or within the circle.

Let the perpendicular CF be let fall from a point C in the chord ACB upon a diameter DE; the rectangle BC, CA, is greater or less than the rectangle EF, FD, by the square of the perpendicular CF, according as this lies without or within the circle. -

First, let the perpendicular CF lie without the circle, and join CE and DG.

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The square of the hypotenuse CE is equivalent to the squares of FE and CF (II. 10.). But the square of CE is composed of the rectangles CE, EG, and CE, CG (II. 14.); and the square of FE is composed of the rectangles FE, ED, and FE, FD : Wherefore the rectangles CE, EG and CE, CG are equivalent to the rectangles FE, ED and FE, FD, together with the square of CF. And since EGD, standing in a semicircle, is a right angle (III. 19.), its adjacent angle CGD is also right, and the angle opposite to this at F is right; consequently (III. 17, cor. 1.) a circle might be described through the four points C, G, D, F. Whence (III. 26.) the rectangle CE, EG is equivalent to FE, ED: and taking these from the terms of the former equality, there remains the rectangle CE, CG, that is, (III. 26.) AC, CB, equivalent to the rectangle FE, FD, together with the square of CF. Next, let the perpendicular CF lie within the circle. The same construction being made, the rectangle CE, EG is still equivalent to the rectangle FE, ED. But the rectangle CE, EG is (II. 14.) equivalent to the rectangle CE, CG, and the square of CE, or the squares of FE and CF; and the rectangle FE, ED is equivalent to the rectangle FE, FD and the square of FE. From these equal quantities, therefore, take away the common square of FE, and there remains the rectangle CE, CG, or AC, CB, with the square of CF, equivalent to the rectangle FE, FD. Lastly, if the perpendicular CF lie partly without and partly within the circle, the Proposition must be slightly modified. The former construction being retained: Because the square of CE is equivalent to the squares of CF and

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FE, the rectangles CE, EG and CE, C
CG are together equivalent to the -
square of CF and the difference be-
tween the rectangle FE, ED and FE,
FD; but the rectangle CE, EG is e-
quivalent to the rectangle FE, ED,
and consequently the rectangle CE,
CG, or the rectangle AC, CB, is
equivalent to the difference between
the square of CF and the rectangle

FE, FD.

In the first case, if the square of FH be equivalent to the rectangle FD, FE, the square of CH will be likewise equivalent to the rectangle CG, CE; for the rectangle AC, CB, being equivalent to the rectangle FD, FE, or the square of FH, together with the square of CF, must (II. 10. El.) be equivalent to the square of CH.

PROP. VI. THEOR.

A straight line drawn from the verter of a triangle through the intersection of two perpendiculars from the extremities of the base to the opposite sides, is likewise perpendiculur to the base.

In the triangle ABC, the straight line BFG drawn from the vertex B through F, the intersection of the perpendiculars AE and CD from A and C upon the opposite sides CB and AB is perpendicular to the base AC.

For join DE. Because BDF and BEF are right angles, the quadrilateral figure ADEC(III. 17. cor. 1.) is A: G- C contained in a circle; and for the same reason, the quadrilateral ADEC is contained in a circle. Wherefore the exterior angle BDE (III. 17. cor. 2.) is equal to ACE; but (III. 16.) BDE is equal to the angle BFE in the same segment, which is therefore equal to ACE or GCE, and con

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sequently the quadrilateral CEFG is also contained in a circle. Whence (III. 17.) the opposite angles CEF and CGF are equal to two right angles, and CEF, being a right angle by hypothesis, EGF must likewise be right; or the straight me BFG is perpe dicular to the base AC.

PROP. VII. PROB.

Through a given point, between two diverging straight lines, to draw a straight line that shall have equal segments terminated by them.

Let AB and AC be two diverging straight lines given in a position, and Fan intermediate point, through which it is required to draw GFH, such that the intercepted segments FG and FH shall be equal. This may be easily effected, by drawing a parallel from F to AB, and doubling the portion so cut off, from A to G, to mark the position of GFH. But the problem may be constructed in another way, which, though more complex, is important in its application to the Theory of Lines of the Second Order.

Draw AD bisecting the angle BAC, and upon it let fall the perpendicular FE, which produce both ways to B and C; from B erect BD perpendicular to AB, join DF; and EFH, being drawn perpendicular to it, is the line required.

For join DC, DG and DH. The right-angled triangles ABD and ACD are (s. 20.) equal, and consequently BDC is isosceles. But GBD and GFD being right angles, and therefore equal, the quadrilateral figure GB, FD (III. 16.) is contained in a circle, and hence the angle DGF is equal to DBF ; for the same reason, since DCH and DFH are right angles, the quadrilateral figure DCHF is likewise contained in a circle, and hence the angle DHF is equal to DCF. Consequently the angle DGF is equal to DHF, and the right-angled triangles DFG and DFH are equal, and the base FG equal to FH.

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If the point F were taken in the extension of the line EB, the perpendicular to DF may then be shown to have equal segments intercepted by the sides of the exterior angle formed by AG and the production of CA beyond the vertical point A. -

BOOK IV.

1. THE equilateral triangle, the square, the pentagon, the hexagon, and other polygons derived from these, were the only regular figures known to the Greeks. The inscription of all the rest has for ages been supposed absolutely to transcend the powers of elementary geometry. But a curious and most unexpected discovery was lately made by Mr Gauss, now Professor in the University of Göttingen, who has demonstrated, in a work entitled Disquisitiones Arithmetica, and published at Brunswick in 1801, that certain very complex polygons can yet be described merely by help of circles. Thus, a regular polygon containing 17, 257, 65537, &c. sides, is capable of being inscribed, by the application of elementary geometry; and in general, when the number of sides may be denoted by 2" + 1, and is at the same time a prime number. The investigation of this principle is rather intricate, being founded on the arithmetic of sines and the theory of equations; and the constructions to which it would lead are hence, in every case, unavoidably and most excessively complicated. Thus the cosine of the several arcs arising from the division of the circumference of a circle into seventeen equal parts, are all contained in this very involved expression: - — 's-H * M 17-F ov/(34–24/17)— #y (17+ 34/17–W (34–24/17)—2A/(34+24/17))

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