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be to the of the given side (by obser. 3. b. 2,) take half the produced part and add to it the given difference, their sum shall be the hypothenuse, and the remaining half will be the third side.

Given the rectangle under any two lines and their difference to find them.

Fig. 6.

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2

Take CF to half the given difference, from its extremity C erect a perpendicular CD to the side of a which is to the given rectangle, join CF; and from F as centre with the radius FD describe a circle and produce FC both ways to meet it in A and B: it is evident that AC and CB are the lines required to be found.

33. If the sides about an obtuse angle be produced, and perpendiculars be drawn to them from the opposite angle, it is evident from prop. 12, b. 2, that the greatest intercept is conterminous with the least side.

34. If a right line be drawn, cutting the circumferences of tro concentric circles, the intercepts are equal.

If it passes through the centre it is evident, but if not let fall a perpendicular on it from the centre: then it is evident that the entire drawn line is bisected by it and also that part of it within the inner circle, if .. this latter part be taken away the remainders, viz. the intereepts shall be =

* Concentric circles are those which lying one within another, have a common centre.

25. To draw a chord in a circle, whose square shall be equal to half the square of the diameter.

Fig. 13.

Bisect the semicircular arch AB by CD, join DB: the of DB is to half the of AB.

2

For it is to the sum of the rs of BC, CD, and they are together to half the of AB, (cor. prop. 4, b. 2,)

36. The converse of the second part of prop. 15, b. 3, is added by Robert Simpson, which is as follows: And the greater line is nearer to the centre than the less, the demonstration of which is evident.

37. It is evident from prop. 22, b. 3, that any external angle of a triangle, is equal to the angle in the remote segment, whose base is the subtense of the internal angle adjacent to the external.

For either of them with this adjacent internal is = to two right angles.

38. If two diameters in a circle intersect one another at right angles, they divide the circumference into four equal arches; (prop. 26, b. 3;) and by drawing the subtenses of the right angles a square will be inscribed in the circle.

If those right angles be bisected, and the subtenses of the halves be drawn, an octagon will be inscribed; and if bisected a duodecagon.

By trisecting the right angles also an hexagon may be inscribed, and also an equilateral triangle.

For suppose the circumference to be 360 degrees, by trisecting the right angles it is divided into arches of 60 and 120, . by connecting the subtenses of those of 60 an hexagon will be formed, and the sides of the hexagon are evidently to the radii of the circle.

By connecting the subtenses of those of 120, an equilateral triangle will be formed.

It is evident that the of the side and an equilateral triangle inscribed in a circle is to three times the 2 of the radius.

For it is the difference between the 2 of the diameter bisecting one of the angles of the equilateral, and the

of the line connecting the extremities of the diameter and base.

39. If two parallel chords be drawn in a circle, they intercept equal arches, (prop. 26, b. 3.)

40. In prop. 35, b. 8, if right lines be drawn, joining the extremities of the sides of the angles vertically opposite, there shall be formed two similar triangles.

This is evident, from prop. 32, b. 1, and prop. 21, b. 3.

40. To find the area of an squilateral triangle, by having the difference of the perpendicular and half side.

2

Take half the given difference and extract the root of three times its 2, to which root add the same half and multiply their sum by the difference and said sum. (See prop. 115, b. 1.) For finding the side take the of half the excess from the 2 of the excess, and extract the root of the difference, this root + half the excess is half the side, (prop. 115, b. 1.)

A VARIETY

OF

THE MOST IMPORTANT

Problems and Theorems,

DEDUCIBLE

FROM THE SECOND & THIRD BOOKS.

DONE ANALYTICALLY.

PROP. 1. THEOR.

The difference between the hypothenuse [BC] and sum of the sides [BA,AC] of a right angled triangle [ABC] is equal to two radii of the inscribed circle.

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Suppose it to be the case; then since BC with FG, GE is to BA,AC together, BC with FA, AE is to BA,AC; for it is evident that FA,AE are respectively

to GE,FG, since each of the angles F, A, E, is a right angle, .. BC is to BF and CE together, if this can be proved the proposition is evident.

Join BG; then since in the triangles BGD,BGF the sides BG and GD are respectively to BG and GF, and the angles BDG,BFG are as being right and the angles DBG,FBG of the same affection, the triangles

are in every way,., BD is reason CD is to CE, .. BC is and CE, . &c.

to BF; for the same

to the sum of BF

PROP. 2, PROB.

To produce a given right line [AB] so that the rectangle under the whole and produced part may be equal to a given quantity.

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Suppose it done, and that AC is the line produced, let F be to a side of the 2 of the given quantity; then it is evident from prop. 37, b. 3, that CA shall be to the secant drawn from the extremity of the tangent (to F,) through the centre of the circle, the diameter of which is to the given line.

=

Describe a circle of which the diameter is the given line AB, draw to it a tangent IA to the given quantity, and from its extremity I draw through the centre the secant ID, it is evidently to the whole produced line, .. make BC to GI, and the rect. ACB is to

the given quantity.

Cor. 1. By being given the tangent and internal part of the secant passing in any way, we can find the

secant.

Cor. 2. By being given the tangent and external part of the secant, we can find the secant.

2

Take the of the given part from the of the tangent, and then produce the given part so that the rectangle under the parts may be to the remainder ; this produced line shall be the required tangent.

PROP. 3. THEOR.

The perpendicular (AI) of an equilateral triangle (ABC) is equal to three times the radius of the inscribed circle.

Fig. 16.

It can be proved that each of the radii from the centre to the points of contact bisects the side to which it is drawn, (prop. 32, b. 1,) and if they be produced they

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