33. If the sides about an obtuse angle be produced, and perpendiculars be drawn to them from the opposite an gle, 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 two 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 intercepts, shall be =.

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

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

Fig. 13.

Bisect the semicircular arc AB by CD; join DB; the of DB is to half the 2 of AB.

For it is = to the sum of the □rs of BC and 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 arcs, (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 trisected, 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 arcs of 609 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 of an equilateral triangle inscribed in a circle is to three times the of the radius.

For it is the difference between the □ of the diameter bisecting one of the angles of the equilateral, and the ☐ of the line connecting the extremity of the diameter and base; but that connecting line = radius ; .'., &c.

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

40. In prop. 35, b. 3, 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.

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

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 2 of half the excess from the ☐ 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 AND THIRD BOOKS,

DONE ANALYTICALLY.

PROP. 1. THEOR.

The difference between the hypotenuse (BC) and sum of the sides (BA and AC) of a right angled triangle (ABC) is equal to two radi of the inscribed circle.

Fig. 14.

ANALYSIS.

Suppose it to be the case; then since BC together with FG and GE are to BA and AC together, BC with FA and AE are to BA and AC; for it is evident that FA and AE are respectively to GE and FG, since each of the angles F, A, and 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 and BGF, the sides BG and GD are respectively to BG and GF, and the angles BDG and BFG are as being right, and the angles DBG and FBG of the same affection, the triangles are in every respect; .. BD is to BF; for the same reason CD is to CE; ... BC is of BF and CE; .., &c.

to the sum

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.

Suppose it done, and that AC is the line produced; let F be to a side of the □ 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.

Take the of the given part from the 2 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 will bisect the angles; also produce DE until EF is = to it, and draw FC; then the triangles ADE and FEC are evidently, and have also their sides and angles respectively to one another; and also the triangles CDE and CEF are in every way; . the angle FCD is the angle of an equilateral triangle; and since the angles FDC and DFC are, they must be the angles of an equilat. ; ... DF is to FC or DA; but DF is double

..

of DE or DI; DA is double of DI or DE; .. AI is three times DI or DE; .., &c.

PROP. 4. PROB.

Given of any triangle the base, line bisecting the base, and sum of the sides, to find the triangle.

Take from the 2 of the sum two ars of half the base, plus two ars of the bisecting line; take half the remainder and find the side of a 2 to it; then cut the given sum'so that the rectangle under its segments may be to this o2, &c.

=

PROP. 5. THEOR.

If a tangent (FH) be drawn to a circle, parallel to a chord (AB), the point of contact (F) will be the middle point of the arc cut off by that chord.

Suppose it to be the case; join FA and FB; then, since F is the middle point of the arc, FA and FB are = ; .'. the angles FAB and FBA are; if this can be proved, the proposition is true.

Since FH is parallel to AB, the angle ABF is = to BFH; but BFH is to BAF, (prop. 32, b. 3,); . ABF is to BAF; .. FA is to FB, &c.

PROP. 6. PROB.

To draw a common tangent, to two given circles, in a transverse direction.

Suppose it done, and that OL is the required tangent; join FO and LE, (i. e. the centres with the points of contact); produce FO till OH is to LE; join HE; because OL is a tangent, the angles FOL and ELO are right; .. FH is parallel to LE; and, since OH is = to LE, HE is parallel to OL; .. the angle FHE is a right angle.

Then join the centres F and E; and on FE describe a semicircle FHE; in it inflect a right line FH to the sum of the radii FO and LE; join HE; draw EL parallel

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