39. If. two parallel chords draron 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.

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10.' 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 , 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 o of half the excess from the O? of the excess, and extract the root of the difference, this root + half the excess is

half the side, (prop: 115, b. 1.)

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Problems and Theorems,





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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Fig. 14.

ANALYSIS. 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 apgle, . Bo 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 = to BF; for the same reason CD is = to CE, .. BC is = to the sum of BF and CE, .', &c.


To produce a given right line [AB] so that the rectangle

under the whole and produoed part may be equal to

a given quantity. Fig. 15.

ANALYSIS. Suppose it done, and that AC is the line produced, let F be to a side of the a? of the given quantity; then it is evident from prop. 37, b. 3, that ÇA shall be = to the secant drawn from the extremity of the tangent

to F) through the centre of the circle, the diameter of wḥich 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 secapt, we can find the secant.

Take the O’ of the given part from the O’ 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.


The perpendicular (AI) of an equilateral triangle (ABC)

is equal to three times the radius of the inscribed a

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 apgle of an equilateral triangle, and since the angles FDC,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, ,. Al 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 O’ of the sum two *rs of half the base, plus two o’rs of the bisecting line, take half the remainder and find the side of a O’= to it, then cut the given sum so that the rectangle under its segments may be = to this o’, &c,


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 arch cut off by that chord. Fig. 13.

ANALYSIS, Suppose it to be the case ; join FA,FB; then since F is the middle point of the arch FA,FB are =, .. the angles FAB,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 isto FB, &o.

PROP. 6, PROB. To draw a common tangent, to two given circles, in a

transverse direction: Fig. 17.

ANALYSIS 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, ÉLO 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, E; and on FE describe a semicircle FHE, in it inflect a right liné FH = to the sum of the radii FO, LE, join HE, draw EL parallel to HF and join LO: LO is the tangent; the demonstration is evident.

Also a direct tangent can be drawn to two given circles.

1. If the given circles be = connect their centres, draw a radius in each on the same side and at right angles to the connecting line, and connect their extremities at the circumference; this connecting line is the required tangent.

For the circles being the radii are =, and they being

and at right angles to the line connecting the diameter, they are parallel, :. the connecting lines are = and par. . the radii are at right angles to the one at the circum: ferences, ... it is a tangent.

2. If the given circles be unequal Fig. 18.

ANALYSIS. Suppose IH is the required tangent, TI and HE are evidently par. to one another, make IF = to HE and join FE, it is = and par. to IH, , the angle IFE is right and IF is the difference between TI and HE; with FI a radits describe a circle FA, EF is evidently a tangent to it.

Therefore with the centre į and the difference of the radii IF describe a circle FA, and from the centre E of the lesser circle draw a tangent EF to it, produce IF to T, draw EH par. to IT and join TĦ; it is evidently the required tangent.

PROP. 7. THEOR. The distance of the chord (AB), whose square is three

fourths of the square of the diameter from the centre

(CD,) is half the radiús. Fig. 19

Bisect AB in F, and join EF, EA and EB

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