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Bisect FE in &, and put AB or BC=a, FG or GE=b, and BG=x; then will BE=x+b and Br=x-b,

But since, by right angled triangles, AEBE3 —AB3, we shall have

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And, because the triangles BCF, EAB, are similar,

BF BC :: BE: AE, or

a(x+b)=(x—b)√/{(x+b)3 —a2 }

Whence, by squaring each side of this equation, and arranging the terms in order, there will arise

x2 -2(a2+b2)x2=b2 (2a2 —b2).

Which equation, being resolved after the manner of a quadratic, will give

· x=√{a3+b3±a√(a2+4b2)}

And, consequently, by adding b to, or subtracting it from this last expression, we shall have

BE=√{a2+b2±a✅✅(a3+4b3)}+b, or
BF=√{a2+b2±a√(a2 +462)}—b.

Which values, by determining the point E, or F, will satisfy the problem.

Where it may be observed, that the point & lies in the circumference of a circle, described from the centre D, with the radius FG, or half the given linc.

PROBLEM XVI.

The perimeter of a right angled triangle ABC, and the radius of its inscribed circle being given, to determine the triangle.

F

E

D

Let the perimeter of the triangle = p, the radius OD, OE, or of the inscribed circle =r, AE=x, and BD=y.

Then, since in the right angled triangles AEo, afo, oe is equal to of, and oA is common, AF will also be equal

AE, or x.

And, in like manner, it may be shown, that BF is equal to BD, or y.

But, by the question, and Euc. 1, 47, we have

(x+r)+(y+r)+(x+y)=p,
(x+r)2+(y+r)2=(x+y)2·

and'

Or, by adding the terms of the first, and squaring those of the second,

x+y=p-r, and
r(x+y)=xy--2.

Hence, since, in the first of these equations, y=(}p—r) -x, if this value be substituted for y in the second, there will arise

x2 - (}p-r)x=-r(tp -r.

Which equation, being resolved in the usual manner,

gives

x=}({p−r)±√{}({p− r)2 — r({p−'r) },

and

y={(}p—r)=√{{(}p − r)2 — r( }p − r)}.

And, consequently, if r be added to each of these last expressions, we shall have

ac=}{}p+r)±√{{{{p—r)2 — r({p—r)},

and

BC=}{} p+r)=√{}(}p - r)2 —r(}p—r)}, for the values of the perpendicular and base of the triangle, as was required.

PROBLEM XVII.

From one of the extremities A, of the diameter of a given semicircle ADB, to draw a right line AE, so that the part DE, intercepted by the circumference and a perpendicular drawn from the other extremity, shall be of a given length.

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Let the diameter AB=d, DE α, and AE=x; and join BD.

Then, because the angle ADB is a right angle, (Euc. 11, 31,) the triangles ABE ABD, are similar.

And, consequently, by comparing their like sides, we shall have

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Whence, multiplying the means and extremes of these proportionals, there will arise

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Which equation, being resolved after the usual manner, gives

x=‡a+√(}a2+d2).

PROBLEM XVIII.

To describe a circle through two given points A, B, that shall touch a right line CD given in position.

B

COF GH D

Join AB; and through o, the assumed centre of the required circle, draw FE perpendicular to AB; which will bisect it in E (Euc. 111, 3).

Also, join oв; and draw EH, OG, perpendicular to CD;

the latter of which will fall on the point of contact & (Euc. 111. 18).

Hence, since A, E, B, H, F, are given points, put EB=α, EF=b, Eн=c, and so=x; which will give OF=b-x.

Then, because the triangle OEB is right angled at E, we shall have

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But, by similar triangles, FE: EH :: FO: OG or ob1⁄2 or b c :: b-: -x: оB; whence, also,

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And, consequently, if these two values of oв be put equal to each other, there will arise

√(x2+a2)={(b−x).

Or, by squaring each side of this equation, and simplifying the result,

(b2 —c2)x2+2bc2x=b2 (c2 —a2).

Which last equation, when resolved in the usual manner, gives

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for the distance of the centre o from the chord AB; where b must, evidently, be greater than c, and c greater than a.

PROBLEM XIX.

The three lines AO, BO, co, drawn from the angular points of a plane triangle ABC, to the centre of its inscribed circle, being given, to find the radius of the circle, and the sides of the triangle.

Z

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Let o be the centre of the circle, and, on Ao produced, let fall the perpendiculars CD; and draw OE, OF, OG, to the points of contact E, F, G.

Then, because the three angles of the triangle ABC are, together, equal to two right angles, (Euc. 1, 32,) the sum of their halves oAc+00A+оBE will be equal to one right angle.

But the sum of the two former of these, OAC +oca, is equal to the external angle Doc; whence the sum of DOC OBE, as also of Doc+OCD, is equal to a right angle; and, consequently, OBE=OCD.

Let, therefore, Aо=α, во=b, co=c, and the radius

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Then, since the triangles BOE, COD are similar, Bo: OE :: co : oD, or b:x:: c: oD; which gives

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Also, because the triangle AOC is obtuse angled at o, we shall have (Euc. 11, 12)

AC2=A02+co2+2A0XOD; or

Ac=√(a2+c2+ 2acx) or√

b

̧b(a2+c2)+2ácx).

b

But the triangles ACD, AOF, being likewise similar,

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